Postulates, Theorems, and Corollaries

Postulates, Theorems, and Corollaries

Chapter 2 Reasoning and Proof Postulate 2.1 Through any two points, there is exactly one line. (p. 89)

Postulate 2.2 Through any three points not on the same line, there is exactly one plane. (p. 89)

Postulate 2.3 A line contains at least two points. (p. 90)

Postulate 2.4 A plane contains at least three points not on the same line. (p. 90)

Postulate 2.5 If two points lie in a plane, then the entire line containing those points lies in that plane. (p. 90)

Postulate 2.6 If two lines intersect, then their intersection is exactly one point. (p. 90)

Postulate 2.7 If two planes intersect, then their intersection is a line. (p. 90)

Theorem 2.1 Midpoint Theorem If M is the midpoint of AB, then AM MB. (p. 91)

Postulate 2.8 Ruler Postulate The points on any line or line segment can be paired with real numbers so that, given any two points A and B on a line, A corresponds to zero, and B corresponds to a positive real number. (p. 101)

Postulate 2.9 Segment Addition Postulate If B is between A and C, then AB BC AC. If AB BC AC, then B is between A and C. (p. 102)

Theorem 2.2 Congruence of segments is reflexive, symmetric, and transitive. (p. 102) P

ostulates, Theorems, Postulate 2.10 Protractor Postulate Given AB and a number r between 0 and 180, there is exactly and Corollaries one ray with endpoint A, extending on either side of AB, such that the measure of the angle formed is r. (p. 107)

Postulate 2.11 Angle Addition Postulate If R is in the interior of PQS, then mPQR mRQS mPQS. If mPQR mRQS mPQS, then R is in the interior of PQS. (p. 107)

Theorem 2.3 Supplement Theorem If two angles form a linear pair, then they are supplementary angles. (p. 108)

Theorem 2.4 Complement Theorem If the noncommon sides of two adjacent angles form a right angle, then the angles are complementary angles. (p. 108)

Theorem 2.5 Congruence of angles is reflexive, symmetric, and transitive. (p. 108)

Theorem 2.6 Angles supplementary to the same angle or to congruent angles are congruent. (p. 109) Abbreviation: suppl. to same or are .

Theorem 2.7 Angles complementary to the same angle or to congruent angles are congruent. (p. 109) Abbreviation: compl. to same or are .

Theorem 2.8 Vertical Angle Theorem If two angles are vertical angles, then they are congruent. (p. 110)

Theorem 2.9 Perpendicular lines intersect to form four right angles. (p. 110)

Theorem 2.10 All right angles are congruent. (p. 110)

Postulates, Theorems, and Corollaries R1 Postulates, Theorems, and Corollaries R2 Theorem 4.2 Theorem 4.1 CongruentTriangles Chapter 4 Theorem 3.9 Theorem 3.8 Theorem 3.7 Theorem 3.6 Theorem 3.5 Postulate 3.5 Postulate 3.4 Postulate 3.3 Postulate 3.2 Theorem 3.4 Theorem 3.3 Theorem 3.2 Theorem 3.1 Postulate 3.1 Perpendicular andParallel Lines Chapter 3 Theorem 2.13 Theorem 2.12 Theorem 2.11 Postulates, Theorems,and Corollaries eodtinl,te h hr nlso h rage r ogun.(p.186) second triangle, thenthethird anglesofthetriangles are congruent. Third Angle Theorem Angle SumTheorem (p.161) parallel toeachother. In aplane,iftwolinesare eachequidistantfrom athird line,thenthetwolinesare (p. 152) In aplane,iftwolinesare perpendiculartothesameline,thentheyare parallel. (p.152) angles iscongruent, thenthelinesare parallel. If twolinesinaplaneare cutbyatransversalsothatpairofalternateinterior angles issupplementary, thenthelinesare parallel.(p.152) If twolinesinaplaneare cutbyatransversalsothatpairofconsecutiveinterior (p.152) angles iscongruent, thenthetwolinesare parallel. If twolinesinaplaneare cutbyatransversalsothatpairofalternateexterior (p.152) exactly onelinethrough thepointthatisparalleltogivenline. Parallel Postulate (p.151) congruent, thenthelines are parallel. If twolinesinaplaneare cutbyatransversalsothatcorresponding anglesare is T Tw (p.134) two parallellines,thenitisperpendiculartotheother. Perpendicular Transversal Theorem (p.134) each pairofalternateexterioranglesiscongruent. Alternate ExteriorAnglesTheorem (p.134) then eachpairofconsecutiveinterioranglesissupplementary. Consecutive InteriorAnglesTheorem (p.134) each pairofalternateinterioranglesiscongruent. Alternate InteriorAnglesTheorem (p.133) each pairofcorresponding anglesiscongruent. Corresponding AnglesPostulate (p.110) If twocongruent anglesformalinearpair, thentheyare rightangles. (p.110) If twoanglesare congruent andsupplementary, theneachangle isarightangle. Abbreviation: Ifalt.int. Abbreviation: Ifcons.int. Abbreviation: Ifalt.ext. epniua ie omcnretajcn nls (p.110) Perpendicular linesformcongruent adjacentangles. wo nonverticallinesare perpendicularifandonlytheproduct oftheirslopes ovria ie aetesm lp fadol fte r aall (p.141) o nonverticallineshavethesameslopeifandonlytheyare parallel. .(p.141) 1. Abbreviation: If2linesare If there isalineandpointnotontheline,thenthere exists are are h u ftemaue fteage fatinl s10 (p.185) The sumofthemeasures oftheangles ofatriangleis180. are suppl.,thenlines If twoangles ofonetriangleare congruent totwoanglesofa , thenlinesare , thenlinesare to thesameline,thenlinesare If twoparallellinesare cutbyatransversal,then If twoparallellinesare cutbyatransversal,then If twoparallellinesare cutbyatransversal,then In aplane,iflineisperpendiculartooneof . If twoparallellinesare cutbyatransversal, . . Abbreviation: Ifcorr. . are , linesare . Theorem 4.3 Exterior Angle Theorem The measure of an exterior angle of a triangle is equal to the sum of the measures of the two remote interior angles. (p. 186)

Corollary 4.1 The acute angles of a right triangle are complementary. (p. 188)

Corollary 4.2 There can be at most one right or obtuse angle in a triangle. (p. 188)

Theorem 4.4 Congruence of triangles is reflexive, symmetric, and transitive. (p. 193)

Postulate 4.1 Side-Side-Side Congruence (SSS) If the sides of one triangle are congruent to the sides of a second triangle, then the triangles are congruent. (p. 201)

Postulate 4.2 Side-Angle-Side Congruence (SAS) If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the triangles are congruent. (p. 202)

Postulate 4.3 Angle-Side-Angle Congruence (ASA) If two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, the triangles are congruent. (p. 207)

Theorem 4.5 Angle-Angle-Side Congruence (AAS) If two angles and a nonincluded side of one triangle are congruent to the corresponding two angles and side of a second triangle, then the two triangles are congruent. (p. 208)

Theorem 4.6 Leg-Leg Congruence (LL) If the legs of one right triangle are congruent to the corresponding legs of another right triangle, then the triangles are congruent. (p. 214)

Theorem 4.7 Hypotenuse-Angle Congruence (HA) If the hypotenuse and acute angle of one right triangle are congruent to the hypotenuse and corresponding acute angle of

another right triangle, then the two triangles are congruent. (p. 215) P ostulates, Theorems, and Corollaries Theorem 4.8 Leg-Angle Congruence (LA) If one leg and an acute angle of one right triangle are congruent to the corresponding leg and acute angle of another right triangle, then the triangles are congruent. (p. 215)

Postulate 4.4 Hypotenuse-Leg Congruence (HL) If the hypotenuse and a leg of one right triangle are congruent to the hypotenuse and corresponding leg of another right triangle, then the triangles are congruent. (p. 215)

Theorem 4.9 Isosceles Triangle Theorem If two sides of a triangle are congruent, then the angles opposite those sides are congruent. (p. 216)

Theorem 4.10 If two angles of a triangle are congruent, then the sides opposite those angles are congruent. (p. 218) Abbreviation: Conv. of Isos. Th.

Corollary 4.3 A triangle is equilateral if and only if it is equiangular. (p. 218)

Corollary 4.4 Each angle of an equilateral triangle measures 60°. (p. 218)

Chapter 5 Relationships in Triangles Theorem 5.1 Any point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment. (p. 238)

Theorem 5.2 Any point equidistant from the endpoints of a segment lies on the perpendicular bisector of the segment. (p. 238)

Postulates, Theorems, and Corollaries R3 Postulates, Theorems, and Corollaries R4 Theorem 6.3 Theorem 6.2 Theorem 6.1 Postulate 6.1 ProportionsandSimilarity Chapter 6 Theorem 5.14 Theorem 5.13 Corollary 5.1 Theorem 5.12 Theorem 5.11 Theorem 5.10 Theorem 5.9 Theorem 5.8 Theorem 5.7 Theorem 5.6 Theorem 5.5 Theorem 5.4 Theorem 5.3 Postulates, Theorems,and Corollaries iiaiyo rage srfeie ymti,adtastv.(p.300) Similarity of triangles isreflexive, symmetric, andtransitive. (p.299) included anglesare congruent, thenthetrianglesare similar. proportional tothemeasures oftwocorresponding sidesofanothertriangle andthe Side-Angle-Side (SAS)Similarity (p.299) triangles are proportional, thenthetriangles are similar. Side-Side-Side (SSS)Similarity (p.298) angles ofanothertriangle,thenthetrianglesare similar. Angle-Angle (AA)Similarity (p.268) the corresponding angleinthesecond triangle. then theanglebetweenpairofcongruent sidesinthefirsttriangleisgreater than triangle andthethird sideinonetriangleislongerthanthethird sideintheother, SSS Inequality (p.267) is longerthanthethird sideofthesecondtriangle. measure thantheincludedangleinother, thenthethird sideofthefirsttriangle sides ofanothertriangleandtheincludedangleinonehasagreater SAS Inequality/HingeTheorem (p.263) point totheplane. The perpendicularsegmentfrom apointtoplaneistheshortestsegmentfrom the (p.262) point totheline. The perpendicularsegmentfrom apointtolineistheshortestsegmentfrom the (p.261) greater thanthelengthofthird side. T (p. 250) side oppositethegreater angleislongerthanthesideoppositelesserangle. If oneangleofatrianglehasgreater measure thananotherangle,thenthe (p. 249) the longersidehasagreater measure thantheangleoppositeshorterside. If onesideofatriangleislongerthananotherside,thentheangleopposite (p.248) interior angles. then itsmeasure isgreater thanthemeasure ofeitheritscorresponding remote Exterior AngleInequalityTheorem (p.240) from avertextothemidpointofsideoppositeonmedian. Centroid Theorem (p.240) triangle. Incenter Theorem (p.239) Any pointequidistantfrom thesidesofananglelieson anglebisector. (p.239) Any pointontheanglebisectorisequidistantfrom thesides oftheangle. (p.239) vertices ofthetriangle. Circumcenter Theorem riangle InequalityTheorem If twosidesofatriangleare congruent totwosidesofanother The incenterofatriangleisequidistantfrom eachsideofthe The centroid ofatriangleislocatedtwo-thirds ofthedistance The circumcenter ofatriangleisequidistantfrom the The sumofthelengthsanytwosidesatriangleis If thetwoanglesofonetriangleare congruent totwo If themeasures ofthecorresponding sidesof two If twosidesofatriangleare congruent totwo If themeasures oftwosidesa triangleare If anangleisexteriorofatriangle, Theorem 6.4 Triangle Proportionality Theorem If a line is parallel to one side of a triangle and intersects the other two sides in two distinct points, then it separates these sides into segments of proportional lengths. (p. 307)

Theorem 6.5 Converse of the Triangle Proportionality Theorem If a line intersects two sides of a triangle and separates the sides into corresponding segments of proportional lengths, then the line is parallel to the third side. (p. 308)

Theorem 6.6 Triangle Midsegment Theorem Amidsegment of a triangle is parallel to one side of the triangle, and its length is one-half the length of that side. (p. 308)

Corollary 6.1 If three or more parallel lines intersect two transversals, then they cut off the transversals proportionally. (p. 309)

Corollary 6.2 If three or more parallel lines cut off congruent segments on one transversal, then they cut off congruent segments on every transversal. (p. 309)

Theorem 6.7 Proportional Perimeters Theorem If two triangles are similar, then the perimeters are proportional to the measures of corresponding sides. (p. 316)

Theorem 6.8 If two triangles are similar, then the measures of the corresponding altitudes are proportional to the measures of the corresponding sides. (p. 317) Abbreviation: s have corr. altitudes proportional to the corr. sides.

Theorem 6.9 If two triangles are similar, then the measures of the corresponding angle bisectors of the triangles are proportional to the measures of the corresponding sides. (p. 317) Abbreviation: s have corr. bisectors proportional to the corr. sides.

Theorem 6.10 If two triangles are similar, then the measures of the corresponding medians are proportional to the measures of the corresponding sides. (p. 317) Abbreviation: s have corr. medians proportional to the corr. sides. P ostulates, Theorems,

Theorem 6.11 Angle Bisector Theorem An angle bisector in a triangle separates the opposite side and Corollaries into segments that have the same ratio as the other two sides. (p. 319)

Chapter 7 Right Triangles and Trigonometry Theorem 7.1 If the altitude is drawn from the vertex of the right angle of a right triangle to its hypotenuse, then the two triangles formed are similar to the given triangle and to each other. (p. 343)

Theorem 7.2 The measure of the altitude drawn from the vertex of the right angle of a right triangle to its hypotenuse is the geometric mean between the measures of the two segments of the hypotenuse. (p. 343)

Theorem 7.3 If the altitude is drawn from the vertex of the right angle of a right triangle to its hypotenuse, then the measure of a leg of the triangle is the geometric mean between the measures of the hypotenuse and the segment of the hypotenuse adjacent to that leg. (p. 344)

Theorem 7.4 Pythagorean Theorem In a right triangle, the sum of the squares of the measures of the legs equals the square of the measure of the hypotenuse. (p. 350)

Theorem 7.5 Converse of the Pythagorean Theorem If the sum of the squares of the measures of two sides of a triangle equals the square of the measure of the longest side, then the triangle is a right triangle. (p. 351)

Theorem 7.6 In a 45°-45°-90° triangle, the length of the hypotenuse is 2 times the length of a leg. (p. 357)

Postulates, Theorems, and Corollaries R5 Postulates, Theorems, and Corollaries R6 Theorem 8.18 Theorem 8.17 Theorem 8.16 Theorem 8.15 Theorem 8.14 Theorem 8.13 Theorem 8.12 Theorem 8.11 Theorem 8.10 Theorem 8.9 Theorem 8.8 Theorem 8.7 Theorem 8.6 Theorem 8.5 Theorem 8.4 Theorem 8.1 Quadrilaterals Chapter 8 Theorem 7.7 Theorem 8.3 Theorem 8.2 Postulates, Theorems,and Corollaries ohpiso aeage fa ssee rpzi r ogun.(p. 439) Both pairsof baseanglesofanisosceles trapezoidare congruent. (p.431) Each diagonalofarhombus bisects apairofoppositeangles. r If thediagonalsofaparallelogram are perpendicular, thenthe parallelogram isa r If thediagonalsofaparallelogramare congruent, thentheparallelogramisa (p.424) If aparallelogramisrectangle, thenthediagonalsare congruent. (p.418) the quadrilateralisaparallelogram. If onepairofoppositesidesaquadrilateralisboth parallelandcongruent, then (p.418) parallelogram. If thediagonalsofaquadrilateralbisecteachother, thenthequadrilateralisa (p.418) is aparallelogram. If bothpairsofoppositeanglesaquadrilateralare congruent, thenthequadrilateral (p.418) is aparallelogram. If bothpairsofoppositesidesaquadrilateralare congruent, thenthequadrilateral (p.414) triangles. The diagonalofaparallelogramseparatestheintotwocongruent (p.413) The diagonalsofaparallelogrambisecteachother. (p.412) If aparallelogramhasonerightangle,itfourangles. (p.412) Consecutive anglesinaparallelogramare supplementary. (p.412) Opposite anglesofaparallelogramare congruent. the measures ofitsinteriorangles,then Interior AngleSumTheorem leg, andthelengthoflongerlegis triangle,thelengthofhypotenuseistwiceshorter In a30°-60°-90° Abbreviation: If Abbreviation: Ifonepairofopp.sidesis Abbreviation: Diag.of Abbreviation: If Abbreviation: Cons. Abbreviation: Opp. Abbreviation: Opp.sidesof poiesdso aallga r ogun.(p.412) Opposite sidesofaparallelogramare congruent. (p.406) of theexteriorangles,oneateachvertex,is360. Exterior AngleSumTheorem h ignl farobsaepredclr (p.431) The diagonalsofarhombus are perpendicular. obs (p.431) hombus. (p.426) ectangle. is rectangle,diag.are has 1rt. of in Abbreviation: Diag.of Abbreviation: Ifdiagonalsof bisect eachother. are are suppl. Abbreviation: Ifdiag.bisecteachother, thenquad.is , it has 4 rt. , ithas4rt. are Abbreviation: Ifbothpairsofopp. . Abbreviation: Ifbothpairsofopp.sidesare . If aconvexpolygonhas If apolygonisconvex,thenthesumofmeasures and . . , thenthequad.isa separates S 3 ie h egho h hre e.(p.359) times thelengthofshorterleg. 180( are n , into 2 ) (p.404) 2). is arectangle. n are sides and . s. , thenquad.is , thenquad.is S is thesumof . . . Theorem 8.19 The diagonals of an isosceles trapezoid are congruent. (p. 439)

Theorem 8.20 The median of a trapezoid is parallel to the bases, and its measure is one-half the sum of the measures of the bases. (p. 441)

Chapter 9 Transformations Postulate 9.1 In a given rotation, if A is the preimage, A’ is the image, and P is the center of rotation, then the measure of the angle of rotation, APA’ is twice the measure of the acute or right angle formed by the intersecting lines of reflection. (p. 477)

Corollary 9.1 Reflecting an image successively in two perpendicular lines results in a 180˚ rotation. (p. 477)

Theorem 9.1 If a dilation with center C and a scale factor of r transforms A to E and B to D, then ED r(AB). (p. 491)

Theorem 9.2 If P(x, y) is the preimage of a dilation centered at the origin with a scale factor r, then the image is P’(rx, ry). (p. 492)

Chapter 10 Circles Theorem 10.1 Two arcs are congruent if and only if their corresponding central angles are congruent. (p. 530)

Postulate 10.1 Arc Addition Postulate The measure of an arc formed by two adjacent arcs is the sum of the measures of the two arcs. (p. 531) P ostulates, Theorems,

Theorem 10.2 In a circle or in congruent circles, two minor arcs are congruent if and only if their and Corollaries corresponding chords are congruent. (p. 536) Abbreviations: In , 2 minor arcs are , iff corr. chords are . In , 2 chords are , iff corr. minor arcs are .

Theorem 10.3 In a circle, if a diameter (or radius) is perpendicular to a chord, then it bisects the chord and its arc. (p. 537)

Theorem 10.4 In a circle or in congruent circles, two chords are congruent if and only if they are equidistant from the center. (p. 539)

Theorem 10.5 If an angle is inscribed in a circle, then the measure of the angle equals one-half the measure of its intercepted arc (or the measure of the intercepted arc is twice the measure of the inscribed angle). (p. 544)

Theorem 10.6 If two inscribed angles of a circle (or congruent circles) intercept congruent arcs or the same arc, then the angles are congruent. (p. 546) Abbreviations: Inscribed of same arc are . Inscribed of arcs are .

Theorem 10.7 If an inscribed angle intercepts a semicircle, the angle is a right angle. (p. 547)

Theorem 10.8 If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. (p. 548)

Theorem 10.9 If a line is tangent to a circle, then it is perpendicular to the radius drawn to the point of tangency. (p. 553)

Postulates, Theorems, and Corollaries R7 Postulates, Theorems, and Corollaries R8 Theorem 13.1 Volume Chapter 13 Postulate 11.2 Postulate 11.1 Area ofPolygons AndCircles Chapter 11 Theorem 10.17 Theorem 10.16 Theorem 10.15 Theorem 10.14 Theorem 10.13 Theorem 10.12 Theorem 10.11 Theorem 10.10 Postulates, Theorems,and Corollaries of If twosolidsare similarwithascalefactorof (p.619) The area ofaregion isthesumofareas ofallitsnonoverlappingparts. (p.603) Congruent figures haveequalareas. (p.571) of themeasures ofthesecantsegmentanditsexternalsegment. point, thenthesquare ofthemeasure ofthetangentsegmentisequaltoproduct If atangentsegmentandsecantare drawntoacircle from anexterior (p.570) secant segment. equal totheproduct ofthemeasures oftheothersecantsegmentanditsexternal product ofthemeasures ofonesecantsegmentanditsexternalis If twosecantsegmentsare drawn toacircle from anexteriorpoint,thenthe (p.569) of thechords are equal. If twochords intersectinacircle, thentheproducts ofthemeasures ofthesegments (p.563) measures oftheintercepted arcs. circle, thenthemeasure oftheangleformedisone-halfpositivedifference ofthe If twosecants,asecantandtangent,ortangentsintersectintheexteriorof (p.562) angle formedisone-halfthemeasure ofitsintercepted arc. If asecantandtangentintersectatthepointoftangency, thenthemeasure ofeach (p.561) its verticalangle. formed isone-halfthesumofmeasure ofthearcs intercepted bytheangleand If twosecantsintersectintheinteriorofacircle, thenthe measure ofanangle (p.554) congruent. If twosegmentsfrom thesameexteriorpointare tangenttoacircle, thentheyare (p.553) line isatangenttothecircle. If alineisperpendiculartoradiusofcircle atitsendpoint onthecircle, thenthe a 2 : b 2 , andthevolumeshavearatioof a 3 : b a 3 : . b (p. 709) , thenthesurfaceareas havearatio Glossary/Glosario

English Español A acute angle (p. 30) An angle with a ángulo agudo Ángulo cuya medida en degree measure less than 90. grados es menos de 90.

A 0 mA 90 acute triangle (p. 178) A triangle in triángulo acutángulo Triángulo cuyos which all of the angles are acute 80û ángulos son todos agudos. angles. 40û 60û three acute angles tres ángulos agudos adjacent angles (p. 37) Two angles that lie in the ángulos adyacentes Dos ángulos que yacen sobre same plane, have a common vertex and a el mismo plano, tienen el mismo vértice y un common side, but no common interior points. lado en común, pero ningún punto interior. alternate exterior angles (p. 128) In the t ángulos alternos externos En la figura, figure, transversal t intersects lines 5 6 la transversal t interseca las rectas 8 7 and m. 5 and 3, and 6 and 4 1 2 m y m. 5 y 3, y 6 y 4 son are alternate exterior angles. 4 3 ángulos alternos externos. alternate interior angles (p. 128) In the figure ángulos alternos internos En la figura anterior, above, transversal t intersects lines and m. la transversal t interseca las rectas y m. 1 and 7, and 2 and 8 are alternate 1 y 7, y 2 y 8 son ángulos alternos interior angles. internos . altitude 1. (p. 241) In a triangle, a segment from a altura 1. En un triángulo, segmento trazado desde vertex of the triangle to the line containing the el vértice de un triángulo hasta el lado opuesto y opposite side and perpendicular to that side. que es perpendicular a dicho lado. 2. El seg- 2. (pp. 649, 655) In a prism or cylinder, a segment mento perpendicular a las bases de prismas y perpendicular to the bases with an endpoint in cilindros que tiene un extremo en cada plano. each plane. 3. (pp. 660, 666) In a pyramid or 3. El segmento que tiene un extremo en el vértice cone, the segment that has the vertex as one de pirámides y conos y que es perpendicular a la Glossary/Glosario endpoint and is perpendicular to the base. base. ambiguous case of the Law of Sines (p. 384) caso ambiguo de la ley de los senos Dadas Given the measures of two sides and a las medidas de dos lados y de un ángulo nonincluded angle, there exist two possible no incluido, existen dos triángulos triangles. posibles. angle (p. 29) The intersection of two noncollinear ángulo La intersección de dos semirrectas no rays at a common endpoint. The rays are called colineales en un punto común. Las semirrectas se sides and the common endpoint is called the llaman lados y el punto común se llama vértice. vertex. angle bisector (p. 32) A ray that Q bisectriz de un ángulo Semirrecta divides an angle into two congruent W que divide un ángulo en dos angles. P ángulos congruentes.

R PW is the bisector of P. PW es la bisectriz del P. Glossary/Glosario R9 Glossary/Glosario R10 circle chord centroid central angle center ofrotation biconditional between axis arc apo angle ofrotation angle ofelevation angle ofdepression Glossary/Glosario p 3)Apartofacircle thatisdefinedbytwo A (p. 530) AC if andonly a line,there isanotherpoint conditional statementanditsconverse. medians ofatriangle. center ofthecircle. circle intwopointsand hasitsvertexatthe endpoints thatare onthesphere. 2. endpoints thatare onthecircle. polygon. polygon perpendiculartoasideofthe drawn from endpoints. that are thevertexandcenterofbase. 2. endpoints thatare thecentersofbases. a preimage isrotated toformtheimage. point calledthe a planeequidistant from a given position. which shapesmoveinacircular motiontoanew observer looksdownward. the lineofsightandhorizontalwhenan looks upward. line ofsightandthehorizontalwhenanobserver them p 6)Inacone,thesegment withendpoints (p. 666) p 2)Thelocus ofallpointsin (p. 522) (p. 671)Foragivensphere, asegmentwith 1. 1. (p. 655)Inacylinder, thesegmentwith p 4)Thepointofconcurrency ofthe (p. 240) CB p 4 Foranytwopoints (p. 14) (p. 522)Foragivencircle, asegmentwith p 1)Asegmentthatis A (p. 610) p 2)Ananglethatintersectsa (p. 529) AB

p 1 Theconjunctionofa (p. 81) the centerofaregular p 7)Theanglethrough which (p. 476) . A center p 7)Theanglebetweenthe (p. 371) p 7)Afixedpointaround A (p. 476) , p 7)Theanglebetween (p. 372) B , and of thecircle. C C are collinearand between A and A P P and es el centro del círculo. del centro el es B is thecenterof circle. on B B C P apotema apothem eje arco ángulo derotación ángulo deelevación ángulo dedepresión cuerda centroide ángulo central centro derotación bicondicional ubicado entre AC r que mirahaciaarriba. horizontal ylalíneadevisiónunobservador extremos deunarecta. determinada. gira unafigurahastaalcanzarposición A de unarecta, existeunpunto de untriángulo. forman elvérticeycentro delabase. 2. forman elcentro delasbases. 2. círculo. condicional ysurecíproco. que mirahaciaabajo. horizontal ylalíneadevisiónunobservador del círculo. dos puntosycuyovérticeselocalizaenelcentro ota unapreimagen paraformarlaimagen. 1. Segmento cuyosextremos estánenunaesfera. y Parte deuncírculo definidaporlosdos El segmentoenunconocuyosextremos El segmentoenuncilindro cuyosextremos B 1. CB apotema si ysólo Segmento cuyosextremos estánenun Punto deintersecciónlasmedianas círculo lados. polígono regular hastaunodesus trazado desdeelcentro deun AB dado llamado plano, equidistantes deunpunto por elconjunto depuntosenun Para cualquierpardepuntos La conjunciónentre unenunciado Ángulo queintersecauncírculo en . Punto fijoalrededor delcual El ánguloatravésdelcualse Lugar geométricoformado Segmento perpendicular A Ángulo formadoporla , Ángulo formadoporla B y centro C son colinealesy C . ubicado entre A y B circumcenter (p. 238) The point of concurrency of circuncentro Punto de intersección de las the perpendicular bisectors of a triangle. mediatrices de un triángulo. circumference (p. 523) The distance around a circle. circunferencia Distancia alrededor de un círculo. circumscribed (p. 537) A circle is A circunscrito Un polígono está B circumscribed about a polygon if circunscrito a un círculo si todos the circle contains all the vertices E sus vértices están contenidos en el of the polygon. D C círculo.

E is circumscribed about quadrilateral ABCD. E está circunscrito al cuadrilátero ABCD.

(p. 6) Points that lie on Puntos que yacen en la collinear R colineal the same line. P Q misma recta. P, Q, and R are collinear. P, Q y R son colineales. column matrix (p. 506) A matrix containing one matriz columna Matriz formada por una sola column often used to represent an ordered pair columna y que se usa para representar pares x ordenados o vectores como, por ejemplo, or a vector, such as x, y . y x x, y . y complementary angles (p. 39) Two angles with ángulos complementarios Dos ángulos cuya measures that have a sum of 90. suma es igual a 90 grados. component form (p. 498) A vector expressed as an componente Vector representado en forma de par ordered pair, change in x, change in y. ordenado, cambio en x, cambio en y. composition of reflections (p. 471) Successive composición de reflexiones Reflexiones reflections in parallel lines. sucesivas en rectas paralelas. compound statement (p. 67) A statement formed enunciado compuesto Enunciado formado por la by joining two or more statements. unión de dos o más enunciados.

concave polygon (p. 45) A polygon for which there polígono cóncavo Polígono para el cual existe una Glossary/Glosario is a line containing a side of the polygon that also recta que contiene un lado del polígono y un contains a point in the interior of the polygon. punto interior del polígono. conclusion (p. 75) In a conditional statement, the conclusión Parte del enunciado condicional que statement that immediately follows the word then. está escrita después de la palabra entonces. concurrent lines (p. 238) Three or more lines that rectas concurrentes Tres o más rectas que se intersect at a common point. intersecan en un punto común. conditional statement (p. 75) A statement that can enunciado condicional Enunciado escrito en la be written in if-then form. forma si-entonces.

vertex cone (p. 666) A solid with a circular base, a vértice cono Sólido de base circular cuyo vértice vertex not contained in the same plane as the no se localiza en el mismo plano que la base, and a lateral surface area composed of base base y cuya superficie lateral está base all points in the segments connecting the formada por todos los segmentos que vertex to the edge of the base. unen el vértice con los límites de la base.

Glossary/Glosario R11 Glossary/Glosario R12 coplanar coordinate proof convex polygon converse contrapositive construction consecutive interiorangles conjunction conjecture congruent triangles congruent solids congruent arcs congruent congruence transformations Glossary/Glosario geometric concepts. in thecoordinate planeandalgebratoprove conditional statement. exchanging thehypothesisandconclusionofa the converseofaconditionalstatement. negating boththehypothesisandconclusionof consecutive interiorangles: known information. word by joiningtwoormore statementswiththe congruent. for whichageometricfigure anditsimageare straightedge, andcompassare used. measuring tools.Generally, only apencil, geometric figures withoutthebenefitof 4. 3. 2. 1. if allofthefollowingconditionsare met. their corresponding partscongruent. lines In thefigure, transversal and apointintheinterior of thepolygon. is nolinethatcontainsbothasideofthepolygon congruent circles thathavethesamemeasure. 1, and The volumesare congruent. Corresponding facesare congruent. Corresponding edgesare congruent. The corresponding anglesare congruent. p )Pointsthat lie inthe sameplane. (p. 6) and and p 5 Havingthesamemeasure. (p. 15) p 7 Thestatementformedby (p. 77) p 2 Aneducatedguessbasedon (p. 62) (p. 68) A compound statementformed compound A (p. 68) . 7 and m p 5 methodofcreating A (p. 15) p 3)Arcs ofthesamecircle or (p. 530) p 7 Thestatementformedby (p. 77) p 5 polygonforwhichthere A (p. 45) . There are twopairsof p 0)Two solidsare congruent (p. 707) p 2)Aproof thatusesfigures A (p. 222) p 9)Triangles thathave (p. 192) 2. t p 9)Amapping A (p. 194) intersects (p. 128) 8 and 4 1 3 8 2 5 7 6 t conjetura triángulos congruentes sólidos congruentes arcos congruentes congruente transformación decongruencia coplanar prueba decoordenadas polígono convexo recíproco antítesis construcción conjunción m enunciado condicionaldado. intercambiar lahipótesis ylaconclusióndeun al unirdosomásenunciadosconlapalabra para demostrarconceptos geométricos. álgebra yfigurasenelplano decoordenadas enunciado condicionaldado. hipótesis ylaconclusióndelrecíproco deun 4. 3. 2. 1. si cumplentodaslassiguientescondiciones: una regla sinescalayuncompás. medición. Engeneral,sólorequiere deunlápiz, geométricas sinelusodeinstrumentos de imagen soncongruentes. en unplanolaquefigurageométricaysu un puntoenelinteriordel polígono. r círculos congruentes, quetienenlamismamedida. correspondientes soncongruentes. ecta algunaquecontengaunladodelpolígonoy Los volúmenessoncongruentes. Las carascorrespondientes soncongruentes. Las aristascorrespondientes soncongruentes. Los ánguloscorrespondientes soncongruentes. ángulos internosconsecutivos Enunciado formadoporlanegaciónde Puntos queyacen enunmismoplano. Juicio basadoeninformaciónconocida. Enunciado queseobtieneal Que midenlomismo. Enunciado compuestoqueseobtiene r internos: pares deángulosconsecutivos figura, latransversal ectas Método paradibujarfiguras Arcos deunmismocírculo, ode Polígono paraelcualnoexiste y Dos sólidossoncongruentes m 8 y . Lafigurapresenta dos T Demostración queusa riángulos cuyaspartes 1, y T 7 y t ransformación interseca las 2. En la y . corner view (p. 636) The view from a corner of a vista de esquina Vista de una figura three-dimensional figure, also called the tridimensional desde una esquina. También se perspective view. conoce como vista de perspectiva. corollary (p. 188) A statement that can be easily corolario La afirmación que puede demostrarse proved using a theorem is called a corollary of fácilmente mediante un teorema se conoce como that theorem. corolario de dicho teorema. corresponding angles (p. 128) In t ángulos correspondientes En la 5 6 the figure, transversal t intersects figura, la transversal t interseca las 8 7 lines and m. There are four pairs 1 2 m rectas y m. La figura muestra cuatro of corresponding angles: 5 and 1, 4 3 pares de ángulos correspondientes: 5 8 and 4, 6 and 2, and 7 and 3. y 1, 8 y 4, 6 y 2, y 7 y 3. cosine (p. 364) For an acute angle of a right coseno Para un ángulo agudo de un triángulo triangle, the ratio of the measure of the leg rectángulo, la razón entre la medida del cateto adjacent to the acute angle to the measure of the adyacente al ángulo agudo y la medida de la hypotenuse. hipotenusa de un triángulo rectángulo. counterexample (p. 63) An example used to show contraejemplo Ejemplo que se usa para demostrar that a given statement is not always true. que un enunciado dado no siempre es verdadero.

a c a c cross products (p. 283) In the proportion , productos cruzados En la proporción, , b d b d where b 0 and d 0, the cross products are ad donde b 0 y d 0, los productos cruzados son and bc. The proportion is true if and only if the ad y bc. La proporción es verdadera si y sólo si los cross products are equal. productos cruzados son iguales. cylinder (p. 638) A figure with bases that base cilindro Figura cuyas bases son círculos are formed by congruent circles in base congruentes localizados en planos parallel planes. base paralelos. base

D deductive argument (p. 94) A proof formed by a argumento deductivo Demostración que consta group of algebraic steps used to solve a problem. del conjunto de pasos algebraicos que se usan para resolver un problema. Glossary/Glosario deductive reasoning (p. 82) A system of reasoning razonamiento deductivo Sistema de razonamiento that uses facts, rules, definitions, or properties to que emplea hechos, reglas, definiciones y reach logical conclusions. propiedades para obtener conclusiones lógicas. degree (p. 29) A unit of measure used in measuring grado Unidad de medida que se usa para medir angles and arcs. An arc of a circle with a measure ángulos y arcos. El arco de un círculo que mide 1 1 of 1° is of the entire circle. 1° equivale a del círculo completo. 360 360 P Q diagonal (p. 404) In a polygon, a diagonal Recta que une vértices no segment that connects nonconsecutive consecutivos de un polígono. vertices of the polygon. SR SQ is a diagonal. SQ es una diagonal. diameter 1. (p. 522) In a circle, a chord that passes diámetro 1. Cuerda que pasa por el centro de un through the center of the circle. 2. (p. 671) In a círculo. 2. Segmento que incluye el centro de sphere, a segment that contains the center of the una esfera y cuyos extremos se localizan en la sphere, and has endpoints that are on the sphere. esfera.

Glossary/Glosario R13 Glossary/Glosario R14 flow proof extremes exterior angle exterior equilateral triangle equiangular triangle equal vectors disjunction direction direct isometry dilation Glossary/Glosario CP joining twoormore statementswiththeword the pointonrayopposite CP horizontal line. vector formswiththepositive the image center point the extensionofanotherside. formed byonesideofatriangleand intact withintheplane. image ofafigure isfoundbymovingthefigure with allsidescongruent. with allanglescongruent. magnitude anddirection. indicate theorder ofthestatements. written below thebox. Arrows are usedto box withthe reason verifying thestatement given statements.Eachstatement iswrittenina statements inlogicalorder, starting withthe the angle. on theanglenorininteriorof exterior ofanangleifitisneither p 9)Atransformationdeterminedbya A (p. 490) p 9 pointisinthe A (p. 29) p 8)In (p. 283) p 9)Themeasure oftheanglethata (p. 498) k  k p 8 compoundstatementformedby A (p. 68)  CP p 8)Aproof thatorganizes A (p. 187) P p 9)Vectors thathavethesame (p. 499) CP . C p 8)Anisometryinwhichthe (p. 481) of p 8)Anangle (p. 186) . When and ascalefactor P p 7)Atriangle A (p. 179) is thepointon b a p 7)Atriangle A (p. 178) k d c , thenumbers 0, theimage x -axis oranyother k CP . When CP such that such that a P A A and

of está en el exterior del del exterior el en está is intheexteriorof k A 1 or 1 1 P d Y

. es un ángulo externo. ángulo un es is anexteriorangle. . 0, is X F E vectores iguales disyunción dirección isometría directa dilatación demostración deflujo extremos Z XYZ k es elpuntoenlasemirrecta opuesta CP unir dosomásenunciadosconlapalabra junto consuplano. imagen deunafigura,almoverlafiguraintacta punto central CP horizontal. con elejepositivo magnitud ydirección. orden delosenunciados seindicamedianteflechas. escribe elargumento queverificaelenunciado. El escribe enuna casillaydebajodecada casilla se do conlosenunciadosdados. Cadaenunciadose ordenan losenunciadosenorden lógico, empezan- XYZ. . 0, laimagen ángulo externo triángulo equilátero triángulo equiangular exterior k  Los números k Medida delánguloqueformaunvector ángulo. ni enelángulointeriordel de otro desuslados. lados soncongruentes entre sí. exterior deunángulosinoselocaliza un ladodetriánguloylaextensión ángulos soncongruentes entre sí.  CP Tr Enunciado compuestoqueseformaal ansformación determinadaporun CP . C . Cuando Un puntoyaceenel Isometría enlacualseobtiene P V y unfactordeescala ectores queposeenlamisma de x a o concualquierotrarecta P y Demostración enquese es elpuntoen k Ángulo formadopor d en 0, laimagen b a T T riángulo cuyos riángulo cuyos d c . CP k CP . Cuando o P . tal que tal que de P fractal (p. 325) A figure generated by repeating a fractal Figura que se obtiene mediante la repetición special sequence of steps infinitely often. Fractals infinita de una sucesión particular de pasos. Los often exhibit self-similarity. fractales a menudo exhiben autosemejanza. G geometric mean (p. 342) For any positive numbers media geométrica Para todo número positivo a x a x a and b, the positive number x such that . a y b, existe un número positivo x tal que . x b x b geometric probability (p. 622) Using the probabilidad geométrica El uso de los principios principles of length and area to find the de longitud y área para calcular la probabilidad probability of an event. de un evento. glide reflection (p. 475) A composition of a reflexión de deslizamiento Composición que consta translation and a reflection in a line parallel to de una traslación y una reflexión realizadas sobre the direction of the translation. una recta paralela a la dirección de la traslación. great circle (p. 671) For a given sphere, the círculo máximo La intersección entre una esfera intersection of the sphere and a plane that dada y un plano que contiene el centro de la contains the center of the sphere. esfera.

H height of a parallelogram (p. 595) AB altura de un paralelogramo The length of an altitude of a La longitud de la altura de un h parallelogram. paralelogramo. D C h is the height of parallelogram ABCD. H es la altura del paralelogramo ABCD. hemisphere (p. 672) One of the two congruent hemisferio Cada una de las dos partes congruentes parts into which a great circle separates a sphere. en que un círculo máximo divide una esfera. hypothesis (p. 75) In a conditional statement, the hipótesis El enunciado escrito a continuación de la statement that immediately follows the word if. palabra si en un enunciado condicional.

I Glossary/Glosario if-then statement (p. 75) A compound statement enunciado si-entonces Enunciado compuesto de of the form “if A, then B”, where A and B are la forma “si A, entonces B”, donde A y B son statements. enunciados. incenter (p. 240) The point of concurrency of the incentro Punto de intersección de las bisectrices angle bisectors of a triangle. interiores de un triángulo. included angle (p. 201) In a triangle, the angle ángulo incluido En un triángulo, el ángulo formed by two sides is the included angle for formado por dos lados cualesquiera del those two sides. triángulo es el ángulo incluido de esos dos lados. included side (p. 207) The side of a triangle that is lado incluido El lado de un triángulo que es a side of each of two angles. común a de sus dos ángulos. indirect isometry (p. 481) An isometry that cannot isometría indirecta Tipo de isometría que no se be performed by maintaining the orientation of puede obtener manteniendo la orientación de los the points, as in a direct isometry. puntos, como ocurre durante la isometría directa.

Glossary/Glosario R15 Glossary/Glosario R16 irregular polygon irregular figure inverse interior in inscribed inductive reasoning indirect reasoning indirect proof tercepted Glossary/Glosario 3. 2. 1. only ifeachofthefollowingconditionsare met. deductive reasoning. the logicalcertaintyofthosearrivedatby Conclusions arrivedatbyinductivereasoning lack plausible generalizationorprediction. number ofspecificexamplestoarriveata conditional statement. both thehypothesisandconclusionofa not regular. assumed falsemustinfactbetrue. obtained, oneconcludesthatthestatement of theassumptions.Onceacontradictionis statement contradictsapostulate,theorem, orone One thenuseslogicalreasoning todeducethata assumes thatthestatementtobeproved isfalse. vertices lieonthecircle. inscribed inacircle ifeachofits classified asasinglepolygon. on thesidesofangle. segment withendpointsthatare on theangleitselfanditliesa interior ofanangleifitdoesnotlie must betrue. assumption hasbeenproved false,theconclusion postulate, theorem, orcorollary. Then,sincethe hypothesis orsomeotheracceptedfact,likea this assumptionleadstoacontradictionofthe that theconclusionisfalseandthenshows the arc. Each sideoftheanglecontainsanendpoint the interiorofcircle. All pointsofthearc excepttheendpointsare in The endpointsofthearc lieontheangle. p 7 Thestatementformedbynegating (p. 77) p 9 pointisinthe A (p. 29) p 3)Apolygonis A (p. 537) p 4)Anangleintercepts anarc ifand (p. 544) p 5)Inanindirect proof, one (p. 255) p 1)Afigure thatcannotbe A (p. 617) p 1)Apolygonthatis A (p. 618) p 5)Reasoningthatassumes (p. 255) p 2 Reasoningthatusesa (p. 62) M M

está en el interior del del interior el en está is intheinteriorof LMN LMN LMN N is inscribedin está inscrito en en inscrito está L J M L P razonamiento inductivo razonamiento indirecto demostración indirecta figura irregular inversa in tersecado M JKL como unsolopolígono. condicional. hipótesis ylaconclusióndeunenunciado 3. 2. 1. se cumplentodaslassiguientescondiciones. entonces laconclusióndebeserverdadera. se hademostradoquelaconjeturaesfalsa, un teorema ouncorolario. Finalmente,dadoque hipótesis ounhechoaceptadocomopostulado, después, sedemuestraqueestocontradicela primero seasumequelaconclusiónesfalsay, debe ser, enrealidad, verdadero. concluye queelenunciadosesuponíafalso conjeturas. Unavezhalladaunacontradicción,se contradice unpostulado,teorema ounadelas lógicamente queexisteunenunciado demostrar esfalso.Después,sededuce indirecta, seasumequeelenunciadopor deductivo. aquellas obtenidasmedianteelrazonamiento inductivo carecen delacertidumbre lógicade conclusiones generalización ounapredicción creíble. Las varios ejemplosespecíficosparalograruna P JKL. K P . del arco. Cada ladodelángulocontieneunextremo extremos, yacenenelinteriordelcírculo. T Los extremos delarco yacenenelángulo.

. . polígono irregular odos lospuntosdelarco, exceptuandosus Enunciado queseobtienealnegarla r egular. inscrito Un ángulointersecaunarco siysólo interior obtenidas medianteelrazonamiento yacen enelcírculo. un círculo sitodossusvértices Figura quenosepuedeclasificar yacen enlosladosdelángulo. un segmentocuyosextremos en elángulomismoysiestá interior deunángulo,sinoyace Un polígonoestáinscritoen Un puntoselocalizaenel En unademostración Razonamiento enque Razonamiento queusa Polígono que noes isometry (p. 463) A mapping for which the original isometría Transformación en que la figura original figure and its image are congruent. y su imagen son congruentes. WX isosceles trapezoid (p. 439) A trapecio isósceles Trapecio cuyos trapezoid in which the legs are catetos son congruentes, ambos pares congruent, both pairs of base angles de ángulos son congruentes y las are congruent, and the diagonals are Z Y diagonales son congruentes. congruent. isosceles triangle (p. 179) A vertex angle triángulo isósceles Triángulo que triangle with at least two sides ángulo del vértice tiene por lo menos dos lados congruent. The congruent sides leg leg congruentes. Los lados are called legs. The angles cateto cateto congruentes se llaman catetos. opposite the legs are base angles. base angles Los ángulos opuestos a los The angle formed by the two ángulos de la base catetos son los ángulos de la base. base legs is the vertex angle. The side base El ángulo formado por los dos opposite the vertex angle is the catetos es el ángulo del vértice. Los base. lados opuestos al ángulo del vértice forman la base. iteration (p. 325) A process of repeating the same iteración Proceso de repetir el mismo procedure over and over again. procedimiento una y otra vez. K kite (p. 438) A quadrilateral with cometa Cuadrilátero que tiene exactly two distinct pairs of adjacent exactamente dospares de lados congruent sides. congruentes adyacentes distintivos.

L lateral area (p. 649) For prisms, pyramids, área lateral En prismas, pirámides, cilindros y cylinders, and cones, the area of the figure, not conos, es el área de la figura, sin incluir el área de including the bases. las bases. lateral edges 1. (p. 649) In a prism, the intersection aristas laterales 1. En un prisma, la intersección of two adjacent lateral faces. 2. (p. 660) In a de dos caras laterales adyacentes. 2. En una pyramid, lateral edges are the edges of the lateral pirámide, las aristas de las caras laterales que Glossary/Glosario faces that join the vertex to vertices of the base. unen el vértice de la pirámide con los vértices de la base. lateral faces 1. (p. 649) In a prism, the faces that are caras laterales 1. En un prisma, las caras que no not bases. 2. (p. 660) In a pyramid, faces that forman las bases. 2. En una pirámide, las caras intersect at the vertex. que se intersecan en el vértice.

Law of Cosines (p. 385) Let ABC be any triangle ley de los cosenos Sea ABC cualquier triángulo with a, b, and c representing the measures of sides donde a, b y c son las medidas de los lados opuestos opposite the angles with measures A, B, and C a los ángulos que miden A, B y C respectivamente. respectively. Then the following equations are true. Entonces las siguientes ecuaciones son ciertas. a2 b2 c2 2bc cos A a2 b2 c2 2bc cos A b2 a2 c2 2ac cos B b2 a2 c2 2ac cos B c2 a2 b2 2ab cos C c2 a2 b2 2ab cos C

Law of Detachment (p. 82) If p → q is a true ley de indiferencia Si p → q es un enunciado conditional and p is true, then q is also true. condicional verdadero y p es verdadero, entonces q es verdadero también.

Glossary/Glosario R17 Glossary/Glosario R18 major arc magnitude logically equivalent locus linear pair line segment line ofsymmetry line ofreflection line Law ofSyllogism Law ofSines Glossary/Glosario r true conditionals,then opposite theangleswithmeasures a the sametruth values. condition. on theoppositeside. is thereflection imageofthefigure figure sothatthefigure ononeside can bedrawnthrough aplane that separatesthefigure intotwomirror images. ACB measure greater than180. rays. common sidesare opposite adjacent angleswhosenon- all ofthepointsbetweenthem. that consistsoftwopoints,calledendpoints,and arrow overthepairofletters. letters fortwopointsontheline,withadouble by lowercase scriptlettersorbywritingcapital arrowhead ateachend.Linesare usuallynamed width. Inafigure, alineisshownwithan line ismadeupofpointsandhasnothicknessor espectively. Then, p )Abasicundefined term ofgeometry. A A (p. 6) , p 1 Thesetofpointsthatsatisfyagiven (p. 11) b , and is amajorarc. p 9)Thelengthofavector. (p. 498) p 3)Anarc witha (p. 530) p 7)Let (p. 377) c p 7 pairof A (p. 37) p 3 measurablepartofaline A (p. 13) r epresenting themeasures ofsides p 6)Alinethat A (p. 466) p 6)Alinethrough afigure A (p. 463) p 3 If (p. 83) p 7 Statementsthathave (p. 77) sin a A p ABC → r p sin is alsotrue. be anytrianglewith → b B q and sin A c , C q B PSQ PSQ PSR . → , and D A AC AC y y r and

C es un eje de simetría. de eje un es is alineofsymmetry. are QSR C B QSR QSR C P M forman un par lineal. par un forman A are alinearpair. A B línea dereflexión recta ley delsilogismo ley delossenos magnitud equivalente lógico lugar geométrico segmento derecta B también esverdadero. condicionales verdaderos, entonces doble sobre elpardeletrasmayúsculas. de dospuntossobre lalínea.Seescribeunaflecha letras minúsculasoconlasdosmayúsculas en cadaextremo. Porlogeneral,sedesignancon una figura,recta serepresenta conunaflecha formada porpuntosycarece degrosor oancho.En mismo valordeverdad. dos imágenesespeculares. todos lospuntoslocalizadosentre ellos. Consta dedospuntos,llamadosextremos, y satisfacen unacondicióndada. donde re lados opuestosalosángulos Q spectivamente. Entonces, C Término primitivoengeometría.Unarecta está arco mayor a , La longituddeunvector. eje desimetría b ACB 180°. y par lineal imagen reflejada delladoopuesto. modo queunladodelafiguraes través deunafiguraplana, c r Sea es unarco mayor. Si epresentan lasmedidasdelos opuestas. comunes formansemirrectas adyacentes cuyosladosno Línea quedivideunafiguraen Sección medibledeunarecta. Conjunto depuntosque p Enunciados queposeenel → Arco quemide másde ABC q y q sin Recta quesetrazaa cualquier triángulo Par deángulos → a A r son enunciados sin A b , B B sin p y c → C C . r matrix logic (p. 88) A method of deductive lógica matricial Método de razonamiento deductivo reasoning that uses a table to solve problems. que utiliza una tabla para resolver problemas.

a c a c means (p. 283) In , the numbers b and c. medios Los números b y c en la proporción . b d b d median 1. (p. 240) In a triangle, a line segment with mediana 1. Segmento de recta de un triángulo endpoints that are a vertex of a triangle and the cuyos extremos son un vértice del triángulo y el midpoint of the side opposite the vertex. punto medio del lado opuesto a dicho vértice. 2. (p. 440) In a trapezoid, the segment that joins 2. Segmento que une los puntos medios de los the midpoints of the legs. catetos de un trapecio. midpoint (p. 22) The point halfway between the punto medio Punto que es equidistante entre los endpoints of a segment. extremos de un segmento. midsegment (p. 308) A segment with endpoints segmento medio Segmento cuyos extremos son los that are the midpoints of two sides of a triangle. puntos medios de dos lados de un triángulo. minor arc (p. 530) An arc with a measure A arco menor Arco que mide menos de 180° . less than 180. AB is a minor arc. AB es un arco menor. P

B N negation (p. 67) If a statement is represented by p, negación Si p representa un enunciado, entonces then not p is the negation of the statement. no p representa la negación del enunciado. net (p. 644) A two-dimensional figure that when red Figura bidimensional que al ser plegada folded forms the surfaces of a three-dimensional forma las superficies de un objeto object. tridimensional. n-gon (p. 46) A polygon with n sides. enágono Polígono con n lados. non-Euclidean geometry (p. 165) The study of geometría no euclidiana El estudio de sistemas geometrical systems that are not in accordance geométricos que no satisfacen el Postulado de las

with the Parallel Postulate of Euclidean Paralelas de la geometría euclidiana. Glossary/Glosario geometry. O oblique cone (p. 666) A cone that is not a cono oblicuo Cono que no es un cono recto. right cone.

oblique cylinder (p. 655) A cylinder that cilindro oblicuo Cilindro que no es un is not a right cylinder. cilindro recto.

oblique prism (p. 649) A prism in which prisma oblicuo Prisma cuyas aristas the lateral edges are not perpendicular laterales no son perpendiculares a to the bases. las bases.

Glossary/Glosario R19 Glossary/Glosario R20 perpendicular bisector perimeter parallelogram parallel vectors parallel planes parallel lines paragraph proof orthogonal drawing orthocenter ordered triple opposite rays obtuse triangle obtuse angle Glossary/Glosario same oroppositedirection. an obtuseangle. lengths ofthesidesapolygon. that side. of asideand isperpendicularto that passesthrough themidpoint a triangle,line,segment, orray called a side ofaparallelogrammaybe with paralleloppositesides. Any that donotintersect. specific order usedtolocatepointsinspace. BA less than180. degree measure greater than90and conjecture foragivensituationistrue. in theformofaparagraphthatexplainswhy a three-dimensional object. top view, leftview, front view, andrightviewof between the altitudesofatriangle. and p 6 Thesumofthe (p. 46) base A p 4)Thepointofconcurrency of (p. 240) BC p 2)Coplanarlines (p. 126) and p 1)Aquadrilateral A (p. 411) p 2)Planesthatdonotintersect. (p. 126) . p 1)Three numbersgivenina (p. 714) p 9 Two rays (p. 29) p 0 Ananglewith (p. 30) p 7)Atrianglewith A (p. 178) p 0 Aninformalproof written (p. 90) such that p 9)Vectors thathavethe (p. 499) C . p 3)Thetwo-dimensional (p. 636) p 3)In (p. 238) B is A BD DC D D AB AB es el punto medio de medio punto el es C AB is themidpointof A 90 A un ángulo obtuso ángulo un one obtuseangle 120 43

A DC DC AB AB û û m ; B P AD AD

A A CD mediatriz bisector perpendicular ortocentro triple ordenado vectores paralelos planos paralelos demostración depárrafo vista ortogonal

180 BC 17 BC C û D B

un triángulo. verdadera. una conjeturaacerca deunasituacióndadaes escrita enformadepárrafoqueexplicaporqué dirección oladirección opuesta. derecha deuncuerpotridimensional. desde laizquierda, desdeelfrente odesdela espacio. específico quesirvenparaubicarpuntosenel BC. . C ángulo obtuso triángulo obtusángulo paralelogramo rectas paralelas Punto deintersecciónlasalturas 0 ymenosde180°. 90° mediatriz perímetro tiene unánguloobtuso. ogramo puedeserla sí. Cualquierladodelparalel- lados opuestossonparalelosentre que noseintersecan. semirrectas opuestas V Tr lado. que esperpendicular adicho medio dellado deuntriánguloy de losladosunpolígono. semirrecta queatraviesaelpunto Planos quenoseintersecan. ista bidimensionaldesdearriba, es números dadosenunorden semirrectas B V ectores quetienenlamisma se localizaentre Ángulo quemidemásde La sumadelalongitud Recta, segmentoo Demostración informal Cuadrilátero cuyos Rectas coplanares BA base y T riángulo que BC A . y tales que C . Dos perpendicular lines (p. 40) Lines that m rectas perpendiculares Rectas que form right angles. n forman ángulos rectos.

line m line n recta m recta n perspective view (p. 636) The view of a three- vista de perspectiva Vista de una figura dimensional figure from the corner. tridimensional desde una de sus esquinas. pi () (p. 524) An irrational number represented by pi () Número irracional representado por la razón the ratio of the circumference of a circle to the entre la circunferencia de un círculo y su diameter of the circle. diámetro. plane (p. 6) A basic undefined term of geometry. A plano Término primitivo en geometría. Es una plane is a flat surface made up of points that has superficie formada por puntos y sin profundidad no depth and extends indefinitely in all directions. que se extiende indefinidamente en todas direccio- In a figure, a plane is often represented by a nes. Los planos a menudo se representan con un shaded, slanted 4-sided figure. Planes are usually cuadrilátero inclinado y sombreado. Los planos en named by a capital script letter or by three general se designan con una letra mayúscula o con noncollinear points on the plane. tres puntos no colineales del plano. plane Euclidean geometry (p. 165) Geometry geometría del plano euclidiano Geometría based on Euclid’s axioms dealing with a system basada en los axiomas de Euclides, los que of points, lines, and planes. integran un sistema de puntos, rectas y planos.

Platonic Solids (p. 637) The five regular polyhe- sólidos platónicos Cualquiera de los siguientes dra: tetrahedron, hexahedron, octahedron, cinco poliedros regulares: tetraedro, hexaedro, dodecahedron, or icosahedron. octaedro, dodecaedro e icosaedro. point (p. 6) A basic undefined term of geometry. punto Término primitivo en geometría. Un punto A point is a location. In a figure, points are representa un lugar o localización. En una figura, represented by a dot. Points are named by se representa con una marca puntual. Los puntos capital letters. se designan con letras mayúsculas.

point of concurrency (p. 238) The point of punto de concurrencia Punto de intersección de Glossary/Glosario intersection of concurrent lines. rectas concurrentes. point of symmetry (p. 466) The punto de simetría El punto común common point of reflection for all R de reflexión de todos los puntos de points of a figure. una figura.

R is a point of symmetry. R es un punto de simetría. point of tangency (p. 552) For a line that intersects punto de tangencia Punto de intersección de una a circle in only one point, the point at which they recta que interseca un círculo en un solo punto, el intersect. punto en donde se intersecan. point-slope form (p. 145) An equation of the forma punto-pendiente Ecuación de la forma form y y m(x x ), where (x , y ) are the y y m(x x ), donde (x , y ) representan 1 1 1 1 1 1 1 1 coordinates of any point on the line and m is las coordenadas de un punto cualquiera sobre la the slope of the line. recta y m representa la pendiente de la recta.

Glossary/Glosario R21 Glossary/Glosario R22 postulate polyhedrons polygon pyramid proportion proof bycontradiction proof prism precision Glossary/Glosario 3. 2. 1. the followingconditionsare met. number ofcoplanarsegmentscalled that statestworatiosare equal. that isacceptedastrue. statement youmakeissupportedbya measuring tool. depends onthesmallestunitavailable statement assumedfalsemustinfactbetrue. contradiction isobtained,oneconcludesthatthe theorem, oroneoftheassumptions.Once a deduce astatementthatcontradictspostulate, proved isfalse.Onethenuseslogicalreasoning to which oneassumesthatthestatementtobe intersect are called called are called re figures madeupofflatpolygonalregions. Theflat 2. 1. following characteristics. 1. following characteristics. without proof. of geometry. Postulates are acceptedastrue fundamental relationship betweenthebasicterms 3. 2. gions formedbythepolygonsandtheirinteriors r called The facesmeetingatthevertex are polygonal region. only attheirendpoints,calledthe Each sideintersectsexactlytwoothersides,but noncollinear. The sidesthathaveacommonendpointare vertex iscalledthe The facethatdoesnotcontain the intersect atapointcalledthe All ofthefaces,except one face, T edges lateral facesare called The intersectionsoftwoadjacent parallelograms. called The facesthatare notbases, that lieinparallelplanes. formed bycongruent polygons p 0 logicalargument inwhich each A (p. 90) egions. wo faces,called p 3)Asolidwiththe A (p. 637) p 5 closedfigure formedbyafinite A (p. 45) p 4 Theprecision ofany measurement (p. 14) edges p 3)Asolidwiththe A (p. 637) p 9 statementthatdescribesa A (p. 89) and are parallelsegments. p 8)Anequationoftheform (p. 283) aea faces lateral lateral faces faces p 3)Closedthree-dimensional (p. 637) . Pointswhere three ormore edges . Pairsoffacesintersectinsegments vertices , are formedby and are triangular p 5)Anindirect proof in (p. 255) base bases . and isa lateral , are vertex sides . vertices such that base base cara face lateral lateral . vértice prisma triangular prisma prism triangular b a vertex arista lateral arista lateral edge pirámide rectangular pirámide rectangular pyramid d c precisión postulado poliedro polígono proporción demostración porcontradicción demostración cara lateral cara face lateral base base más aristassellaman que dosrazonessoniguales. supuso falsoes,enrealidad, verdadero. dicción, seconcluyequeelenunciado conjeturas. Unavezqueseobtieneunacontra- tradiga unpostulado,teorema ounadelas lógicamente paradeducirunenunciadoquecon- va ademostraresfalso.Después,serazona indirecta enqueseasumeelenunciado de medición. unidad demedidamáspequeñadelinstrumento acepta comoverdadero. enunciado estábasadoenunquese llama llaman definidas porunpolígonoysusinteriores se r verdaderos sinnecesidaddedemostración. geometría. Lospostuladosseaceptancomo fundamental entre lostérminosprimitivosde finito desegmentoscoplanares llamados 2. que satisfacelassiguientescondiciones: 1. egiones poligonalesplanas.Lasregiones planas sólo ensusextremos, formandolos Cada ladointersecaexactamentedoslados,pero colineales. Los ladosquetienenunextremo comúnsonno pirámide arista Figura tridimensionalcerradaformadapor caras prisma La precisión deunamedidadependela Figura cerradaformadaporunnúmero Enunciado quedescribeunarelación Ecuación delaforma 2. 3. 1. características: . Lospuntosdondeseintersecantres o 1. siguientes características: 3. 2. . Cadaintersecciónentre doscarasse intersecan enunpuntollamado T r llama vértices sellaman La caraquenocontieneel vérticese Las carasqueseencuentran enlos Argumento lógicoenquecada egiones triangulares. odas, exceptounadelas caras,se T laterales laterales adyacentessellaman Las interseccionesdedosaristas por paralelogramos. llamadas Las carasquenosonlasbases, que yacenenplanosparalelos. formadas porpolígonoscongruentes iene doscarasllamadas Sólido queposeelas Sólido conlassiguientes base vértices y sonsegmentosparalelos. y esunaregión poligonal. caras laterales . a b caras laterales d c Demostración , sonformadas que establece vértices lados vértice aristas . y son bases , y , . Pythagorean identity (p. 391) The identity identidad pitagórica La identidad cos2 cos2 sin2 1. sin2 1.

Pythagorean triple (p. 352) A group of three whole triplete de Pitágoras Grupo de tres números numbers that satisfies the equation a2 b2 c2, enteros que satisfacen la ecuación a2 b2 c2, where c is the greatest number. donde c es el número más grande. R radius 1. (p. 522) In a circle, any segment with radio 1. Cualquier segmento cuyos extremos están endpoints that are the center of the circle and a en el centro de un círculo y en un punto cual- point on the circle. 2. (p. 671) In a sphere, any quiera del mismo. 2. Cualquier segmento cuyos segment with endpoints that are the center and a extremos forman el centro y en punto de una point on the sphere. esfera. rate of change (p. 140) Describes how a quantity is tasa de cambio Describe cómo cambia una changing over time. cantidad a través del tiempo. ratio (p. 282) A comparison of two quantities. razón Comparación entre dos cantidades.

ray (p. 29) PQ is a ray if it is the set of PQS semirrecta PQ es una semirrecta si consta points consisting of PQ and all points S del conjunto de puntos formado por PQ y for which Q is between P and S. todos los S puntos S para los que Q se localiza entre P y S. reciprocal identity (p. 391) Each of the three identidad recíproca Cada una de las tres razones trigonometric ratios called cosecant, secant, and trigonométricas llamadas cosecante, secante y cotangent, that are the reciprocals of sine, cosine, tangente y que son los recíprocos del seno, el and tangent, respectively. coseno y la tangente, respectivamente rectangle (p. 424) A quadrilateral with four rectángulo Cuadrilátero que tiene cuatro right angles. ángulos rectos. reflection (p. 463) A transformation representing a reflexión Transformación que se obtiene cuando se flip of the figure over a point, line, or plane. "voltea" una imagen sobre un punto, una línea o un plano. reflection matrix (p. 507) A matrix that can be matriz de reflexión Matriz que al ser multiplicada Glossary/Glosario multiplied by the vertex matrix of a figure to find por la matriz de vértices de una figura permite the coordinates of the reflected image. hallar las coordenadas de la imagen reflejada. regular polygon (p. 46) A convex polígono regular Polígono convexo en polygon in which all of the sides are el que todos los lados y todos los congruent and all of the angles are ángulos son congruentes entre sí. congruent.

regular pentagon pentágono regular regular polyhedron (p. 637) A poliedro regular Poliedro cuyas caras polyhedron in which all of the faces are son polígonos regulares congruentes. regular congruent polygons.

regular prism (p. 637) A right prism with bases prisma regular Prisma recto cuyas bases son that are regular polygons. polígonos regulares.

Glossary/Glosario R23 Glossary/Glosario R24 rotational symmetry rotation matrix rotation right triangle right prism right cylinder right cone right angle rhombus resultant remote interiorangles relative error related conditionals regular tessellation Glossary/Glosario center ofrotation and direction aboutafixedpoint,called the point ofapreimage through aspecifiedangle are called legs. the opposite therightangleiscalled with arightangle.Theside four sidescongruent. an altitude. is alsoanaltitude. based onagivenconditionalstatement. converse, inverse,andcontrapositivethatare the coordinates oftherotated image. multiplied bythevertexmatrix ofafigure tofind measure of90. are alsoaltitudes. by onlyonetypeofregular polygon. angle. triangle thatare notadjacenttoagivenexterior the figure has rotational symmetry. image andthe preimage are indistinguishable, r expressed asapercent. difference inprecision totheentire measure, otated lessthan 360 hypotenuse p 7)Atransformationthatturnsevery A (p. 476) p 0)Thesumoftwovectors. (p. 500) p 3)Aquadrilateralwithall A (p. 431) p 6)Aconewithanaxisthatisalso A (p. 666) p 4)Aprismwithlateraledgesthat A (p. 649) p 0 Ananglewithadegree (p. 30) p 5)Acylinderwithanaxisthat A (p. 655) p 9 Theratioofthehalf-unit (p. 19) p 7)Atriangle A (p. 178) . Theothertwosides p 0)Amatrixthatcanbe A (p. 507) . p 8)Atessellationformed A (p. 484) p 7 Statementssuchasthe (p. 77) p 7)Ifafigure canbe (p. 478) ° p 8)Theanglesofa (p. 186) about apoint sothatthe hypotenuse hipotenusa A m A cateto leg A A 90 C resultante ángulos internosnoadyacentes error relativo enunciados condicionalesrelacionados teselado regular simetría derotación matriz derotación rotación prisma recto cilindro recto cono recto B cateto leg punto, conocidocomo una dirección determinadas alrededor deun punto delapreimagen a travésdeunánguloy exterior dado. triángulo quenosonadyacentesaunángulo también sonsualtura. dado. que estánbasadosenunenunciadocondicional tales comoelrecíproco, lainversayantítesis altura. completa, expresada enformadeporcentaje. más precisa delamedicióny calcular lascoordenadas delaimagenrotada. por lamatrizdevértices delafigurapermite presenta simetría derotación. y lapreimagen sonidénticas,entonces lafigura alrededor de unpuntoylaimagen menos de360° tipo depolígonoregular. ángulo recto rombo T grados es90. ransformación enquesehacegirarcada triángulo rectángulo La sumadedosvectores. Cono cuyoejeestambiénsualtura. son congruentes. La razónentre lamitaddeunidad llaman catetos. hipotenusa ángulo recto seconocecomo un ángulorecto. Elladoopuestoal Prisma cuyasaristaslaterales Cilindro cuyoejeestambiénsu Cuadrilátero cuyoscuatro lados T eselado formadoporunsolo Matriz quealsermultiplicada Si sepuederotar unaimagen Ángulo cuyamedidaen centro derotación . Losotros dosladosse Ángulos deun T riángulo con Enunciados . S scalar (p. 501) A constant multiplied by a vector. escalar Una constante multiplicada por un vector. scalar multiplication (p. 501) Multiplication of a multiplicación escalar Multiplicación de un vector by a scalar. vector por una escalar. scale factor (p. 290) The ratio of the lengths of two factor de escala La razón entre las longitudes de corresponding sides of two similar polygons or dos lados correspondientes de dos polígonos o two similar solids. sólidos semejantes. scalene triangle (p. 179) A triangle with triángulo escaleno Triángulo cuyos no two sides congruent. lados no son congruentes.

secant (p. 561) Any line that intersects a secante Cualquier recta que interseca circle in exactly two points. P un círculo exactamente en dos puntos. C D CD is a secant of P. CD es una secante de P. sector of a circle (p. 623) A region of a sector de un círculo Región de un circle bounded by a central angle and círculo que está limitada por un ángulo its intercepted arc. A central y el arco que interseca.

The shaded region is a sector of A. La región sombreada es un sector de A. segment (p. 13) See line segment. segmento Ver segmento de recta. segment bisector (p. 24) A segment, line, or plane bisectriz de segmento Segmento, recta o plano that intersects a segment at its midpoint. que interseca un segmento en su punto medio. segment of a circle (p. 624) The region segmento de un círculo Región de un of a circle bounded by an arc and a círculo limitada por un arco y una A chord. cuerda. Glossary/Glosario The shaded region is a segment of A. La región sombreada es un segmento de A. self-similar (p. 325) If any parts of a fractal image autosemejante Si cualquier parte de una imagen are replicas of the entire image, the image is fractal es una réplica de la imagen completa, self-similar. entonces la imagen es autosemejante. semicircle (p. 530) An arc that measures 180. semicírculo Arco que mide 180°. semi-regular tessellation (p. 484) A uniform teselado semirregular Teselado uniforme com- tessellation formed using two or more regular puesto por dos o más polígonos regulares. polygons. similar polygons (p. 289) Two polygons are polígonos semejantes Dos polígonos son semejan- similar if and only if their corresponding tes si y sólo si sus ángulos correspondientes son angles are congruent and the measures of congruentes y las medidas de sus lados their corresponding sides are proportional. correspondientes son proporcionales.

Glossary/Glosario R25 Glossary/Glosario R26 strictly self-similar statement standard position square spherical geometry sphere space solving atriangle slope-intercept form slope skew lines sine similarity transformation similar solids Glossary/Glosario false, butnotboth. a givenpoint,calledthe points thatare agivendistancefrom given bytheformula right anglesandfourcongruent sides. are notcoplanar. the form greatcircles (lines),andspheres (planes). geometry same figure asthewhole. are located or whatsizeisselected,contain the similar ifany ofitsparts,nomatterwhere they all oftheanglesandsidesatriangle. two points( of avectorisattheorigin. all points. acute angletothemeasure ofthehypotenuse. the ratioofmeasure ofthelegopposite and itstransformationimageare similar. same shape,butnotnecessarilythesize. equation hasslope p 6)Foranacuteangleofarighttriangle, (p. 364) p 3)Fora(nonvertical)linecontaining (p. 139) p )Aboundlessthree-dimensional setof A (p. 8) p 3)Inspace,thesetofall (p. 638) p 3)Aquadrilateralwithfour A (p. 432) p 7 Anysentencethatiseithertrue or (p. 67) p 2)Linesthatdonotintersectand (p. 127) that dealswithasystemofpoints, y p 0)Solidsthathaveexactlythe (p. 707) x 1 , mx p 7)Findingthemeasures of (p. 378) p 9)Whentheinitialpoint (p. 498) y p 2)Afigure isstrictly self- A (p. 325) 1 ) and( (p. 145) A linear equationof linear A (p. 145) m (p. 165) b and m . Thegraphofsuchan center p 9)Whenafigure (p. 491) x y x y 2 -intercept 2 2 , . y 2 x y ), thenumber 1 1 The branchof where b . x C es el centro de la esfera. la de centro el es C C 2 isthecenterofsphere. x m 1 . C forma pendiente-intersección pendiente rectas alabeadas seno transformación desemejanza sólidos semejantes estrictamente autosemejante enunciado posición estándar geometría esférica espacio resolver untriángulo la pendientees triángulo. todos losángulosyladosdeun figura ysuimagentransformadasonsemejantes. necesariamente elmismotamaño. exactamente lamismaforma,pero no inicial deunvectoreselorigen. (rectas) yesferas(planos). estudia lossistemasdepuntos,círculos máximos la forma por lafórmula dos puntos( un triángulorectángulo. al ánguloagudoylamedidadehipotenusa contiene lafigura completa. partes, sinimportar sulocalizacióno tamaño, estrictamente autosemejantesicualquiera desus verdadera, pero noambas. todos lospuntos. no soncoplanares. Es larazónentre lamedidadelcatetoopuesto cuadrado Conjunto tridimensionalnoacotadode r y esfera Para unarecta (novertical)quecontiene ectos ycuatro ladoscongruentes. Una oraciónquepuede ser falsao mx x llamado cierta distanciadeunpuntodado en elespacioqueseencuentrana 1 , m Rectas quenoseintersecany m y Cuadrilátero concuatro ángulos 1 El conjuntodetodoslospuntos y laintersección ) y( b. Ocurre cuandolaposición Rama delageometríaque x y En lagráficadetalecuación, centro 2 2 x Calcular lasmedidasde Sólidos quetienen 2 , x y y 1 1 . 2 ), elnúmero donde Ecuación linealde Aquélla enquela Una figuraes y x es 2 b . x m 1 . dado supplementary angles (p. 39) Two angles with ángulos suplementarios Dos ángulos cuya suma measures that have a sum of 180. es igual a 180°. surface area (p. 644) The sum of the areas of all área de superficie La suma de las áreas de todas faces and side surfaces of a three-dimensional las caras y superficies laterales de una figura figure. tridimensional.

T tangent 1. (p. 364) For an acute angle of a right tangente 1. La razón entre la medida del cateto triangle, the ratio of the measure of the leg opuesto al ángulo agudo y la medida del cateto opposite the acute angle to the measure of the leg adyacente al ángulo agudo de un triángulo adjacent to the acute angle. 2. (p. 552) A line in rectángulo. 2. La recta situada en el mismo the plane of a circle that intersects the circle in plano de un círculo y que interseca dicho círculo exactly one point. The point of intersection is en un sólo punto. El punto de intersección se called the point of tangency. 3. (p. 671) A line that conoce como punto de tangencia. 3. Recta que intersects a sphere in exactly one point. interseca una esfera en un sólo punto. tessellation (p. 483) A pattern that covers a plane teselado Patrón que cubre un plano y que se obtiene by transforming the same figure or set of figures transformando la misma figura o conjunto de so that there are no overlapping or empty spaces. figuras, sin que haya traslapes ni espacios vacíos. theorem (p. 90) A statement or conjecture that can teorema Enunciado o conjetura que se puede be proven true by undefined terms, definitions, demostrar como verdadera mediante el uso de and postulates. términos primitivos, definiciones y postulados. transformation (p. 462) In a plane, a mapping for transformación La relación en el plano en que which each point has exactly one image point and cada punto tiene un único punto imagen y cada each image point has exactly one preimage point. punto imagen tiene un único punto preimagen. translation (p. 470) A transformation that moves traslación Transformación en que todos los puntos all points of a figure the same distance in the de una figura se trasladan la misma distancia, en same direction. la misma dirección. translation matrix (p. 506) A matrix that can be matriz de traslación Matriz que al sumarse a la

added to the vertex matrix of a figure to find matriz de vértices de una figura permite calcular Glossary/Glosario the coordinates of the translated image. las coordenadas de la imagen trasladada. transversal (p. 127) A line that t transversal Recta que interseca en intersects two or more lines in a plane diferentes puntos dos o más at different points. m rectas en el mismo plano.

Line t is a transversal. La recta t es una transversal. trapezoid (p. 439) A quadrilateral base trapecio Cuadrilátero con un sólo with exactly one pair of parallel TRbase par de lados paralelos. Los lados sides. The parallel sides of a paralelos del trapecio se llaman trapezoid are called bases. The leg base angles leg bases. Los lados no paralelos se cateto cateto nonparallel sides are called legs. ángulos llaman catetos. Los ángulos de la base The pairs of angles with their cuyos vértices se encuentran en vertices at the endpoints of the PAbase los extremos de la misma base se same base are called base angles. base llaman ángulos de la base.

Glossary/Glosario R27 Glossary/Glosario R28 volume vertical angles vertex matrix vector uniform tessellations undefined terms two-column proof truth value truth table trigonometry trigonometric ratio trigonometric identity Glossary/Glosario statement. of sidesarighttriangle. two intersectinglines. nonadjacent angles ateachvertex. containing thesamearrangementofshapesand the coordinates ofthevertices intoonematrix. polygon byplacingallofthecolumnmatrices their applications. of trianglesandtrigonometricfunctions enclosed bya three-dimensional figure. line, andplane. basic undefinedtermsofgeometryare point, means ofmore basicwords andconcepts.The understood, thatare notformallyexplainedby all valuesoftheanglemeasure. involving atrigonometricratiothatistrue for direction. quantity thathasbothmagnitude,orlength,and statements. method fororganizing thetruth valuesof r the properties thatjustifyeachstepare called two columns.Eachstepiscalleda contains statementsandreasons organized in easons p 9)Adirected segmentrepresenting a A (p. 498) p 8)Ameasure oftheamount ofspace A (p. 688) . p 0 tableusedasaconvenient A (p. 70) p 7 Thetruth orfalsityofa (p. 67) p 6)Thestudyoftheproperties (p. 364) p 0)Amatrixthat represents a A (p. 506) angles formedby p )Words, usuallyreadily (p. 7) p 7 Two (p. 37) p 6)Aratioofthelengths A (p. 364) p 5 formalproof that A (p. 95) p 9)Anequation (p. 391) p 8)Tessellations (p. 484) statement 2 1 2 and 1 and y y

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B(4, 0) O x

A(3, 2) 51. Sample answer: C(4, 4)

Pages 9–11 Lesson 1-1 1. point, line, plane 3. Micha; the points must be noncollinear to determine a plane. 5. Sample answer: y 7. 6 9. No; A, C, and J lie in plane ABC, but D does not. 53. vertical 55. Sample answer: Chairs wobble because all 11. point 13. n 15. R four legs do not touch the floor at the same time. Answers W 17. Sample answer: PR should include the following. x 19. (D, 9) • The ends of the legs represent points. If all points lie in O 21. the same plane, the chair will not wobble. X B Z • Because it only takes three points to determine a plane, a chair with three legs will never wobble. W Y Q 57. B 59. part of the coordinate plane A above the line y 2x 1. y 61. 23. Sample answer: y 63. 65. R S O x Z

O x

Q

Pages 16–19 Lesson 1-2 1. 25. 27. Align the 0 point on the ruler with the leftmost endpoint a s of the segment. Align the edge of the ruler along the segment. Note where the rightmost endpoint falls on the b C r scale and read the closest eighth of an inch measurement. c D 3 3. 1 in. 5. 0.5 m; 14 m could be 13.5 to 14.5 m 7. 3.7 cm M 4 9. x 3; LM 9 11. BC CD, BE ED, BA DA 1 13. 4.5 cm or 45 mm 15. 1 in. 17. 0.5 cm; 21.5 to 22.5 mm Selected Answers 4 1 1 3 19. 0.5 cm; 307.5 to 308.5 cm 21. ft.; 3 to 3 ft. 29. points that seem collinear; y 8 8 8 1 1 sample answer: (0, 2), 23. 1 in. 25. 2.8 cm 27. 1 in. 29. x 11; ST 22 (1, 3), (2, 4), (3, 5) 4 4 31. x 2; ST 4 33. y 2; ST 3 35. no 37. yes O x 39. yes 41. CF DG, AB HI, CE ED EF EG 43. 50,000 visitors 45. No; the number of visitors to Washington state parks could be as low as 46.35 million or as high as 46.45 million. The visitors to Illinois state parks could be as low as 44.45 million or as high as 44.55 million visitors. The difference in visitors could be as high as 2.0 million.

Selected Answers R29 Selected Answers R30 15. 33. acute; 51. • • the following. 61. 1. P 3. straightedge toconstruct thebisectorofsegment. locate themiddleofsegment.(3)Useacompassand segment andfoldthepapersothatendpointsmatchto you knowthecoordinates oftheendpoints.(2)Drawa 1. P 1. P so thegreatest perimeteris3.5cm 47. 5. answer: A degree is degree answer: A wide. pointed orblunt,sosomethingthatisobtusewouldbe 61. fine tiplikeapen,knife,orneedle. Acute 21. 49c. 27. 53. •Y •W accuracy. Answers shouldincludethefollowing. differentiate between sizeanddistance,aswellfor 51. 49. 37. ABC factor. 53. 45. gs3–6Lesson1-4 ages 33–36 Lesson1-3 ages 25–27 age 19Practice Quiz1 CEA Y 8 Sample answers:(1)UseoneoftheMidpointFormulasif BA See students’work. side oftheangleintersects scaleoftheprotractor. of theprotractor. Observe thepointatwhichother protractor andthevertexofangle atthecenterpoint Place onesideoftheangleto coincidewith0onthe shown than thatstated. Therefore, theactualmeasure couldbegreater orless precision oftheinstrument usedtomakethemeasure. PR Selected Answers C 1.7% 3 120°, obtuse C Sample answer: 15.5 cm;Eachmeasurement isaccuratewithin0.5cm, 5 Sample answer:Unitsofmeasure are usedto (10, 3) Sample answer:Theperimeterincreases bythesame T es; theyallhavethesamemeasure. ou canassumeequalmeasures whensegmentsare E hen ameasurement isstated,youdonotknowthe 353.4 and , 5. FEA can meansomethingthatissharporhavingavery 61 3 8 BC 17. 63. A 43. 53. 3. , 11 10 55. 3, obtuse 25. BCD to becongr , PR 55. 11 31; 59 7.8 7. 39. ( 49d. 45 7. 4 D 80 0.08% 47. 135°, obtuse 1, 2 19. 5. ( 61. B 29. 23. 6 65. 420.3 8.35 35. LaFayette, LA 27. 10, 45. 3) 10 8.9; (2,2) 13. 5 3 17.3 units 9. uent. 8 60 1 65 30, 30 57. 1, 3,6,10,15 AED 57. ( 21. 63. B 3) of acircle. Answers shouldinclude 49e. 3 2.5, 4) D ( 37. AB B 15. 13 22 , 41. 67. 63. 9. 37.4 ) 29. 59. 4 F 59. 55. A DEA 47 (5.6, 2.85) 23. 65. 10 1 31. 3 3 39. 60°, acute Sample answer:planes 11. 4 9 49a. 17. 15 1 4 1 3 2 , 11. 49f. 65. 47. 5.5 cm 2 4 in. in. (3, 5) 3 ( A Obtuse CD AEB 1 25. B 9 21, 45 11.8 2092.9 41. 3. 33. , 1, right; ) 57. AD 43. , m Sample answer: 13. 2.5 31. 13 means not 6.5 cm. 90 49b. BEA R A 49. 19. 2 (2, 7) 51. 90°, right 35. 59. , Sample 212.0 9.5 m AD 2, 1 F AEC 72.1 , , Z L AE , , J 25. 13 units,5units 1 13. 11. 19. 13. • • Answers shouldincludethefollowing. Others havepolygon-shapedpiecesthatconnecttogether. 37. concave; irregular their measures is 31. m or 11. the twoanglescannotbecomplementaryorsupplementary. .3. 1. P 1. P 1. P 1. P adjacent. Since let 39. 37. 9. 59. 53. n m creates measures thatare halfoftheoriginalangle,so know 23. 5. 25. denotes that 2 x gs4–2Lesson1-5 ages 41–62 age 36Practice Quiz2 gs5–6Chapter1StudyGuide andReview ages 53–56 Lesson1-6 ages 48–50 WTU XWY Y Divide theperimeterby10. Sample answer: d triangles, quadrilaterals,pentagons 180 m AAB 125 m 82 ft octagon; convex;regular Sample answer:Sometoysusepiecestoformpolygons. sometimes s hysaeacmo ieadvre,sotheyare es; theyshare acommonsideandvertex, 70°, acute Because Sample answer: 24 YUZ YUT 70 2 1 180 WUT 2 3. m 13 173 , 1 WUT 61. 2 f x 45. 17. 41. ; 21. . Then ABC , x 27. 5. 90, 3.6 2 1 40 . Thissumsimplifiesto obtuse 13.2 m 53, 37 9 65 40 units b VUX . WUT PR DAB Y 30 units x C 33. 27. U 25. 43. m WUT and m 33. 7. falls between 8.1 55. 7. 3.75 50°, acute is arightangle. p and (3, not apolygon U ABC 19. YUZ ADC 4 in.,17in. 13. m m 47. or 33 ft Z or 110 23. . 3. 148 m 29. , right TUV 5) 20 , which90. 2 x 29. (0, 0); 41. 1 The perimeteristripled. TUV UWT and 9. CBE All are 15 cm. m 9. 15. 27. 114 21. A 16 units 4.5 43. 41. 39. 15. 19. F 21. 3. PQ 35. m , are supplementary, 49. YUT pentagon 180 (0.6, 84, 96 P 2 43. 63 sometimes D 7. s x not enoughinformation and 31. 2000 TWY 36 57. TUZ obtuse 18 45. x 33. 101 2 straight line. of anglesforma sides ofalinearpair The noncommon 3 35. 0 Y 3, s 6, PS x n 6.35) es; thesymbol Sample answer: 37. m 11. or 90.Because bisector . A Y PB 23. 24, PB , AB 5. es; theirsumof No; wedonot 15. m 44.7 22.5 units 3, 40 10.0 2 1 31. TUZ 4605 ft pentagon; 17. y always , 51. m 35. QR m QPR 12 29. 18 13 units, triangle WTY 39. 8 52 units 5. TUV or AAB 20 FE 17. 34; 135 20 , , , TWY 90, so FG yes , Chapter 2 Reasoning and Proof 11. Sample answer: pqp q TTT Page 61 Chapter 2 Getting Started 18 1. 10 3. 0 5. 50 7. 21 9. 9 11. 13. 16 TFF 5 FTF Pages 63–66 Lesson 2-1 FFF 1. Sample answer: After the news is over, it’s time for dinner. 3. Sample answer: When it’s cloudy, it rains. 13. Sample answer: pr p p r Counterexample: It is often cloudy and it does not rain. TTF F 5. 7 7. A, B, C, and D are noncollinear. A D TFF F P FTT T C B FFT F

15. 14 17. 3 19. 64 8 or an equilateral triangle has 11 9. true 11. 13. 32 15. 17. 162 21. 3 three congruent sides; true. 0 0 and an obtuse angle 19. 30 measures greater than 90° and less than 180°; false. 23. An equilateral triangle has three congruent sides and an obtuse angle measures greater than 90° and less than 180°; true. 25. An equilateral triangle has three congruent sides and 0 0; false. 27. An obtuse angle measures greater than 90° and less than 180° or an equilateral triangle has three congruent sides; true. 29. An obtuse angle measures 21. Lines and m form 23. 3 and4 are greater than 90° and less than 180°, or an equilateral four right angles. supplementary. triangle has three congruent sides and 0 0; true.

31. pq p q p q 34 TTF F F

TFF T F m FTT F F 25. PQR is a scalene triangle. 27. PQ SR, QR PS FFT T T y P Q P 33. Sample answer: 35. Sample answer: R(6, 5) (–1, 7) qrq and r prp or r S R TT T TT T TF F TF T 29. false; FT F FT T 2 O x 1 FF F FF F

Q(6, –2) 37. Sample answer: qr rq r TTF F 31. false; W X Y Z 33. true 35. False; JKLM may not have a TFT T right angle. 37. trial and error, a process of inductive FTF F reasoning 39. C7H16 41. false; n 41 43. C 45. hexagon, convex, irregular 47. heptagon, concave, FFT F irregular 49. No; we do not know anything about the 39. Sample answer: angle measures. 51. Yes; they form a linear pair. pq r p rq r p (q r) 53. (2, 1) 55. (1, 12) 57. (5.5, 2.2) 59. 8; 56 61. 4; 16 Selected Answers TT TF F F F 63. 10; 43 65. 4, 5 67. 5, 6, 7 TT FF T T T

Pages 71–74 Lesson 2-2 TF TF F F F 1. The conjunction (p and q) is represented by the TF FF T F F 3. intersection of the two circles. A conjunction is a FT TT F F T compound statement using the word and, while a disjunction is a compound statement using the word or. FT FT T T T 5. 9 5 14 and a square has four sides; true. FF TT F F T 7. 9 5 14 or February does not have 30 days; true. FF FT T F T 9. 9 5 14 or a square does not have four sides; false.

Selected Answers R31 Selected Answers R32 be usedtodescribe howtogetadiscount,rebate, or refund. daylight, then it iswinter. summer. In Alaska, ifthere are more hoursofdarkness than there are more hoursofdaylightthandarkness,then itis not arectangle; true. Contrapositive: Ifafigure isnotaquadrilateral,thenit r r 43. may exercise alot,butstillnotbeingoodshape. False;anillperson Ifyouare not ingoodshape, then youdonotexercise regularly. Contrapositive: shape; true. If youdonotexercise regularly, thenyou are not ingood Inverse: in goodshape,thenyouexercise regularly; true. 35. least 13yearsold are prevalent. Ifyouare inVermont, thenmapletrees from the swamps. Ifyouare inFlorida,thencypress trees rise mountains. Colorado, thenaspentrees coverhighareas ofthe killed byoverwatering. then theydonothavewater. False;theymayhavebeen Contrapositive:Ifplantsdonotgrow, will notgrow; true. Inverse:Ifplantsdonothavewater, thenthey water; true. 11. formed byperpendicularlines,thenitisarightangle. then itisnotacute;true. Contrapositive: Ifanangle’smeasure isnotlessthan90, Ifan angle isnotacute,thenitsmeasure is notlessthan90;true. Inverse: measure lessthan90,thenitisacute; true. then youlovetosolveproblems. and 90;C:theangleisacute points are collinear pitcher, thenitholdsaquartofliquid. 5. negate thehypothesisandconclusionofconverse. conclusion oftheconditional.Incontrapositive,you 3. hypothesis andconclusionare easilyrecognizable. 55. • • the following. choices onamultiplechoicetest. Answers shouldinclude 53. they are equilateral. 27. two anglesare adjacent,thentheyhaveacommonside. 1. P 47. 45. 41. acute ectangle, thenitisnotaquadrilateral;false,rhombus. Inverse:Ifafigure isnota ectangle; false,rhombus. gs7–0Lesson2-3 ages 78–80 H: In theinverse,younegatebothhypothesisand W See students’work. favorite activity. Math ismyfavoritesubjectanddramaclub Selected Answers Converse: Ifafigure isaquadrilateral,thenit true true C Sample answer:Logiccanbeusedtoeliminatefalse Sample answer:Iftwotrianglesare equiangular, then 42 riting aconditionalinif-thenformishelpfulsothatthe x Level ofParticipation Among 310Students 69. 57. Sports 43. 95 3 37. 13. 222 feet 81 25 false Converse: Ifplantsgrow, thentheyhave 7; C: 59. 17. 20 x 19. 1 H: youare ateenager;C:youare at Academic 39. 71. 21. Clubs 29. 45. 60 61. 10 H: three pointslieonaline;C:the true 44 H: themeasure ofanisbetween0 15. true Converse: Ifananglehas 47. 405 49. 7. Sample answer:Ifyouare in 73. If apitcheris32-ounce 23. Sample answer:In Alaska, if 41. Conditional statements can 31. 63. 51. 184 If youare amathteacher, ovre Ifyouare Converse: 25. true 34.4 135 Sample answer:If C 9. 33. 65. 49. If anangleis false 29.5 true B 67. A 55º, 53. W not 100%satisfied,thenreturn theproduct forafullrefund. Ifyouare Sample answersshouldincludethefollowing. 55. the following. and theirdoses forpatients. Answers should include and nursesuse charts toassistindetermining medications obtuse. 21. triangle are noncollinear, andtherefore determineaplane. not necessarilynoncollinear. are adjacent.False; a commonvertex,thentheangles 5. United Statesand60 57. then theyare notadjacent;true. Contrapositive: Iftwoanglesdonothaveacommonvertex, are notadjacent. 63. is 180. of theanglesinatriangle adjacent to then shewouldwintherace. Le MayDoanskatedhersecond 500metersin37.45seconds, 27. 15. the LawofDetachmentindicatesthattheirsumiseven. 1. P $30,000 oflifeinsurance. man orawoman.Shecouldbe45andhavepurchased with equalmeasures. 7 5. mean thatyouare seasickandthushaveanupsetstomach. 3. b: Itisrainy;c:Thegamewillbecancelled. 1. P 59. common vertex.False; adjacent, thentheydonothavea Inverse: Iftwoanglesare not 67. . g 0Practice Quiz1 age 80 gs8–7Lesson2-4 ages 84–87 earing aseatbeltreduces theriskofinjuries. DBE The midpointofasegmentdividesitintotwosegments WX Invalid; congruent anglesdonothavetobevertical. Lakeisha; ifyouare dizzy, thatdoes notnecessarily Sample answer:a:Ifitisrainy, thegamewillbecancelled; Converse: Iftwoangleshave false If themeasure ofanangleislessthan 90,thenitisnot yes; LawofDetachment Invalid; thesumiseven. A A George Washington wasnotthefirstpresident ofthe The sumofthemeasures Multiply eachsideby2. hexagon hasfivesidesor60 hexagon doesn’thavefivesidesor60 41 have acommonvertexand F 23. or 6.4 45 no conclusion ABC 65. . 68 67 ABD Y G 9. 125 3 3. H invalid ABC 13. Sample answer: is not or 11.2 18.; false 25. 29. V 17. 19. pq F F TFF F TTF F F FFT T FTT 33. and alid; since5and7are odd, 61. yes; LawofDetachment invalid Invalid; V Sample answer:Doctors 11. alid; theverticesofa P Q PQR 3 No; Terry couldbea A A B E 18.; false DE 31. is arightangle. , F , and If Catriona 3 p B 51. C G 18.; true D R B are p C q • Doctors need to note a patient’s symptoms to determine would be obtuse. Inverse: If ABC is not a right triangle, which medication to prescribe, then determine how none of its angle measures are greater than 90. False; it much to prescribe based on weight, age, severity of the could be an obtuse triangle. Contrapositive: If ABC does illness, and so on. not have an angle measure greater than 90, ABC is not a • Doctors use what is known to be true about diseases and right triangle. False; mABC could still be 90 and ABC be when symptoms appear, then deduce that the patient has a right triangle. 39. 17 4.1 41. 106 10.3 a particular illness. 35. B 37. They are a fast, easy way to add fun to your 43. 25 45. 12 47. 10 family’s menu. 39. Sample answer: qrq r Pages 97–100 Lesson 2-6 TTT 1. Sample answer: If x 2 and x y 6, then 2 y 6. 3. hypothesis; conclusion 5. Multiplication Property TFF 2 7. Addition Property 9a. 5 x 1 9b. Mult. Prop. 3 FTF 9c. Dist. Prop. 9d. 2x 12 9e. Div. Prop.

FFF 11. Given: Rectangle ABCD, 10 A B AD 3, AB 10 33 41. Sample answer: Prove: AC BD D C pq rq rp (q r) 10 Proof: TTT T T Statement Reasons TTF T T 1. Rectangle ABCD, 1. Given TFT T T AD 3, AB 10 2. Draw segments AC 2. Two points determine TFF F F and DB.a line. FTT T F 3. ABC and BCD are 3. Def. of rt FTF T F right triangles. 4. AC 32 102, 4. Pythagorean Th. FFT T F DB 32 102 FFF F F 5. AC BD 5. Substitution 43. HDG 45. Sample answer: JHK and DHK 47. Yes, slashes on the segments indicate that they are 13. C 15. Subt. Prop. 17. Substitution 19. Reflexive Property 21. Substitution 23. Transitive Prop. congruent. 49. 10 51. 130 11.4 1 1 25a. 2x 7 x 2 25b. 3(2x 7) 3x 2 53. 55. 3 3 W r A 25c. Dist. Prop. 25d. 5x 21 6 25e. Add. Prop. n 25f. x 3 s 3 27. Given: 2y 8 B 2 Prove: y 13 4 Proof: Statement Reasons 3 1. 2y 8 1. Given 57. Sample answer: 1 and 2 are complementary, 2 3 m1 m2 90. 2. 2(2y ) 2(8) 2. Mult. Prop. 2 Pages 91–93 Lesson 2-5 3. 4y 3 16 3. Dist. Prop. 1. Deductive reasoning is used to support claims that are 4. 4y 13 4. Subt. Prop. 13 made in a proof. 3. postulates, theorems, algebraic 5. y 5. Div. Prop. properties, definitions 5. 15 7. definition of collinear 4 2 9. 29. Given: 5 z 1 Through any two points, there is exactly one line. 3 11. 15 ribbons 13. 10 15. 21 17. Always; if two points Prove: z 6 lie in a plane, then the entire line containing those points Proof: Selected Answers lies in that plane. 19. Sometimes; the three points cannot Statement Reasons be on the same line. 21. Sometimes; and could be 2 m 1. 5 z 1 1. Given skew so they would not lie in the same plane R. 23. If two 3 2 points lie in a plane, then the entire line containing those 2. 35 z 3(1) 2. Mult. Prop. 3 25. points lies in that plane. If two points lie in a plane, 3. 15 2x 3 3. Dist. Prop. then the entire line containing those points lies in the plane. 4. 15 2x 15 3 15 4. Subt. Prop. 27. Through any three points not on the same line, there is 5. 2x 12 5. Substitution exactly one plane. 29. She will have 4 different planes and 2x 12 6. 6. Div. Prop. 6 lines. 31. one, ten 33. C 35. yes; Law of Detachment 2 2 37. Converse: If ABC has an angle with measure greater 7. x 6 7. Substitution than 90, then ABC is a right triangle. False; the triangle

Selected Answers R33 Selected Answers R34 7. 5. exactly onepoint. between of losing. answer: Ifapersonischampion,thentheafraid happy, thentheyrarely correct theirfaults. which isnotaninteger. students’ work. 33. 3. is thesameasdistancefrom ClevelandtoChicago. 1. P 1. P 31. gs1316Lesson2-7 ages 103–106 Practice Quiz2 age 100 Given: Given: If Sample answer:Thedistancefrom ClevelandtoChicago invalid a. 1. b. Proof: Prove: Prove: c. 2. Proof: d. 4. 3. f. e. 6. 5. 8. 7. ttmnsReasons Statements Reasons Statement Selected Answers 1. 5. 4. 3. 2. All oftheanglemeasures wouldbeequal. Given: Prove: ttmn Reasons Statement Proof: A P 2( PQ PS 2 PQ 2 2 P PS n 2 n n n , m m m m m m m m S Q n 1 1 B , and A n 2 1 XCA XCA XCA XCA YBA YBA YBA ACB 6 1 PQ P n P 2( R RT 3) 3 n RS R QS and 47. m 3. m S Q T n S , , 2 If twolinesintersect,thentheirintersectionis C 5 3 2 n Q QS n 3 2 1 C XCA n 5 ACB R QS are collinearand 3) R S . RS ft m m m m T m m m 37. m S 3 3 , 3 , 3( n 3 5. n RT Q ST S 49. n ACB ABC ABC ABC ACB ACB ACB B 5 YBA T S Symmetric ST m m 3 3 3 0.5 in. 39. 1) 43. 3( RS S YBA ABC n 2 T 180 180 n 6 Sample answer:Ifpeopleare ST 1) CBY XC 41. 1. 3. 2. 5. 4. 8. 7. 6. 51. AB 4. 3. 5. 1. 2. Given Substitution Dist. Prop. Substitution Subt. Prop. Symmetric Prop. Substitution Add. Prop. Invalid; 27 11 f. e. a. c. d. b. Substitution Substitution Subt. Prop. Given Def. ofsupp. Def. of Substitution Segment Addition Def. of Given Addition Property Post. BC 53. A R 45. 47 AC 35. Sample segments , then segments 6 S P Q See 4.5, B T is 23. 21. 13. 11. 19. 9. Given: 1. 2. Proof: Prove: 3. 4. 5. 6. 7. 9. 8. ttmnsReasons Statements 4. 3. 1. a. 1. c. b. 2. 2. d. 4. 3. e. f. g. h. i. Given: Given: Substitution Helena isbetweenMissoulaandMilesCity. Given: Prove: Prove: ttmnsReasons Statements Reasons: Statements: Proof: Reasons Statements Proof: Prove: Proof: H HI HI TU TU TU TU I IJ J X XY AB W X WA WY A A XY AC WY AC AC ZX WA WA 2 WA W I WA Y Y A Y is themidpointof is themidpointof U UV TU T IJ I H IJ UV IJ TU V J W X AC W A A AB X U W WZ BC I AB A AB 2 BC ZA AY AY WA ZA ZX Z WA Z Y Y BC , , A Y 2 B is themidpointof is themidpointof Z X A H HJ HJ U ZA TV TU , T and and J V ZA U TV BC A 2 W 15. BC BC Z Z AX ZA BC , ZA AY B X A Z T TV H W WZ V UV T J and , Z ransitive AX AX ZA T W Z W A AB V X B Z Y BC AB . . 9. 1. 2. 3. 4. 5. 6. 8. 7. W Z A 3. 2. i. a. 4. 1. b. 1. c. 4. 3. 2. d. e. f. g. h. 17. X B Def. of Given Def. of Seg. Add. Post. Seg. Add. Y Substitution Seg. Add. Post. Seg. Add. Substitution Subt. Prop. Reflexive Prop. . T Def. of Def. of Given Def. of Given Def. of Given Definition ofmidpoint Substitution Substitution Seg. Add. Post. Seg. Add. Segment Addition Post. Segment Addition Substitution Substitution Substitution Division Property . H ransitive Prop. Subtraction A I X ZX J UV TU segs. segs. segs. W B segs. segs. segs. Y A Y W Z 25. Given: AB DE, C is the 27. sometimes 29. always 31. sometimes midpoint of BD. 33. Given: m Prove: AC CE Prove: 2, 3, and 4 are rt. . 12 m ABCDE 34 Proof: Statements Reasons Proof: 1. AB DE, C is the 1. Given Statements Reasons midpoint of BD. 1. m 1. Given 2. BC CD 2. Def. of midpoint 2. 1 is a right angle. 2. Def. of lines 3. AB DE 3. Def. of segs. 3. m 1 90 3. Def. of rt. 4. AB BC CD DE 4. Add. Prop. 4. 1 4 4. Vert. are . 5. AB BC AC 5. Seg. Add. Post. 5. m 1 m 4 5. Def. of CD DE CE 6. m 4 90 6. Substitution 6. AC CE 6. Substitution 7. 1 and 2 form 7. Def. of 7. AC CE 7. Def. of segs. a linear pair. 3 and linear pair 4 form a linear pair. 27. Sample answers: LN QO and LM MN RS 8. m1 m2 180, 8. Linear pairs are 29. 31. 33. ST QP PO B Substitution Addition m4 m3 180 supplementary. 35. Property Never; the midpoint of a segment divides it 9. 90 m2 180, 9. Substitution 37. into two congruent segments. Always; if two planes 90 m3 180 39. intersect, they intersect in a line. 3; 9 cm by 13 cm 10. m2 90, m3 90 10. Subt. Prop. 41. 15 43. 45 45. 25 11. 2, 3, and 4 11. Def. of rt. Pages 111–114 Lesson 2-8 are rt. . (steps 6, 10) 1. Tomas; Jacob’s answer left out the part of ABC 35. Given: m represented by EBF. 3. m2 65 5. m11 59, Prove: 1 2 m12 121 12 m 7. Given: VX bisects WVY. W 34 VY bisects XVZ. X Proof: Prove: WVX YVZ V Statements Reasons Y 1. m 1. Given 2. 1 and 2 rt. 2. lines intersect to Z form 4 rt. . Proof: 3. 1 2 3. All rt. . Statements Reasons 37. Given: ABD CBD, 1. VX bisects WVY, 1. Given ABD and DBC form a linear pair. VY bisects XVZ. Prove: ABD and CBD are rt. . D 2. WVX XVY 2. Def. of bisector 3. XVY YVZ 3. Def. of bisector ACB 4. WVX YVZ 4. Trans. Prop. Proof: 9. sometimes Statements Reasons 11. Given: ABC is a right angle. C 1. ABD CBD, 1. Given B Prove: 1 and 2 are 2 ABD and CBD form 1 complementary angles. a linear pair. A 2. ABD and CBD are 2. Linear pairs are supplementary. supplementary. Proof: 3. ABD and CBD 3. If are and suppl., Statements Reasons are rt. . they are rt. . 1. ABC is a right angle. 1. Given 39. Given: mRSW mTSU 2. m ABC 90 2. Def. of rt. Prove: m RST m WSU T 3. mABC m1 m2 3. Angle Add. Post. R 4. m1 m2 90 4. Substitution W

5. 1 and 2 are 5. Def. of complementary Selected Answers U complementary angles. S Proof: 13. 62 15. 28 17. m4 52 19. m9 86, m10 94 Statements Reasons 21. m13 112, m14 112 23. m17 53, m18 53 1. mRSW mTSU 1. Given 25. Given: A 2. mRSW mRST 2. Angle Addition Prove: A A mTSW, mTSU Postulate A Proof: mTSW mWSU Statements Reasons 3. mRST mTSW 3. Substitution 1. A is an angle. 1. Given mTSW mWSU 2. mA mA 2. Reflexive Prop. 4. mTSW mTSW 4. Reflexive Prop. 3. A A 3. Def. of angles 5. mRST mWSU 5. Subt. Prop.

Selected Answers R35 Selected Answers R36 13. 9. true. sum ofthemeasures oftwosupplementaryanglesis180; 47. • should includethefollowing. supplementary tothesameangleare congruent. Answers 1. P Therefore, formed are rightangles. All rightanglesare congruent. 41. definition ofcongruent segments, measures oftwosupplementaryanglesis180,and 33. with rightangle supplementary anglesis180;false. C • • then itisnotMarch; true. days. Contrapositive:Ifamonthdoesnothave31days, March, thenitdoesnothave31days.False;Julyhas March. False; Julyhas31days.Inverse:Ifamonthisnot false. if therightanglesformalinearpair. 27. by definition,adjacentangleshaveacommonvertex. 43. not. angles mustshare acommonside,andverticalanglesdo 31. 55. 49. gs1510Catr2StudyGuideandReview Chapter2 ages 115–120 , m conjecture the anglesare congruent. If twoanglesare complementarytothesameangle,then supplementary. a Selected Answers Because thelinesare perpendicular, theangles 2. 1. 1. 2. 3. 3. 4. Always; if 2 In arighttrianglewithangle Given: Sometimes; ifthepointsare collinear. yes; LawofDetachment Given: Proof: Prove: ttmnsReasons Statements Reasons Statements Prove: Proof: 1 and 1 and NML ONM 39. A WX 5 X 5 XY 3 WX 15. 19. b is themidpointof 2 45 Distributive Property 2 2 m , Converse: Ifamonthhas31days,thenitis , x WX X 5 2 are supplementary; 3 are verticalangles,andare therefore congruent. 1 iscongruent to YZ c 1 2 1 A XY YZ 2 x NMP is themidpointof , orthesumofmeasures oftwo MNR B P 2 2 1 3. 2 x is themidpointof 135 0, andinarighttrianglewithangle C B compound 6 , XZ YZ 180 2 1 XZ , a x 2 1 2 x 51. NMO b 2 XZ 2 N W 21. , or 45. Y c 2 29. 5. . R 2. RNM, true and thesumof 11. B hypothesis W 1. 3. 1. 2. 2. 3. 4. 41. yes; LawofSyllogism XP 53. 43. X 2 and Y M LMNO W Given Substitution Given Def. ofmidpoint Subt. Prop. Postulate Segment Addition Substitution Y . 17. Subtraction Property 23. , then T obtuse L 37. C ONM wo anglesthatare , PY In arighttriangle X a false Never; adjacent 2 3 are is asquare. . 35. Y X 7. b P Sometimes; 2 N 25. Postulates O P c Z Va 2 Y 1 or the . By lid; 0; 45. 53. 51. 47. 49. memorial “cuts”thepillars. parallel planes. interior C 31. 11. r tracks 3. parallel planes. of acylinderare containedin 1. P 1. P Lines Chapter 3Parallel andPerpendicular interior 55. 45. APU interior 35. planes iftheyare extended. gs1811Lesson3-1 ages 128–131 GettingStarted Chapter3 age 125 D and PQ Sample answer:lookingdownrailroad Sample answer:Thebottomandtop , 1. 2. 1. 4. 3. 8. 5. 4. 5. 6. 2. 3. 4. 5. 7. 6. Given: Division orMultiplicationProperty Reflexive Property C alternate interior q a 145 Given: corresponding Prove: ttmnsReasons Statements ttmnsReasons Statements Proof: Prove: Proof: D , ; alternateinterior G and DET BA t E AC BC BC BC BC EC x AB x 4 4 4 4 , x x x x , 3. 2 D 6 2(3) 2 5. 2 E x 9. x 57. 19. 15. 41. H ST b F 7 A , 1 6 , 1 6 , p BA x AC AB BC DE CA EC CA CA CD 6 x 6 B AB x a Q x 25. E 90 x Sample answer:Theroof andthefloorare p p and , I R and 5. ; alternateinterior ; consecutiveinterior J , 1 14 1 , K 6 2 , 7 CA 4 6 A AC 2 x 14 R 51. , 21. 13 x EC DE 6 AB r P EC BA EC L DE S c x , 4, 6 , M , , p x 37. , B No; plane 13 , S b Sample answer:Thetopofthe 2 1 14 CA Q and AC T CD x 4 13. and 13 , 6, , x 7. CA CD alternate interior 13 49. T C 47. U R q t consecutive 6 c , 8 x CD , 1, 4 Addition Property and r m F x 1 53. 23. 29. and U 33. ; consecutiveinterior 7. ADE , r 1. 2. 4. 3. 8. 7. 4. 1. 5. 6. 5. 2. 3. 4. 6. 5. P 1, infinite number , ABC a alternate exterior t U q 43. 1, and Given Add. Prop. Seg. Add. Post. Seg. Add. Substitution Substitution Div. Prop. Mult. Prop Given Substitution Symmetric Prop. Substitution Substitution Subt. Prop. Substitution Substitution Subt. Prop. , B and will intersectall the Q D 17. , R ABQ 5, ; alternateexterior c , t , q , R a ; alternate AB 39. S 7 , and , C PQR 4 T x consecutive

E U 9. 6 x r 1

, 9 , r CDS 27. A 13 and 11. B , C C c , 2 3 55. Sample answer: Parallel lines and planes are used in 11.y 13. (1500, 120) or architecture to make structures that will be stable. Answers (1500, 120) 4 should include the following. A(6, 4) 1 15. 17. 5 • Opposite walls should form parallel planes; the floor 7 may be parallel to the ceiling. O 4 8 12 x 19. perpendicular • The plane that forms a stairway will not be parallel to 4 21. neither 23. parallel some of the walls. 25. 3 27. 6 29. 6 57. 16, 20, or 28 8 31. undefined 59. Given: PQ ZY, QR XY P Q R Prove: PR XZ 33. 35. XY Z y y Proof: Since PQ ZY and QR XY, PQ ZY and P(2, 1) O QR XY by the definition of congruent segments. By x the Addition Property, PQ QR ZY XY. Using the Segment Addition Postulate, PR PQ QR and XZ M(4, 1) XY YZ. By substitution, PR XZ. Because the measures are equal, PR XZ by the definition of O x congruent segments. 61. mEFG is less than 90; Detachment. 63. 8.25 65. 15.81 67. 10.20 69.T 71. 90, 90 73. 72, 108 37.y 39. Sample answer: 0.24 R 75. 76, 104 41. 2016 S O x Pages 136–138 Lesson 3-2 1. Sometimes; if the transversal is perpendicular to the parallel lines, then 1 and 2 are right angles and are Q(2, 4) congruent. 3. 1 5. 110 7. 70 9. 55 11. x 13, y 6 13. 67 15. 75 17. 105 19. 105 21. 43 23. 43 25. 137 27. 29. 31. 33. 35. 60 70 120 x 34, y 5 113 19 43. ; y 45. 2001 37. x 14, y 11, z 73 39. (1) Given (2) Corresponding 2 (4, 8) 1 11 Angles Postulate (3) Vertical Angles Theorem (4) Transitive 8 47. y x Property 2 2 49. C 51. 131 53. 49 41. 4 (–1–9, 2) Given: m, m n (4, 5) 2 55. 49 57. ; alternate Prove: n 12 exterior m 4 O 4 8 x 59. p; alternate interior 3 4 (2, 1) n 4 61. m; alternate interior

Proof: Since m, we know that 1 2, because 63. H, I, and J are H perpendicular lines form congruent right angles. Then by noncollinear. the Corresponding Angles Postulate, 1 3 and 2 I 4. By the definition of congruent angles, m1 m2, J m1 m3, and m2 m4. By substitution, m3 m4. Because 3 and 4 form a congruent linear pair, 65. R, S, and T are collinear. 67. obtuse 69. obtuse they are right angles. By definition, n. y 71. y 1x 5 43. 2 and 6 are consecutive interior angles for the same 2 4 transversal, which makes them supplementary because WX YZ. 4 and 6 are not necessarily supplementary O x Selected Answers because XY may not be parallel to WZ. 45. C 47. FG 49. CDH 51. m1 56 53. H: it rains this evening; C: I 2 3 4 will mow the lawn tomorrow 55. 57. 59. STR 3 8 5 Page 138 Practice Quiz 1 1. p; alternate exterior 3. q; alternate interior 5. 75

Pages 142–144 Lesson 3-3 Pages 147–150 Lesson 3-4 1. horizontal; vertical 3. horizontal line, vertical line 1. Sample answer: Use the point-slope form where 1 2 5. 7. 2 9. parallel (x , y ) (2, 8) and m . 2 1 1 5

Selected Answers R37 Selected Answers R38 should beequidistantalongthe entire line. court hasparallellines,asdoes anewspaper. Theedges int. 10 days price ofappliancessold so thelinesare notparallel. is interior 35. 23. 19. 15. 1. P 9. 1. P 3. and 39. corresponding 21. consecutive interior are 43. 25. 55. •W • should includethefollowing. the equation ofaline,the 65. 29. 67. gs1417Lesson3-5 ages 154–157 Practice Quiz2 age 150 y neither 8 1 Sample answer:Useapairofalternateexterior the planswouldbeequal. equation. The feeforairtimecanbeconsidered theslopeof Sample answer: Selected Answers 1. 2. 3. 4. mx , andtheslopeofline y KN A y y y y B y 26.69 ttmnsReasons Statements Proof: Prove: Given: e canfindwhere theequationsintersecttoseewhere C and cutbyatransversal;showthatpairof AC AC DF AB BC 8 7 value indicatescharges basedonusage. Answers 57. m 1 17.12 2 5 2 7. E x ; x PR 3 51. 71. 5 1 G p y are x 69. undefined 3. x BC AC ; 2( EF BC DE 4 1 DF, AB AB ; suppl.cons.int. corr. 4 ( y 2 2 7 x q x 5 4 ; 5 4 alt. int. O 2 and 0.48( ; showtwolinesare 31. 1 and y 5. EF x are

3) DF DE 5) 45. 37. BC EF alt. 41. y 4 5

y x x , 180 y y 23. AB ext. DE b 17. no slope-intercept form, x 7. are suppl.;showthatalternate . 59. value indicatesthefixedrate,while EF 8, 5) 5, y y 47. AB A 0.05 3. x x 58 53. 21. E DE 5 3 apease:Abasketball Sample answer: A 3 and 27. is y x 13. 2 and 5 x 5 9. 7. 9. 19. 17. 15. 11. 13. 5. Sample answer:Inthe B HS + 3 5 4 61. 7 1 F x y 1 1. 2. 4. 3. . Theslopesare notequal, a y y y 33. ; 1.375 750, where y y y y y 750 75 5 4 Given Segment Addition Postulate Subtraction Property Substitution Property JT b 1 5 5 corr. y ( 1 137.5 6 ; 6, x 39.95, 3 8 6 5 1 ; x D A to sameline;show x 11. x x 63. 5 3 x x alt. int. x 2 3 12) corr. 4 and 10,800 ( The slopeof 8 1 73 2 x 6 4 y 3 x 5. E B 2 x 1.25( 2 x 4) 0.95 x total 8, m 49. x that ; 6 20) in 3 4.95 CD alt. C F 47. line are parallel. sideline andtwoormore linesperpendiculartothesame will beparallelbecausetheyare allperpendiculartothe 57. 55. 35. 39. 37. 25. 33. 27. 5. 2. 1. 1. 6. 4. 3. 2. 3. y undefined No, theslopesare notthesame. Given: Given: 15 Property ofCongruence, vertical anglestheyare congruent. BytheTransitive ttmnsReasons Statements Proof: Proof: ttmnsReasons Statements Proof: are corresponding angles,andtheyare congruent, Prove: Prove: Prove: Given: 3. 2. 4. 1. 180 m are supplementary. P supplementary. m 180 m All rt. Definition ofperpendicular If corresponding Given A B A S pqp FF FF F FT F TF T TT 0.3 C 29. QRS QRS RSP D D QRS QRS RSP x We Q P B are supplementary. A C C B R C S D D and and 8 D 1 QRS RSP 4 C

know that 6 m , are Q C 31. m C m m PQR 49. D 43. R D 1 and RSP PQR 2 6 21.6 PQR RSP PQR . y See students’work. n q and PQR are 2 PQR are 4 2 1 3. 2. 1. x 4 Perpendicular Transversal Th. If alternateinterior Given lines are 4. 5. 1. 6. 3. 2. , thenlinesare 6. Because 1 2 9 Substitution Def. ofsuppl. Given If consecutiveinterior Def. ofsuppl. Def. of R C 7. Since 41. are suppl.,lines B 51. 7 The 10-yard lines 4 . 2 6 45. Q 4 5 6 and 4 and B 53. . S 1 m 1 are D 7 are . 7 A m P . , 59. pq p p q • After marking several points, a slope can be calculated, TTF F which should be the same slope as the original brace. • Building walls requires parallel lines. TFF F 1 35. D 37. DA EF; corresponding 39. y x 3 2 FTT T 2 2 11 41. y x 2 43. y x 3 3 3 FFT F

Pages 167–170 Chapter 3 Study Guide and Review 61. complementary angles 63. 859.22 1. alternate 3. parallel 5. alternate exterior 7. consecutive 9. alternate exterior 11. corresponding Pages 162–164 Lesson 3-6 13. consecutive. interior 15. alternate interior 17. 53 1. Construct a perpendicular line between them. 19. 127 21. 127 23. neither 25. perpendicular 3. Sample answer: Measure distances at different parts; 27.y 29. y 2x 7 compare slopes; measure angles. Finding slopes is the most 2 readily available method. 31. y x 4 ( ) 7 2, 3 33. y 5x 3 5.C 7. 0.9 35. AL and BJ, alternate 9. 5 units; O x exterior B D 3 1 y y –x – 37. CF and GK, 2 lines 4 4 P(2, 5) same line 39. CF and GK, A E consecutive interior (1, 1) suppl. 41. 5 O x

Chapter 4 Congruent Triangles

Pages 177 Chapter 4 Getting Started 11. 13. R 1 3 A B S 1. 6 3. 1 5. 2 7. 2, 12, 15, 6, 9, 3, 13 2 4 9. 6, 9, 3, 13, 2, 8, 12, 15 11. 11.2 13. 14.6 D C Pages 180–183 Lesson 4-1 Q P 1. Triangles are classified by sides and angles. For example, a triangle can have a right angle and have no two sides 15.M G 17. d 3; congruent. 3. Always; equiangular triangles have three y acute angles. 5. obtuse 7. MJK, KLM, JKN, LMN 9. x 4, JM 3, MN 3, JN 2 11. TW 125, WZ L H P(4, 3) 74, TZ 61; scalene 13. right 15. acute 17. obtuse 19. equilateral, equiangular 21. isosceles, acute 23. BAC, CDB 25. ABD, ACD, BAC, K J O x CDB 27. x 5, MN 9, MP 9, NP 9 29. x 8, JL 11, JK 11, KL 7 31. Scalene; it is 184 miles from Lexington to Nashville, 265 miles from Cairo to Lexington, and 144 miles from Cairo to Nashville. 33. AB 106, BC 233, AC 65; scalene 75 19. 4 21. 5 23. 5 35. AB 29, BC 4, AC 29; isosceles 25. 1; 27. 13; 37. AB 124, BC 124, AC 8; isosceles y y 39. Given: MN (2, 5) m NPM 33 33 Prove: P Selected Answers (2, 4) y 5 y –2x 3 3 (0, 3) RPM is obtuse. R

O x O (2, 0) x Proof: NPM and RPM form a linear pair. NPM and RPM are supplementary because if two angles form a linear pair, then they are supplementary. So, mNPM mRPM 180. It is given that mNPM 29. It is everywhere equidistant from the ceiling. 31. 6 33. By substitution, 33 mRPM 180. Subtract to 33. Sample answer: We want new shelves to be parallel so find that mRPM 147. RPM is obtuse by they will line up. Answers should include the following. definition. RPM is obtuse by definition.

Selected Answers R39 Selected Answers R40 angles ofexterior 1. P 57. 49. AD 41. 41. 3. three: 45. • •T structural support. Answers shouldincludethefollowing. 43. 11. 17. 31. 39. gs1811Lesson4-2 ages 188–191 3 are theremote interior 11 12, 6, architecture andconstruction. Isosceles trianglesseemtobeusedmore oftenin If allofthesidesare congruent, thetriangleisequilateral. scalene. Ifatleasttwosidesare congruent, itisisosceles. equiangular. Ifnotwosidesare congruent, thetriangleis right. Ifeachanglehasthesamemeasure, thetriangleis triangle isobtuse.Ifoneangleequals90˚,the acute. Ifthemeasure ofoneangleisgreater than90,the measure ofeachangleislessthan90,thetriangle Sample answer: 43 Selected Answers 2. 1. 3. riangles canbeclassifiedbysidesandangles.Ifthe ttmnsReasons Statements Prove: Proof: Proof: Prove: 15 AD Given: B Sample answer:Triangles are usedinconstruction as 93 49 103 Given: 2, r upeetr.pair,they are suppl. are supplementary. form alinearpair. CD 55. 4 and 5. 7 and ABC 51. 13. 19. 2 and CBD CBD 33. 55 , so 5, and m m 44 65, 65 53 a 6, 4 0 50 2 and and A ABC F FGI 7. 7, 2 a are rightangles. GI All rt. D A CBD 10, 53. 21. 147 1 9, and GIF GIF 2 b 2

a Given ⊥ 35. 1, ⊥ 2 linesformrt. 8

2 3 and 15. C FH 1. ABC ABC m H any 32 and

( 2 and 40 D 8 and are 3 and 9. IGH, G

. ( b 76 0 C ) 25 23. 2 GIH 12 37. ADC GIH . b 44 ) I 129 47. 2 is isoscelesbydefinition. . 1. 3. 2. CD 25. F H Given If 2 Def. oflinearpair 8 x x 123 D ;

y 1

Third AngleTheorem Given a C B 4 y a 2 a form alinear 27. 2

F FGI 2

2 O 2 14 2 a

b

FH 2 G ( 2

H 29. b A IGH I 3 ) ( 2 0 ( 53 3, 3 b x ) ) 2 JL 43. U 25. angles are congruent. preserved. by acongruence transformation,socongruence is 1. P 27. that thecorresponding angles are congruent. are congruent; flip. Use aprotractor toconfirmthatthecorresponding angles y 45. 51. Q corrresponding sidesare congruent. that alloftheanglesare congruent andthattheother M gs1518Lesson4-3 ages 195–198 V 9, QV s 3,8,10,and12, H Z R The sidesandtheanglesoftriangleare notaffected 68 N 4. 7. 5. 6. were arightangle,then In Proof: Prove: m m Given: so there cannotbetworightanglesinatriangle. In Proof: Prove: Given: T Tr m , 28, . Useaprotractor toconfirmthatthecorresponding urn; B S BEC ue; Q 3 m m m m 180 m m m m Y C z O R 1 17 5, Z MNO and PQR 8, 13. J , , JK , and RT CBD A ABC C A C A CBD T NP W D 3. 22 right angleinatriangle. There canbeatmostone triangle. one obtuseangleina There can beatmost 48, V 8, 53. 180. 180. Itmustbethat A G X , PQR B , MNO P M , T m 180 m m m R m AFC C J 59. is obtuse. 3, 4 P m X is arightangle. M 2, must beacute. F L S 40 m 20 Z is obtuse.So R CBD C ABC ABC T N 2 r is arightangle. , , eflexive M T units 9. J s 2,4,7,and9 X G X ABC P 15. 7, K , H Z 60, D , DFB 17 E CFH 90, so 3, B U 5 m 2, F . Useaprotractor toconfirm m T B QT 55. C J MP , 61. 40 W 3 O m 5. D m m 6 Z 7. 4. 6. 5. , Y H symmetric 1 KL D , 1 JKL 3 m 72 P N 17 19. 0. Butthatisimpossible, Subtraction Property Def. ofsuppl. Substitution Angle SumTheorem Q W , 34 Q 17. S 21. V 23. units J 68 M , and 90. C 47. m N F 11. m , and We M s 1,5,6,and11, 7. Flip; P 29 1 m S A m O QR

R , Z Q , need toknow 57. 63. K , G WPZ M P MN T 49. T N , X L 90. If U 10, 90. So, P x transitive 5, F m AED X 1 8, 2 12, T Y Q 34 N 29 , R , . O , 29. 12 Pages 203–206 Lesson 4-4 HJ 1. Sample answer: In QRS, R S 10 6 R is the included angle of 12 G RS the sides QR and RS. Q 6 10 3. EG 2, MP 2, FG 4, NP 4, EF 20, and Q MN 20. The corresponding sides have the same 31. K measure and are congruent. EFG MNP by SSS. 64 5. Given: DE and BC bisect each other B Prove: DGB EGC E 36 80 Proof: JL64 DE and BC bisect each other. DE Given G 36 80 DF C DG GE, BG GC 33. Given: RST XYZ Def. of bisector RSXY of segments

T Z DGB EGC DGB EGC SAS Vertical are . Prove: XYZ RST 7. SAS Proof: RST XYZ 9. Given: T is the midpoint of SQ. R Given SR QR Prove: SRT QRT Proof: SQ R X, S Y, T Z, T Statements Reasons RS XY, ST YZ, RT XZ 1. T is the midpoint of SQ. 1. Given CPCTC 2. ST TQ 2. Midpoint Theorem 3. SR QR 3. Given X R, Y S, Z T, 4. RT RT 4. Reflexive Property XY RS, YZ ST, XZ RT 5. SRT QRT 5. SSS Congruence of and 11. JK 10, KL 10, JL 20, FG 2, GH 50, segments is symmetric. and FH 6. The corresponding sides are not congruent so JKL is not congruent to FGH. 13. JK 10, KL XYZ RST 10, JL 20, FG 10, GH 10, and FH 20. Def. of Each pair of corresponding sides have the same measure so they are congruent. JKL FGH by SSS. 35. Given: DEF E Prove: DEF DEF 15. Given: RQ TQ YQ WQ, R RQY WQT DF Proof: DEF Prove: QWT QYR Y Given Q

DE DE, EF EF, D D, E E, DF DF F F W T Proof: Congruence of Congruence of segments is reflexive. is reflexive. RQ TQ YQ WQ RQY WQT Given Given DEF DEF Def. of s QWT QYR 37. Sample answer: Triangles are used in bridge design for SAS structure and support. Answers should include the Selected Answers following. 17. Given: MRN QRP M Q • The shape of the triangle does not matter. MNP QPN R •Some of the triangles used in the bridge supports seem Prove: MNP QPN to be congruent. 39. D 41. 58 43. x 3, BC 10, CD 10, BD 5 Proof: NP 3 45. y x 3 47. y 4x 11 49. 5 51. 13 Statement Reason 2 1. MRN QRP, 1. Given Page 198 Chapter 4 Practice Quiz 1 MNP QPN 1. DFJ, GJF, HJG, DJH 3. AB BC AC 7 2. MN QP 2. CPCTC 5. M J, N K, P L; MN JK, NP KL, 3. NP NP 3. Reflexive Property and MP JL 4. MNP QPN 4. SAS

Selected Answers R41 Selected Answers R42 31. • • should includethefollowing. help landsurveyorsdoublecheckmeasurements. Answers 29. 1. P 19. 23. 21. sides. without corresponding congruent corresponding congruent angles 27. 41. second quartertothethird. gs2023Lesson4-5 ages 210–213 when designingbuildings. Sample answer: Architects alsousecongruent triangles are congruent bySSS. pair ofcorresponding sidesare congruent, thetriangles are congruent, thetrianglesare congruent bySAS.Ifeach two pairsofcorresponding sidesandtheincludedangle congruent, thetrianglesare congruent bydefinition. If If eachpairofcorresponding anglesandsidesare T Selected Answers 2. 1. 1. 1. 4. 3. 9. 8. 3. 4. 3. 2. 2. 4. 5. 5. 7. 6. wo trianglescanhave Prove: Prove: ttmnsReasons Statements Reasons Statements Reason Statement Prove: Proof: Proof: Proof: B Sample answer:Theproperties ofcongruent triangles Given: not possible Given: Given: 1 midpoint of T E F G HJ G E H HJ KJ HJ K KG 33. G S G F G L H J GHJ EFG GHL STH A HTS TSF SHT STH 43. GJ LJ LJ K and of G E T KJ S H F G , G H HL L hr saseprrateofdeclinefrom the There isasteeper F WXZ S F r rightangles. are J G EFG SHT F GHL L GHJ H K TSF , G is themidpoint , L E ; D GJ , SFH H G KJ J HL , KG LKJ HFG SFH F S H . LKG , SHF SFH 25. HTS is the H F E F , ; H B L LJ SFH . SSS orSAS J JG , LKJ HFG FHT, SHF r rightangles. are F LKG YXZ H H , T E and FHT, , and H 45. 35. T 1. 3. 2. 8. 7. 2. 1. 4. 9. 3. 4. 5. 6. 78 5. 3. 1. 2. 4. CBD GL Given Reflexive Property Midpoint Theorem Reflexive Property Def. of CPCTC Given SSS SSS Def. of Addition Property Segment Addition Substitution FG CPCTC All rt. Given Given SAS 37. D HK E T 68 47. A segments segments J C 39. are D S B P H 59 . F H F C E 11. 5. Postulates are acceptedastrue withoutproof. 3. 9. 7. C Proof: Given: Prove: AAS canbeproven usingtheThird Angle Theorem. supplementary. We are giventhat congruent so Angles supplementarytocongruent anglesare Prove: supplementary. Likewise, Prove: Since Proof: Given: Proof: Given: EF Alt. int. V Given: Proof: Prove: ert. E GKH E Given ||

GH

F are

. However, E E E EGD X are H K T V G K F

EGD WXY W DGH V TV Given E R and V . Alt. int. EKF V XW G K and K XWY . H Y Q S H Given H E

Z EGD , QS

R S G || K E VR CPCTC ,

S YZ , F S A

TRV KHD , YZW AAS are DHG DGH B X

K T

G H SR

ZYW EK EF CPCTC H , D . ,

KHD EKF are alinear pair, theanglesare

E Given AAS

WXY Z EGD , so GHK EG QRS KHD

V

GH

KH 1 . Sinceweare giventhat V 1

ert.

AAS X ABC and Given

HKG

R WZ E DGH WY 2 G

2 are KHD YZW Property Reflexive D Z XY DHG S .

by ASA. DEF WY K are DHG . Q . F H 13. Given: MN PQ, M Q L 6. RYM and MYT are 6. Supplement Theorem 2 3 supplementary and Prove: MLP QLN RYN and NYT are supplementary. 1234 7. RYM RYN 7. suppl. to are . MN P Q 8. RYM RYN 8. SAS Proof: MN PQ 21. CD GH, because the segments have the same Given measure. CFD HFG because vertical angles are congruent. Since F is the midpoint of DG, DF FG. It MN PQ cannot be determined whether CFD HFG. The Def. of seg. information given does not lead to a unique triangle. 23. Since N is the midpoint of JL, JN NL. JNK LNK MN NP NP PQ NP NP because perpendicular lines form right angles and right Addition Prop. Reflexive Prop. angles are congruent. By the Reflexive Property, KN KN. JKN LKN by SAS. 25. VNR, AAS or ASA MP NQ MN NP MP 27. MIN, SAS 29. Since Aiko is perpendicular to the Substitution NP PQ NQ ground, two right angles are formed and right angles are Seg. Addition Post. congruent. The angles of sight are the same and her height is the same for each triangle. The triangles are congruent by MP NQ ASA. By CPCTC, the distances are the same. The method is Def. of seg. valid. 31. D 33. MLP QLN M Q Given: BA DE, B D 2 3 DA BE C ASA Given Prove: BEA DAE A E 15. Given: NOM POR, N P Proof: NM MR, DA BE PR MR, Given NM PR MOR Prove: MO OR BA DE BEA DAE Proof: Since NM MR and PR MR, M and R Given ASA are right angles. M R because all right angles are congruent. We know that NOM POR and NM AE AE PR. By AAS, NMO PRO. MO OR by CPCTC. Reflexive Prop. 17. Given: F J, F E H, 35. Turn; RS 2, RS 2, ST 1, ST= 1, RT 1, EC GH G H RT 1. Use a protractor to confirm that the Prove: EF HJ E C corresponding angles are congruent. 37. If people are J happy, then they rarely correct their faults. 39. isosceles 41. isosceles Proof: We are given that F J, E H, and EC GH. By the Reflexive Property, CG CG. Segment addition results in EG EC CG and CH CG GH. Pages 219–221 Lesson 4-6 1. By the definition of congruence, EC GH and CG The measure of only one angle must be given in an CG. Substitute to find EG CH. By AAS, EFG isosceles triangle to determine the measures of the other 3. HJC. By CPCTC, EF HJ. two angles. Sample answer: Draw a line segment. Set your compass to the length of the line segment and draw 19. Given: MYT NYT M an arc from each endpoint. Draw segments from the MTY NTY intersection of the arcs to each endpoint. 5. BH BD Prove: RYM RYN 7. Given: CTE is isosceles with T RT Y vertex C. 60 m T 60 C Selected Answers N Prove: CTE is equilateral. Proof: E Statement Reason Proof: 1. MYT NYT 1. Given Statements Reasons MTY NTY 1. CTE is isosceles 1. Given 2. YT YT, RY RY 2. Reflexive Property with vertex C. 3. MYT NYT 3. ASA 2. CT CE 2. Def. of isosceles 4. MY NY 4. CPCTC triangle 5. RYM and MYT are 5. Def. of linear pair 3. E T 3. Isosceles Triangle a linear pair; RYN and Theorem NYT are a linear pair 4. mE mT 4. Def. of

Selected Answers R43 Selected Answers 31. R44 15. 27. 9. 29. 10. 11. 3. 5. 4. 1. 1. 5. 1. 2. 4. 3. 5. 8. 3. 4. 6. 2. 2. 7. 8. 7. 6. 5. 9. x Prove: ttmnsReasons Statements Reasons Statements Given: Reasons Statements Proof: Prove: Given: Case II: Proof: Prove: Given: Proof: Case I: 20 LTR Selected Answers J A X K equilateral A A equiangular triangle. A triangle. K m m m m m m is themidpointof J X B C B B J 3; ABC ABC ABC ABC XKF KXJ 17. KXJ 1 A A A A CTE CTE bisects y C T C C E T J J equilateral triangle. equiangular triangle. equiangular triangle. equilateral triangle. X 81 F A A A B F is an is themidpointof is equilateral. is an is anequiangular is anequilateral J ABC ABC ABC X XKF ABC C is equiangular. is equilateral. C C C LRT 18 2 bisects B B C B 180 m , A 60 60 60 60 , 19. FXJ FXJ B B B is equilateral. X is an is an is an is an . E C C B 11. . 28 60 C C B C KXF 21. , K LSQ 180 C F , . . 56 K F 5. 3. 2. 4. 3. 8. 4. 1. 4. 3. 2. 7. 2. 5. 1. 1. 5. 6. 23. . 10. 11. LQS 6. 5. 7. 9. 8. Def. of Isosceles Triangle Def. ofequiangular Substitution Theorem Isosceles Triangle Def. ofmidpoint Given Given Theorem Substitution Conv. ofIsos. Def. ofequilateral CPCTC equilateral Definition of Def. of Given Given Def. of ASA AC AC 36.5 Substitution Given Angle SumTheorem Subtraction Substitution Def. ofequiangular equilateral. Equiangular B B KF 13. 25. equiangular equilateral 1 X L bisector 38 S J L 2 R Th. s are congruent so 53. 51. 49. same measure sotheyare congruent. and 35. 33. that willsimplify thecom on thepositive 1. P 1. P 47. SSS. 45. 43. • • following. shapes tocreate visualimages. Answers shouldincludethe acute. gs2426Lesson4-7 ages 224–226 Practice Quiz2 Chapter4 age 221 other row hassixcongruent isoscelestriangles. One row hasthree congruent isoscelestriangles.The There are tworows ofisoscelestrianglesinthepainting. in manyworksofart. Rectangle, squares, rhombi, andotherpolygonsare used Place onevertexattheorigin, place onesideofthetriangle JM 10 1. 5. 4. 3. 2. 6. angles are congruent. U angles are congruent. Therefore, ttmnsReasons Statements Proof: Prove: and Prove: Proof: ( 18 Given: QR Given: D BG , and S Let B A 3. 1, . Perpendicularlinesformfourrightanglesso pq yz F T T FFT T F FTT T T TFF F F TTF F T FFT T FTT F TFF T TTF D B 41. ABD 37. A ABD 52 BD U are rightangles. .Eachpairofcorresponding sideshavethe 5. EH 3) 5 R A We V apease:Artistsuseangles,lines,and Sample answer: 30 C , B 52 S R B VRS 5. ML ABC D B bisect A

x , C are given that QRS 26 -axis. Labelthe coordinates withexpressions RS 39. U CBD C CBD R B S 52 S The trianglesineachsetappeartobe is notcongruent to C , . Thecorresponding sidesare not U ABC TUS 26 p y T 2 , putations. , JL . QS RSV S V U R R rz y or 5, q , 2. 6. 4. 3. 1. 5. BD R 34 Def. of CPCTC Reflexive Property Given Protractor Postulate AAS V UST AC S VRS U , , p or EG JML U because allright T since vertical EGH B 5 , DG D q S bisector . TUS U U 34 BDG R S , and , by ASA. GH by 26 R R S , T y a b c 0 a b c 3. G(b, c) 5. P(0, b) 7. N(0, b), Q(a, 0) Midpoint T is , or , . 2 2 2 2 c c Slope of ST 2 2 0 or 0. a a b b 2 2 2 ( ) ( ) x 0 0 0 F 0, 0 H 2b, 0 y Slope of AB or 0. B(2, 8) a 0 a ST and AB have the same slope so ST AB. 9. Given: ABC Prove: ABC is isosceles. 29. Given: ABD, FBD y AF 6, BD 3 B(3, 4) Prove: ABD FBD

A(0, 1) F(6, 1) A(0, 0) C(4, 0) x D(3, 1) O x Proof: Use the Distance Formula to find AB and BC. 2 2 AB (2 0) (8 0) 4 64 or 68 Proof: BD BD by the Reflexive Property. 2 2 BC (4 2) (0 8) 4 64 or 68 AD (3 0)2 (1 1)2 9 0 or 3 Since AB BC, AB BC. Since the legs are congruent, DF (6 3)2 (1 1)2 9 0 or 3 ABC is isosceles. Since AD DF, AD DF. 11. y 13. y 2 2 P(a, b) –1 AB (3 0) (4 1) 9 9 or 32 X(4b, c) BF (6 3)2 (1 4)2 9 9 or 32 Since AB BF, AB BF. ABD FBD by SSS. M(0, 0) ( ) x 1 N 2a, 0 W(0, 0) – x Z(2b, 0) 31. Given: BPR, BAR PR 800, BR 800, RA 800 15. y 17. Q(a, a), P(a, 0) Prove: PB BA 19. D(2b, 0) 21. P(0, c), X(0, b) Proof: 23. N(2b, 0) J(c, b) PB (8000)2 (800 0)2 or 1,280,000 BA (8001600)2 (800 0)2 or 1,280,000 Y(0, 0) Z(2b, 0) x PB BA, so PB BA. 33. 680,000 or about 824.6 ft 35. (2a, 0) 37. AB 4a; 25. y Given: isosceles ABC C(2a, 2b) AC (0 (2a))2 (2a 0)2 4a2 4a2 or with AC BC 8a2; CB (0 2a)2 (2a 0)2 4a2 4a2 or R and S are R S 2a 0 2a 0 midpoints of legs 8a2; Slope of AC or 1; slope of CB 0 (2a) 0 2a AC and BC. or 1. AC CB and AC CB, so ABC is a right Prove: AS BR A(0, 0) B(4a, 0) x isosceles triangle. 39. C Proof: 2a 0 2b 0 The coordinates of R are , or (a, b). 2 2 41. Given: 3 4 Q 2a 4a 2b 0 The coordinates of S are , or (3a, b). Prove: QR QS 2 2 BR (4a a)2 (0 b)2 (3a)2 (b)2 3 4 12 or 9a2 b2 R S AS (3a 0)2 (b 0)2 (3a)2 (b)2 Proof: 2 2 Statements Reasons or 9a b Selected Answers Since BR AS, AS BR. 1. 3 4 1. Given 2. 2 and 4 form a 2. Def. of linear pair 27. Given: ABC y linear pair. 1 and 3 C(b, c) S is the midpoint form a linear pair. of AC. 3. 2 and 4 are 3. If 2 form a linear S T T is the midpoint supplementary. pair, then they are of BC. 1 and 3 are suppl. Prove: ST AB A(0, 0) B(a, 0) x supplementary. 4. 2 1 4. Angles that are suppl. Proof: b 0 c 0 b c to are . Midpoint S is , or , 2 2 2 2 5. QR QS 5. Conv. of Isos. Th.

Selected Answers R45 Selected Answers R46 corresponding sideshasthesamemeasure. Therefore, lines are 13. triangle are thesamesegment inanequilateraltriangle. 3. the vertexand is notnecessarilyperpendicular totheside. vertex oppositetheside,while amediandoespassthrough of atriangle,anddoesnotnecessarily passthrough the perpendicularbisectoris totheside side. A 1. P 1. P Chapter 5RelationshipsinTriangles 1. P 43. QR C 21. 45. 23. 27. 11. gs2225Lesson5-1 ages 242–245 GettingStarted Chapter5 age 235 StudyGuideandReview Chapter4 ages 227–230 K MNP C Sample answer: An altitude andanglebisectorofa Sample answer:Bothpassthrough the midpointofa ( h Selected Answers 1. 1. 4. 2. 3. 3. 2. , The sumofthemeasures oftheanglesis180. Given: B ttmn Reason Statement Reasons Statements Prove: Proof: Proof: Proof: Given: 40 equiangular, equilateral RKE 4, 5) C B E 3. y A A C ( G R 0, 0 D D D DGC ABD DFC GCF 25. A CFD CDG d , A , 20 K ) D , 3. N to thesameline,theyare A A C QRS 80 ; ifalt.int. 5. NCK RS E C D D E EBC ABD DGC GCF DFC B D ( E a , E , and R E , 0.5, C by SSS. GEF C DFE EBC F K DGE EFD C 7. EDG D ( E 3 E D ( m b 6 , 5 19. , m D KER 5) , n , and GEF DFE , , 0 DGE ) C 9. , ) MN are F 5. , x obtuse, isosceles C A G , QS 68 13. CKN , linesare C 2. 3. 1. 2. 1. 4. 3. 25 7. B .Eachpairof 5. D 20 , 40 Alt. int. Given Given CPCTC Given ASA AAS G , E 15. . NP C ERK 9. B . B 26 D E , 47. N D are C 11. 17. 5 C F G , D MP 14 K m , . E KNC ; if2 , F E E 5, 25. 31. 13. 11. 5. 7. m 13. 12. 14. 15. 10. 11. ttmnsReasons Statements Given: Proof: Prove: 3 2 Prove: 5. 2. 1. 3. 6. 4. 7. 9. 8. Given: Proof: Prove: 1. 3. 7. 5. 2. 4. 8. 6. PR Given: x Proof: ttmnsReasons Statements Reasons Statements WPA 5. 6. 3. 2. 4. 1. , 3 bisector of angle triangle withvertex Y U Y U Y Z XM M N X XY Y XM X XN XM M MZ 2MZ MZ Y C C 3 1 V C UVW 7, Y V Z N M UVY Y M is themidpointof V MYZ UVY D A Z XZY is themidpointof CEA ACD ACD E is themidpointof 18 m are medians. Y Y is amedian. X 9. vertex angle the bisectorof bisector of perpendicular C Y 90 C M M UVW Y Y W Y W X XZ NY C N NY MZ Z V A M MZ MZ MZ C UVW C and V is anisosceles 2 Z 27. Z 2NY Y B N and V D 1 Y E Z is amedian. , , 5 2 17. X , , Y Z A , 2 , (0, 7) WVY C X XN 58 D CEB M Z NZY WVY XYZ BCD N D Y Z XY BCD UVW B is isosceleswith N NY XZ XN always 5 3 V are onthe N , and A is the 15. A are medians. , D B B N NY UVW . D 29. . NY U Y NY x X X W B UVW Y 19. Z D . . . 20, . 3 4 Y V never 1. 6. 4. 2. 8. 3. 7. 5. y . 4. 5. 2. 1. 3. 6. 12. 13. 15. 14. 10. 11. is 1. 4. 2. 5. 3. 6. 7. 9. 8. Given CPCTC Reflexive Property Def. ofisosceles Def. ofmedian Def. ofanglebisector Def. ofmidpoint SAS CPCTC Reflexive Property Reflexive Property Given SSS SAS 4; yes;because Theorem Isosceles Triangle Def. of Given Reflexive Property Def. ofmedian CPCTC Def. ofmedian SAS Def. of Def. of Segment Addition Postulate Substitution Division Property Addition Property Substitution A ZY 21. M C U Y W X 2 E D 23. segs. segs. segs. B N 40 V 7. AE BE 7. CPCTC 3, 1 29. mKAJ mAJK 31. mSMJ mMJS 8. E is the midpoint of AB. 8. Def. of midpoint 33. mMYJ mJMY 9. CEA CEB 9. CPCTC 35. Given: JM JL L 10. CEA and CEB 10. Def. of linear pair JL KL 2 form a linear pair. Prove: m1 m2 11. CEA and CEB are supplementary. 11. Supplement Theorem 1 K M 12. m CEA m CEB 12. Def. of suppl. J 180 Proof: 13. mCEA mCEA 13. Substitution Statements Reasons 180 1. JM JL JL KL 1. , Given 14. 2(m CEA) 180 14. Substitution 2. LKJ LJK 2. Isosceles Theorem 15. m CEA 90 15. Division Property 3. mLKJ mLJK 3. Def. of 16. CEA and CEB are 16. Def. of rt. 4. m 1 m LKJ 4. Ext. Inequality rt. . Theorem 17. CD AB 17. Def. of 5. m1 mLJK 5. Substitution 18. CD is the perpendicular 18. Def. of bisector 6. mLJK m2 6. Ext. Inequality bisector of AB. Theorem 19. C and D are on the 19. Def. of points on a 7. m1 m2 7. Trans. Prop. of perpendicular bisector line Inequality of AB. 37. ZY YR 39. RZ SR 41. TY ZY 43. M, 33. Given: ABC, AD, BE, CF, A L, K 45. Phoenix to Atlanta, Des Moines to Phoenix, KP AB, KQ BC, R P E Atlanta to Des Moines 47. 5; PR, QR, PQ 49. 12; QR, PR, KR AC F K x x 6 Prove: KP KQ KR PQ 51. 2(y 1) , y 53. 3x 15 4x 7 0, D 3 6 B C 7 1 Q x 8 55. A 57. (15, 6) 59. Yes; (3) 1, Proof: 4 3 Statements Reasons and F is the midpoint of BD. 61. Label the midpoints of 1. ABC, AD, BE, CF, 1. Given AB, BC, and CA as E, F, and G respectively. Then the a a b c b c KP AB, KQ BC, coordinates of E, F, and G are , 0, , , and , 2 2 2 2 2 KR AC c respectively. The slope of AF , and the slope of 2. KP KQ, KQ KR, 2. Any point on the a b c c KP KR bisector is equidistant AD , so D is on AF. The slope of BG and a b b 2a c from the sides of the the slope of BD , so D is on BG. The slope of b 2a angle. 2c 2c CE and the slope of CD , so D is on CE. 3. KP KQ KR 3. Transitive Property 2b a 2b a 35. 4 Since D is on AF, BG, and CE, it is the intersection point of 37.C(–6, 12) y 39. The altitude 12 the three segments. 63. C R, D S, G W, E(5, 10) will be the same ( ) A 16, 8 for both triangles, CD RS, DG SW, CG RW 65. 9.5 67. false 8 D(–2, 8) and the bases will F(9, 6) 4 be congruent, so Page 254 Practice Quiz 1 B(2, 4) the areas will be 1. 5 3. never 5. sometimes 7. no triangle 9. mQ equal. 41. C –4 O 4 8 12 16x 56, m R 61, m S 63

Pages 257–260 Lesson 5-3 43. Sample answer: 45. Sample answer: 1. If a statement is shown to be false, then its opposite must y –n y C(2, m) be true. I(0, 3x) 3. Sample answer: ABC is scalene. Given: ABC; AB BC; BC AC; A AB AC Prove: ABC is scalene. A(0, 0) ( ) x B C Selected Answers B n, 0 Proof: Step 1: Assume ABC is not scalene. 47. 5 11 49. ML MN H(0, 0) G(x, 0) x Case 1: ABC is isosceles. 51. 53. If ABC is isosceles, then AB BC, BC AC, or AB AC. This contradicts the given information, so ABC is Pages 251–254 Lesson 5-2 not isosceles. 1. never 3. Grace; she placed the shorter side with the Case 2: ABC is equilateral. smaller angle, and the longer side with the larger angle. In order for a triangle to be equilateral, it must also be 5. 3 7. 4, 5, 6 9. 2, 3, 5, 6 11. mXZY isosceles, and Case 1 proved that ABC is not isosceles. mXYZ 13. AE EB 15. BC EC 17. 1 19. 7 Thus, ABC is not equilateral. Therefore, ABC is scalene. 21. 7 23. 2, 7, 8, 10 25. 3, 5 27. 8, 7, 5. The lines are not parallel.

Selected Answers R47 Selected Answers R48 17. 19. 9. 7. 13. 21. 11. Given: Prove: Given: Step 2: Step 1: Proof: Prove: Step 3: Step 1: Proof: Step 3: Step 2: Selected Answers Points Given: P Given: Prove: Prove: Step 2: Step 2: Step 3: Prove: Step 3: Step 1: Proof: Step 1: Proof: Step 1: Proof: Step 2: Given: Q /S P 1 a assumption that in than oneobtuseangle There canbenomore T angle in Assume thatthere canbemore thanoneobtuse can beatmostoneobtuseanglein of theanglesatriangleequals180.Thus,there The conclusioncontradictsthefactthatsum is greater than180,2 90, The measure ofanobtuseangleisgreater than Assume The conclusionthat1 a 1 a 1 a P of AB a a a 1 P h ReflexiveProperty. the If So, undefined. right triangle. the longestsideofa equals 180.Therefore, thehypotenusemustbe sum ofthemeasures oftheanglesatriangle relationships contradictthefactthat Both 90, if is, righttriangle is notthelongestside.That of a Assume thatthehypotenuse Assume midpoint of If Assume , 1 1 a a ABC Z Q Q is negative. must benegative. ABC AB P AB 1 C 1 1 a x AB 15. , and 1 0 0 m is notamedian Z 0. m ABC 0; 0, sotheassumptionmustbefalse.Thus, sarightangle. is a PQR 0 0 is amedianof BC P a A 90, sothemeasure oftwoobtuseangles sarighttriangle; is a A R C 0 0 BC, . BC 1 a 2 2 byCPCTC. a P R number cannotbeexpressed as BC . 1 a and ABC Z are noncollinear. a , then 90. m then is amedianof or 0. Q 0. 0 AB . R AB a , and A 1 a a m m , 1 0 sincethatwouldmake x AC 0 mustbefalse.Therefore, 8.B h aereasoning, 180. Bythesame AC C C Q PQR 0 Z 0 isfalse,sothe 180. . QZR PZQ m m A , then R PQR Z B A B . . Since A P P Z Z CB . 12 180. PZR is the ABC P m Z by SSS. . C b a by . C 1 a 29. 27. 25. 31. 33. Thus, theassumptionthathewaspresent atthecrimeisfalse. it iscontradictedbyhispresence inChicagoatthattime. 23. Y 1500 1500 Given: Proof: Given: Step 1: ae1Case 2 r Case 1 Step 2: Prove: Step 1: Proof: Step 3: 60 Step 2: tp2 ae1Case2 Case1 Step 2: Step 3: Step 1: Proof: Step 3: Step 3: Step 2: Prove: 60 Step 1: Proof: Step 3: Use es; ifyouassumetheclientwasatsceneofcrime, 60 r 8358.3 58.3 17 0.15 15% 225 3 5 and 2 as 2 eiiino rationalnumbers. Therefore, definition of have acommon factorof2.Thiscontradicts the Thus, 2 2 o rational. not factors, and a Because Because even number. a speed waslessthan60milesperhour. must befalse.Therefore, Ramon’saverage The conclusionsare false,sotheassumption If Therefore, information. Thus,theassumptionisfalse. This conclusioncontradictsthegiven Therefore, the Transitive Property, triangles, This conclusioncontradictsthegivenfact If C than orequalto60milesperhour, Ramon’saveragespeedwasgreater Assume that are equilateral. given fact The conclusionofbothcasescontradictsthe b Assume that Assume that Assume that b a a d t b b b b D , 2 2 2 2 BCD ACD ABC 1 a b t BCD 0, , where aand ba 1 1 2 225 b 225 225 2 4 (2 a D b b 2 sarationalnumber, itcanbewritten is a 3, and r 2 n n 17 2 and is notequilateral. is notequilateral. B n 3 2. Thus, 2 2 a b is anevennumber. So, is anequilateraltriangle,then ) . Since 5 A 2 is evenitcanbewrittenas2 and 0, and a a 60 C 2 ? b . Thus ACD BCD d b a ABD 60 60 b a BCD . Thus, b a A 0. If 2 are bothevennumbers,they P a B 175. 1 is a rational number. rational is a is anequilateraltriangle. b ABC is notanequilateraltriangle. Z . a b are integerswithnocommon 1 2 is notamedianof is anequilateraltriangle. b B is anevennumber, asis C and 2 b a A and C a b 1 , then2 . ABD A A D D b is alsoan r are equilateral A D A C B D 60. n a b . B 2 2 . C , PQR D B . By 2 C B . a is . 35. D 37. P 25. no; 0.18 0.21 0.52 27. 2 n 16 29. 6 n 30 39. Given: CD is an angle bisector. C 31. 29 n 93 33. 24 n 152 35. 0 n 150 CD is an altitude. 37. 97 n 101 Prove: ABC is isosceles. 39. Given: HE EG HE Prove: HE FG EF B D A Proof: G Statements Reasons F Proof: 1. CD is an angle bisector. 1. Given Statements Reasons CD is an altitude. 1. HE EG 1. Given 2. ACD BCD 2. Def. of bisector 2. HE EG 2. Def. of segments 3. CD AB 3. Def. of altitude 3. EG FG EF 3. Triangle Inequality 4. CDA and CDB 4. lines form 4 rt. . 4. HE FG EF 4. Substitution are rt. 41. yes; AB BC AC, AB AC BC, AC BC AB 5. CDA CDB 5. All rt. are . 1 43. no; XY YZ XZ 45. 4 47. 3 49. 51. Sample 6. CD CD 6. Reflexive Prop. 2 7. ACD BCD 7. ASA answer: You can use the Triangle Inequality Theorem to 8. AC BC 8. CPCTC verify the shortest route between two locations. Answers 9. ABC is isosceles. 9. Def. of isosceles should include the following. •Alonger route might be better if you want to collect 41. Given: ABC DEF; BG is A frequent flier miles. an angle bisector of G •Astraight route might not always be available. ABC. EH is an angle C B D 53. A 55. QR, PQ, PR 57. JK 5, KL 2, JL 29, bisector of DEF. H Prove: BG EH F E PQ 5, QR 2, and PR 29. The corresponding sides Proof: have the same measure and are congruent. JKL PQR Statements Reasons by SSS. 59. JK 113, KL 50, JL 65, PQ 58, 1. ABC DEF 1. Given QR 61, and PR 65. The corresponding sides are not 2. A D, AB DE, 2. CPCTC congruent, so the triangles are not congruent. 61. x 6.6 ABC DEF 3. BG is an angle bisector 3. Given Page 266 Practice Quiz 2 of ABC. EH is an 1. The number 117 is not divisible by 13. angle bisector of DEF. 3. Step 1: Assume that x 8. 4. ABG GBC, 4. Def. of bisector Step 2: 7x 56 so x 8 DEH HEF Step 3: The solution of 7x 56 contradicts the 5. mABC mDEF 5. Def. of assumption. Thus, x 8 must be false. Therefore, x 8. 6. mABG mGBC, 6. Def. of mDEH mHEF 5. Given: mADC mADB A 7. mABC mABG 7. Angle Addition Prove: AD is not an altitude mGBC, mDEF Property of ABC. mDEH mHEF 8. mABC mABG 8. Substitution CDB mABG, mDEF Proof: mDEH mDEH Statements Reasons 9. mABG mABG 9. Substitution 1. AD is an altitude 1. Assumption mDEH mDEH of ABC. 10. 2mABG 2mDEH 10. Addition 2. ADC and ADB are 2. Def. of altitude 11.mABG mDEH 11. Division right angles. 12. ABG DEH 12. Def. of 3. ADC ADB 3. All rt are . 13. ABG DEH 13. ASA 4. mADC mADB 4. Def. of angles 14. BG EH 14. CPCTC This contradicts the given information that mADC 43. y 3 2(x 4) 45. y 9 11(x 4) 47. false mADB. Thus, AD is not an altitude of ABC. 7. no; 25 35 / 60 9. yes; 5 6 10 Pages 263–266 Lesson 5-4 Selected Answers 1. Sample answer: If the lines are not horizontal, then the Pages 270–273 Lesson 5-5 segment connecting their y-intercepts is not perpendicular 1. Sample answer: A pair of scissors illustrates the SSS to either line. Since distance is measured along a inequality. As the distance between the tips of the scissors perpendicular segment, this segment cannot be used. decreases, the angle between the blades decreases, allowing 3. Sample answer: 5. no; 5 10 15 7 the blades to cut. 3. AB CD 5. x 6 2, 3, 4 and 1, 2, 3; 7. yes; 5.2 5.6 10.1 3 3 9. 9 n 37 11. 3 n 33 7. Given: PQ SQ P S 2 13. B 15. no; 2 6 11 Prove: PR SR T 4 2 17. no; 13 16 29 19. yes; 1 9 20 21 21. yes; 17 3 30 30 23. yes; 0.9 4 4.1 QR

Selected Answers R49 Selected Answers R50 23. smaller area. ends ofthepliersdecreases, more force isappliedtoa ends oftheplierstodecrease. As thedistancebetween between themdecreases causingthedistancebetween inequality. As force isappliedtothehandles,angle 9. 15. decreases. altitude ofthetriangle base anglesdecrease, the decrease. Thus,asthe increases, thebaseangles 27. increases. and thedistancefrom theendofdoortoframe 25. 21. Sample answer:Thepliersare anexampleoftheSAS 1. 4. 3. 2. 5. ttmnsReasons Statements Proof: Selected Answers 10. 11. 1. 4. 3. 2. As thevertexangle As thedoorisopenedwider, theangle formed increases Given: m Given: Prove: ttmnsReasons Statements Reasons Statements Proof: Proof: Prove: 6. 5. 1. 7. 3. 2. 4. 9. 8. P m m Q PR m P m QR Q Q R A EC E AE C CD m midpoint of AC AF AE AE AE Q AOD S D D E PQR SQR PQR 3 SR 1 S Q AC C D QR E m P Q R Q PS FB FB A AF EC EC EC R D AB D Q BD B B S S is themidpointof m F D ; F 11. m 3 m A ; m m D E D R AB PS 1 AB m AC AF AE 2 S D F is the AOB m SQR PQS , ; B A , m . BDC F 1 . FB FB 1 17. 29. P m m 4 Q 2. 1. 5. 4. 3. 2. 1. 3. 4. 10. 11. tie()Velocity (m/s) Stride (m) 2; FDB 1 5. 4. 2. 6. 3. 1. 7. 9. 8. x Reflexive Property Given Postulate SAS Inequality Def. ofinequality Angle Addition Reflexive Property Given Given SSS Inequality .01.37 1.01 0.70 1.50 0.43 1.25 0.22 1.00 0.07 0.75 0.50 0.25 Inequality Substitution Prop. of Inequality Add. Prop. of SAS Inequality Given Def. ofmidpoint Given Def. of Given Substitution Segment Add. Post. Segment Add. Def. of 10 2 13. B 19. AD F 7 segments segments 2 A 4 R DC x 1 D 3 E 20 S C bisector not findthecross products correctly. 16 13. 1. P 1. P andSimilarity Chapter 6Proportions 1. P 9. FG 41. • • artwork. Answers shouldincludethefollowing. with areas thatwere inproportion asabackground forthis 39. 9 51. 49. 27. 19. 39. 19. •T • through thedirt. Answers shouldincludethefollowing. between thetwoarmsdecreases andtheshovelmoves 31. 23. Detachment 33. 41. g 8–8 Lesson6-1 age 284–287 GettingStarted Chapter6 age 281 Review StudyGuideand Chapter5 ages 274–276 ABC 54, 48,42 Cross multiply anddivideby28. 15 incenter The dimensionsare approximately 24inchesby34inches. the piecesinoutsidecolumn. the leftpiecesasfrom thethird columnare to The centercolumnpiecesare tothethird columnfrom decreases. As theoperatordigs,anglebetweenarms the angledecreases. the anglebetweenarmsincreases, anddecreases as Sample answer:ItappearsthatTiffany usedrectangles 43.2, 64.8,72 1. 3. 2. 4. 5. m D 16.4 lb The trianglesare notcongruent. Given: apease:Abackhoe digswhentheangle Sample answer: A yes; 6 B EF Prove: ttmnsReasons Statements Proof: he distancebetweentheendsofarmsincreases as 13. P A B . 3. 35. C DEF 43. D ( ABC 1, 7,25,79 BCA B 544 5, 10 y 7. bisects 3, yes; 16 always FG O , A A E 18 29. orthocenter 3. EG C 11. ABC B D 5. 4 E m ) T bisects 1.295 riangle InequalityTheorem 2 320 D B 50, DEC 20 21. ECD E FDE E 35; scalene 7. ; . 45. 6 EG A 18 in.,2430in. 25. 13. B DEC B 31. 15 5 6 E x 19 15. BC ; 7 D 9. 5; isosceles 14 6:89 9. E 72 DQ 37. x . yes; MD 33. 45. A 2. 1. 3. 4. 5. 47 11. 15. D B 3 yes, bytheLawof DR Def. ofseg.bisector Given Alt. int. V ASA is anotmedianof alt. int. 3. m ert. 25.3:1 27. 21. 47. 5. A 35. Suki; Madelinedid 43. 17. 1 23. DEF x no; 7 y 1 12 2 EF are 1, O C SR 2 3 7 7. 5. 17. Thm. x 3 2.1275 angle 25. 2 m 11. 5 18 ft,24ft . D SQ 34 145 2: 37. E E DFE 2, 4,8, 20 E 19 ( 2, 2 , 36% x ) 53. Yes; 100 km and 62 mi are the same length, so AB CD. Page 301–306 Lesson 6-3 By the definition of congruent segments, AB CD. 55. 13.0 1. Sample answer: Two triangles are congruent by the SSS, 57. 1.2 SAS, and ASA Postulates and the AAS Theorem. In these triangles, corresponding parts must be congruent. Two Page 292–297 Lesson 6-2 triangles are similar by AA Similarity, SSS Similarity, and 1. Both students are correct. One student has inverted the SAS Similarity. In similar triangles, the sides are ratio and reversed the order of the comparison. 3. If two proportional and the angles are congruent. Congruent polygons are congruent, then they are similar. All of the triangles are always similar triangles. Similar triangles are corresponding angles are congruent, and the ratio of congruent only when the scale factor for the proportional measures of the corresponding sides is 1.Two similar sides is 1. SSS and SAS are common relationships for both figures have congruent angles, and the sides are in congruence and similarity. 3. Alicia; while both have proportion, but not always congruent. If the scale factor corresponding sides in a ratio, Alicia has them in proper is 1, then the figures are congruent. 5. Yes; A E, order with the numerators from the same triangle. AD DC CB B F, C G, D H and EH HG GF 5. ABC DEF; x 10; AB 10; DE 6 7. yes: BA 2 . So ABCD EFGH. 7. polygon ABCD DEF ACB by SSS Similarity 9. 135 ft 11. yes; FE 3 1 QRS TVU by SSS Similarity 13. yes; RST polygon EFGH; 23; 28; 20; 32; 9. 60 m 11. ABCF is 2 JKL by AA Similarity 15. Yes; ABC JKL by similar to EDCF since they are congruent. 13. ABC is SAS Similarity 17. No; sides are not proportional. 1 not similar to DEF. A D. 15. 17. polygon 8 3 3 3 19. ABE ACD; x ; AB 3;AC 9 5 5 5 13 16 10 2 3 ABCD polygon EFGH; ; AB ; CD ; 21. ABC ARS; x 8; 15; 8 23. 25. true 3 3 3 3 2 5 19. ABE ACD; 6; BC 8; ED 5; 21. about 3.9 in. 27. EAB EFC AFD: AA Similarity 9 1 25 29. KP 5, KM 15, MR 13, ML 20, MN 12, by 6.25 in. 23. 3 16 2 PR 16 31. mTUV 43, mR 43, mRSU 47, 25.5 1 in. 27. always 29. never 3 4 33. 31. sometimes 33. always m SUV 47 x y; if BD AE, then BCD ACE 1 BC DC 2 x 3 in. 35. 30; 70 37. 27; 14 by AA Similarity and . Thus, . Cross 8 AC EC 4 x y 39. 71.05; 48.45 41. 7.5 multiply and solve for y, yielding y x. 8 Figure not shown actual size. 43. 108 45. 73.2 47. 5

49. L(16, 8) and P(8, 8) or 51. 18 ft by 15 ft 35. Given: LP MN L P L(16, 8) and P(8, 8) 53. 16:1 55. 16:1 J LJ PJ y 57. 2:1; ratios are the same. Prove: JN JM 8 a b c 59. M N C B 3a 3b 3c Proof: 4 a bc 1 Statements Reasons 3(a b c) 3 D MN 1. LP MN 1. Given O A 4 8 12 x 2. PLN LNM, 2. Alt. Int. Theorem 4 LPM PMN 3. LPJ NMJ 3. AA Similarity 8 LJ PJ 4. 4. Corr. sides of s are JN JM proportional. 61. Sample answer: Artists use geometric shapes in patterns to create another scene or picture. The included objects have the same shape but are different sizes. 37. Given: BAC and EDF B E Answers should include the following. are right triangles. • The objects are enclosed within a circle. The objects AB AC A C D F seem to go on and on DE DF • Each “ring” of figures has images that are approximately Prove: ABC DEF the same width, but vary in number and design. Proof: 63. D Statements Reasons Selected Answers AB AC BC 1 65. 67. 1. BAC and EDF are 1. Given y C AB AC BC 2 69. right triangles. 12 The sides are proportional and the 2. BAC and EDF are 2. Def. of rt. right angles. 8 angles are congruent, so C the triangles are similar. 3. BAC EDF 3. All rt. are . AB AC 4 71. 23 73. OC AO 4. 4. Given DE DF 75. B B m ABD m ADB 5. 5. ABC DEF SAS Similarity A A 4 8 12 16x 77. 91 79. m 1 m2 111 81. 62 83. 118 85. 62 87. 118 39. 13.5 ft 41. about 420.5 m 43. 10.75 m

Selected Answers R51 Selected Answers R52 of slope of3. of and of everytransversalare congruent. the partsofonetransversalare congruent, thentheparts transversals are proportional. Corollary 6.2statesthatif transversals, Corollary 6.1statesthatthepartsof 3. proportional lengths,thenitisparalleltothethird side. and separatessidesintocorresponding segmentsof 1. P 1. P 47. 45. 21. 39. 37. 17. 5. E 1 A 9 6 D g 1–1 Lesson6-4 age 311–315 Practice Quiz1 age 306 B Given three ormore parallellinesintersectingtwo Sample answer:Ifalineintersectstwosidesoftriangle yes; 1947 mi , so D D Selected Answers 18.75 ft 12. 10. 14. 13. x 11. 10 Given: Proof: Prove: ttmnsReasons Statements E E P T 3. 2. 8. 1. 9. 4. 5. 6. 7. A T S are and R D BC D AD A D E AB AB AC AC AB D D A A BC BC 6, ED A A N C 23. D D D E A 8 E E ADE B B is themidpointof A ADE is themidpointof D 2. B D No; segmentsare notproportional; E D Q C 31. 3, C B A EF 2AD AD AD E AE AE 2 DB D P 2, is themidpointof A A A C A are both 0.So is themidpointof 25. F 2 1 . B D B B C E E (3, 8)or(4,4) 9 11. A A A , , B , , O C E 8 4 4 8 and A AE yes C AC ABC F F 19. ABC AC AE E DB AD ; x y C C B DE E 2 , BC , 27. E EC 4 2; 2AE 2 3 1 C , y A D A D 2 1 3. x D C BC 10, , B 4 52 . E . 5 . Both 33. A is notperpendicularto 59. 49. Similarity. similar byAA 53. ABCD A FE 1 ADE 10. 13. 12. 14. C 11. B B 13. ACD ABC 9. 8. 2. 7. 3. 4. 5. 6. 1. 29. . x . C A (5.5, 13) 5 . 5. Def. of Given SAS Similarity Reflexive Prop. Def. of If corr. T Midpoint Theorem Division Prop. Substitution lines are parallel. Def. of Segment Addition Postulate Substitution Substitution D 1 13 ABC The endpoints ; 1.6;1.4;1.1; ransitive Prop. 3, 100 yd 21, 55. E 10 9. 51. 3 1 and , Y y CBE CD 15 2 es; PQRS CBD 7. CBD BC A 61. D 15 ; 2;8;4 The slopes M polygons polygons segments N B are 57. 15. ACD Q PQ A ; N P ; theyare 9, have R 4 and (3.5, 35. No; DE 3 2 1 ; , the E M RQ 7 3 25 ft A B R 2.5) C T 15 . perimeters are inthesameratioassides, or 3.3m bisectors, ormedians. 1 43. 41. 1. P 33. 29. 19. 31. 12, 47. m •A • Answers shouldincludethefollowing. 45. A CD g 1–2 Lesson6-5 age 319–323 D S distance. distance betweentwoparallellinesistheperpendicular developing zoninglaws. City plannersneedtoknowgeometryfactswhen y 16. 15. 10. Given: xy Given: 4 Sample answer:Cityplannersusemapsintheirwork. Prove: ttmnsReasons Statements Proof: Prove: BCD 2 3 or AE ABC city plannerwouldneedtoknowthattheshortest 1. 2. 8. 7. 6. 4. 9. 3. 5. 49. DE 2 CD of TS CB A 21. y 6 z A A A A A RU R R R R DE Y D S z B S B S B S B D ABC ABC ABD B 2 , 11. B 59. ; yes; AA 55. is amedianof 4 RST x z R Q Q A RU BA TU . Thecross CD D T D U P . DB U MNQ S 2 1 D C C ABC ABC ACD U TU TS 63 A R 18 BC 23. m R B BC S D B B S is analtitudeof S B . is amedianof S U ; CD TU B RST Q 11 is amedian A ABD 13. R U D RSU 61. US U D D DB S and S B Z S B 5 1 51. S B , or 20.25 ; R false PQR S,A RST, A CBD 25. 3. products US S A no; anglesnotcongruent 2 2 B D ( ( m 10.8 D U 6 D , B S and X by B ) ) 63. 15. D Y BAD D 27. A Similarity.Thus, AA , C is amedianof RST 5. S is analtitudeof M 16. 10. 15. yield true 78 30; PQR T A 7. 6. 5. 3. 4. 8. 2. 9. 1. 5, 13.5 R 6 C . are altitudes, angle y 57. Division Prop. Def. of Substitution Substitution Substitution Def. of Segment Addition Def. of Mult. Prop. Def. ofmedian Postulate SAS Similarity Given u 17. . Y D xy 7. P 65. Z m 21.6; , 6.75 Y 24; R Q es; the T B z CBD w S 6 3 2 R . 0 0 z polgyons polygons polygons 0 0 9. T X ABC or R 26.4; Z 330 cm 33.6 ABC 53. 2 1 X C BD U . . D , x . x R S Proof: 11. 9 holes 13. Yes, any part contains the A P because of the definition of similar polygons. same figure as the whole, Since BD and QS are perpendicular to AC and PR, 9 squares with the middle BDA QSP. So, ABD PQS by AA Similarity shaded. 15. 1, 3, 6, 10, 15…; QP QS and by definition of similar polygons. Each difference is 1 more than BA BD the preceding difference. 17. The result is similar to a 35. Given: JF bisects EFG. J Stage 3 Sierpinski triangle. EH FG, EF HG 19. 25 E K H Prove: EK GJ KF JF 21. Given: ABC is equilateral. A 1 FG CD CB and 3 Proof: CE 1CA 3 Statements Reasons Prove: CED CAB E 1. JF bisects EFG. 1. Given EH FG, EF HG BC D 2. EFK KFG 2. Def. of bisector Proof: 3. KFG JKH 3. Corresponding Post. Statements Reasons 4. JKH EKF 4. Vertical are . 1. 5. EFK EKF 5. Transitive Prop. ABC is equilateral. 1. Given 1 1 6. FJH EFK 6. Alternate Interior Th. CD CB, CE CA 3 3 7. FJH EKF 7. Transitive Prop. 2. AC BC 2. Def. of equilateral 8. EKF GJF 8. AA Similarity 3. AAC BC 3. Def. of segments GJ EK 1 1 9. 9. Def. of s 4. AC CB 4. Mult. Prop. KF JF 3 3 5. CD CE 5. Substitution CD CE 6. 6. Division Prop. 37. Given: RST ABC, A CB CB W and D are R CD CE 7. 7. Substitution midpoints of TS CB CA B C and CB, respectively. D 8. C C 8. Reflexive Prop. Prove: RWS ADB 9. CED CAB 9. AA Similarity S T W 23. Yes; the smaller and smaller details of the shape have Proof: the same geometric characteristics as the original form. RST ABC S B n 4 25. An 4 ; 65,536 27. Stage 0: 3 units, Stage 1: 3 Given Def. of polygons 3 4 4 4 2 1 4 3 or 4 units, Stage 2: 3 3 or 5 units, Stage 3: 3 3 3 3 3 3 W and D are 1 —RS —TS or 7 units 29. The original triangle and the new triangles AB CB midpoints. 9 are equilateral and thus, all of the angles are equal to 60. By Def. of polygons Given AA Similarity, the triangles are similar. 31. 0.2, 5, 0.2, 5, 0.2; the numbers alternate between 0.2 and 5.0. 33. 1, 2, 4, —RS —2WS 2WS TS AB 2BD 35. 2BD CB 16, 65,536; the numbers approach positive infinity. 0, Substitution 5, 10 37. 6, 24, 66 39. When x 0.00: 0.64, 0.9216, Def. of midpoint 0.2890…, 0.8219…, 0.5854…, 0.9708…, 0.1133…, 0.4019…, —RS WS— 0.9615…, 0.1478…; when x 0.201: 0.6423…, 0.9188…, AB BD RWS ABC 0.2981…, 0.8369…, 0.5458…, 0.9916…, 0.0333…, 0.1287…, Def. of Division SAS Similarity 0.4487…, 0.9894… . Yes, the initial value affected the tenth value. 41. The leaves in the tree and the branches of the LM LN trees are self-similar. These self-similar shapes are repeated 39. 12.9 41. no; sides not proportional 43. yes; MO NP throughout the painting. 43. See students’ work. 45. PQT PRS, x 7, PQ 15 47. y 2x 1 45. Sample answer: Fractal geometry can be found in the

Selected Answers 49. 320, 640 51. 27, 33 repeating patterns of nature. Answers should include the following. Page 323 Practice Quiz 2 •Broccoli is an example of fractal geometry because the 1. 20 3. no; sides not proportional 5. 12.75 7. 10.5 9. 5 shape of the florets is repeated throughout; one floret looks the same as the stalk. • Sample answer: Scientists can use fractals to study the Page 328–331 Lesson 6-6 human body, rivers, and tributaries, and to model how 1. Sample answer: irregular shape formed by iteration of landscapes change over time. self-similar shapes 3. Sample answer: icebergs, ferns, leaf 3 7 1 n 47. C 49. 13 51. 53. 16 55. Miami, Bermuda, veins 5. An 2(2 1) 7. 1.4142…; 1.1892… 9. Yes, the 5 3 4 procedure is repeated over and over again. San Juan 57. 10 ft, 10 ft, 17 ft, 17 ft

Selected Answers R53 Selected Answers R54 parallel to of thehypotenuse. that thealtitudeisgeometricmeanoftwosegments z 2 isnotsimilartoStage1. and So, byTheorem 7.2, vertex oftherightangletohypotenusethattriangle. 1. P 11. 1. P T Chapter 7RightTriangles and 1. P z 11. 23. Additionally, allsidesare ina y 19. 17. Vi 39. 45. 25. 0; 29. rigonometry gs3538Lesson7-1 ages 345–348 Chapter7GettingStarted age 541 Chapter6StudyGuideandReview age 332–336 r PQ Sample answer:2and72 true a ginia Selected Answers 1. 4. 3. 5. 2. 6. 7. Y 15 x 8 Given: yes; ttmnsReasons Statements Prove: Proof: 10 36 OH es, theseare rectangles, soallanglesare congruent. 5 FGH ABC 15 nls perpendicularlines angles. Q Q 16 33 2 , andsoon. 6; S S 6; PQR 1 1 and 2 PQR P R PSQ PQR PSQ 13. G 3. H 7 2 H is analtitudeof y I is arighttriangle. QS 0.8 true R 3. Q 98 11. 1 5.7; . P 95.2 14.1 is arightangle. S 1.3 PQR PSQ PQR PQR PSQ PQR PQR e 4 P R K IK DEC is analtitudeof L 12 2 are right 19. 15. 3; 1 z PQR QSR 5. QSR 24, 3 . 25. is arightangle. 33. 31. 31. , 4 OG false, iteration 23 13. 2 5. 23. 41. 3 f 13. 5 PQR QSR x 6 QSR 1 42 is thegeometricmeanbetween 7 27. 7 yes, 6 5 2.4 yd 26 12 3 8 33. 43. 41. 0.7 6 35. 7. no; lengthsnotproportional 30 O 4.9 15. 3:2 3. 5. 9 G 2 GHI 5.1 6 never 21. 8, Ian; hisproportion shows 13 6, is thealtitudefrom the ratio. 43. 5 3 5.5 2. 3. 1. 4. 5. 6. 7. 35. 29. 3 14.7; 27. 3 7. 20, Definition ofaltitude Definition of Given All right Congruence ofangles is reflexive. ASimilarity AA Statements 4and 5 is transitive. Similarity oftriangles 9.6, 7. 17. yes; Indianaand 24 x 15. 3.5 true PR GJK 21.21 37. x 5 21. y 24 in.and84 Q 56 37. 4 2 9.96 3 1 sometimes 0 S by 2 9. 6.7 6 9. ; 2 36 PQT 15 4 y AA 9. 15 false, 42 are 2 39. 3 Similarity 3 5 7.7 ; 9.4; 2 Stage 38.9; . 6.9 OF RQS ; 5, 25. 49. 5. 57. of 1. P triangles. 33. 13. 3. 39. 27-120-123 63. 47. FE D 6 gs3336Lesson7-2 ages 353–356 RS Maria; Colindoesnothavethelongestsideasvalue c 29 7 3 17 . 1. 1. 3. 2. 4. no, no C Sample answer:Theyconsistofanynumbersimilar 4 Given: y Given: ttmnsReasons Statements Proof: ttmnsReasons Statements Proof: Prove: Prove: , 5, 2 QS D angle at 7. A A 3 51. B D 6, ADC DBC ADC ABC ABD ADC B yes; 4 7 is analtitudeof QS x 15, 18,21 35a. CB d A D A BC A 27. 3 6.9 12 35c. . A 34 A 59. JK with right 6 , D C is arightangle. is arighttriangle. B ADC ADC ABC D C C 58 11 2 no, no , 5 ; 16-30-34; 24-45-51 5; 5 ADC 15. AB D 14-48-50; 21-72-75 ADC ; BC D AC . A ( A 6 2, with rightangleat 9. 3 is arightangle. C x 65. D C C 2 2 8 17 no, no 5 29 53. 7, 13 ft , d D AC 5 x 29. KL 41 2 2 1 C ) 7, 47,2207 2 8 yes, no since A to not formaPythagorean triple whole numbers. factor is Sample answer: and 6 11. C 51.2 O 2 DEF 61. ( D y 29 corresponds to 2 y E 17 1. 4. 2. 3. 21. about 15.1in. 6 , y , C 2 C A D , B Given polygons Definition ofsimilar triangle Definition ofright 1. If thealtitudeisdrawn of art. rt. from thevertexof other. the given formed are similarto 17. 35b. AC JL ( y C ( B 1 2 yes; x 31. x 37. 1 A 5 1 C .N 1 4 Given ) , is analtitudeof 55. corresponds to ,

2 x , and y y 20 1 AB 2 F 5-12-13 to thehypotenuse d 10.8 degrees ) 18-80-82; QR ) o; themeasures do 58 , 8 8 A 9 8 3 34 D 2 , thenthe2 19. B B , 11 , ; D corresponds ( d ABC 23. 5 x and toeach F 2 no; B 9 1 , are not D . Thescale 29 y 17 2 B yes, yes ) QR , 2 E RS E F , , s x 2. (CB)2 + (AC)2 (AB)2 2. Pythagorean Theorem Page 363 Chapter 7 Practice Quiz 1   3. x2 x1 CB 3. Distance on a number 1. 7 3 12.1 3. yes; AB 5, BC 50, AC 45;   2 2 2 y2 y1 AC line 5 45 50 5. x 12; y 63  2 4. x2 x1 4. Substitution y y 2 d2 Pages 367–370 Lesson 7-4 2 1 1. The triangles are similar, so the ratios remain the same. 5. (x x )2 (y y )2 d2 5. Substitution 2 1 2 1 3. All three ratios involve two sides of a right triangle. The 2 2 6. (x2 x1) (y2 y1) d 6. Take square root of sine ratio is the measure of the opposite leg divided by each side. the measure of the hypotenuse. The cosine ratio is the 2 2 7. d (x2 x1) (y2 y1) 7. Reflexive Property measure of the adjacent leg divided by the measure of the hypotenuse. The tangent ratio is the measure of the 41. 43. 45. about 76.53 ft about 13.4 mi Sample answer: opposite leg divided by the measure of the adjacent leg. The road, the tower that is perpendicular to the road, and 14 48 14 48 14 5. 0.28; 0.96; 0.29; 0.96; 0.28; the cables form the right triangles. Answers should include 50 50 48 50 50 the following. 48 3.43 7. 0.8387 9. 0.8387 11. 1.0000 13. mA 54.8 • Right triangles are formed by the bridge, the towers, and 14 3 6 the cables. 15. mA 33.7 17. 2997 ft 19. 0.58; 0.82; •The cable is the hypotenuse in each triangle. 3 3 2 6 3 0.71; 0.82; 0.58; 2 1.41 47. C 49. yes 51. 63 10.4 53. 36 7.3 2 3 3 2 5 25 5 55. 10 3.2 57. 3; approaches positive infinity. 59. 0.25; 21. 0.67; 0.75; 0.89; 0.75; 3 3 5 3 7 3 2 5 alternates between 0.25 and 4. 61. 63. 7 0.67; 1.12 23. 0.9260 25. 0.9974 27. 0.9239 3 3 2 2 65. 122 67. 22 69. 5 526 1 2 29. 5.0000 31. 0.9806 33. 0.2000 1 6 5 26 35. 0.1961 37. 46.4 39. 84.0 41. 83.0 Pages 360–363 Lesson 7-3 26 1. Sample answer: Construct two perpendicular lines. Use a 43. x 8.5 45. x 28.2 47. x 22.6 49. 4.1 mi ruler to measure 3 cm from the point of intersection on the 51. about 5.18 ft 53. about 54.5 55. about 47.9 in. one ray. Use the compass to copy the 3 cm segment. 57. x 17.1; y 23.4 59. about 272,837 astronomical units 22 5 5 Connect the two endpoints to form a 45°-45°-90° triangle 61. 63. C 65. csc A ; sec A ; with sides of 3 cm and a hypotenuse of 32cm. 3. The 5 3 4 4 5 5 3 cot A ; csc B ; sec B ; cot B length of the rectangle is 3 times the width; 3w. 3 4 3 4 23 23 5. x 52; y 52 7. a 4; b 43 67. csc A 2; sec A ; cot A 3; csc B ; 3 3 9. y 3 sec B 2; cot B 69. b 43, c 8 71. a 2.5, D(8 33, 3) B(8, 3) D(8 33, 3) 3 b 2.53 73. yes, yes 75. no, no 77. 117 79. 150 81. 63

O A(8, 0) x Pages 373–376 Lesson 7-5 1. Sample answer: ABC 3. The angle of depression is A 172 B FPB and the angle of 11. 902 or 127.28 ft 13. x ; y 45 15. x 83; 2 elevation is TBP. 52 5. 22.7° 7. 706 ft 9. about y 83 17. x 52; y 19. a 143; CE 21; 2 173.2 yd 11. about 5.3° y 213; b 42 21. 7.53 cm 12.99 cm C 13. about 118.2 yd 15. about 4° 17. about 40.2° 23. 14.83 m 25.63 m 25. 82 11.31 27. (4, 8) 19. 133 100 ft, 300 ft 29. 3 , 6 or about (10.51, 6) 31. a 33, 3 21. about 8.3 in. 23. no 25. About 5.1 mi 27. b 9, c 33, d 9 33. 30° angle Answers should include the following. • Pilots use angles of elevation when they are ascending 35. Sample answer: 37. BH 16 and angles of depression when descending.

Selected Answers 39. 12 3 20.78 cm • Angles of elevation are formed when a person looks 41. 52 4 3 4 6 upward and angles of depression are formed when a units 43. C 45. yes, yes person looks downward. 47. no, no 49. yes, yes 29. A 31. 30.8 33. 70.0 35. 19.5 37. 143; 28 51. 2 21 9.2; 21; 25 39. 31.2 cm 41. 5 43. 34 45. 52 47. 3.75 40 5 53. ; ; 102 14.1 3 3 Pages 380–383 Lesson 7-6 1. Felipe; Makayla is using the definition of the sine ratio for a right triangle, but this is not a right triangle. 3. In 55. mALK mNLO 57. mKLO mALN 59. 15 one case you need the measures of two sides and the 61. 20 63. 28 65. 60 measure of an angle opposite one of the sides. In the other

Selected Answers R55 Selected Answers R56 88.2 Cosines cannotbeused. 3. angle (SAS). all three sidesgiven(SSS)ortwoandtheincluded 1. P 1. P q of aside. case youneedthemeasures oftwoanglesandthemeasure 0.95; m 33. 53. 17. 43a. m 37, 86; 43c. m 63 47. • • following. walls, andfoundations. Answers should includethe 45. 43h. 43e. 19. m 55. 25. x 35. 31. 43. •T • distances inspace. Answers shouldincludethefollowing. 41. 2 gs3730Lesson7-7 ages 387–390 Practice Quiz2 Chapter7 age 383 2 If twoanglesandonesideare given,thentheLawof Sample answer:UsetheLawofCosineswhenyouhave 58.0 sides andanangleoppositeoneofthosesides. The LawofSinesrequires twoanglesandasideor around theedge. The triangularbuildingwasmore efficient withthecells radio wavesemittedbyastronomical objects. radio observatories.Itisusedtomakepictures from the isoneoftheworld’spremier astronomical The VLA Selected Answers p m K L N A riangles are usedintheconstruction oftheantennas. m 36 C Sample answer:Triangles are usedtobuildsupports, 29 Given: m about 536ft 56.9 units Sample answer:Triangles are usedtodetermine Prove: 1 27.5 5.2, 1 25. 51.4 1 Pythagorean Theorem Pythagorean Theorem Def. ofcosine Commutative Property 3 2 2 2 31. 9 1 0.71; 1.00; C X 1 49. 45. 40.9; y 55; 19. m 21. 138.4; 6.0; 1.1, 3. 55. m 11. 0.72; 39. 53.2 5. 33 about 5.97ft m 56; B m 29 19.3, 4.7 m m m QSXZ L 13.1 JFL LFM JFM m n 561.2 units m M 8 1 33. H 51. 2 2 L R 0 99; 1 23. N 9 0 m 29. 5. Q 6 37. 101.9; 2 D 91.8 2 about 14.9mi,13.6mi R yes b m 7. m 57. 78 22.2; 31; 32 W 43f. m 0.69; 79.1 about 1000.7m EFG 215 12 43, 55 EFB 38 GFB 0.71; D 31.3; m X m 1 X 7 29. 53. 15. Cross products 5 11. m m 48.7, 47. 8 5. 2 2 9. G 37. M 0 1 23. 41, 41. 25.6, 159.7 a no about 237.8feet N u 49.6, m 43b. 43d. R 2 2 2 w 2 9 0 m c m 1.05 59.8, 63.4,56.8 109; 36.3; 25.3 R 21.1; 4.9 m 130.8 17, m Substitution Substitution N 0.71; 1.00 0.69; E E 6.3; 45.4 7. g W r 19, 10 m Y 49. AB m 13. 98 39. 27. 38 m 57, 42.8; H Y 2 2 14.7 m 35. 9.5 N 9 1 27. M t 58.4, 42.4, 43g. bu 36mi about 13.6 m D 9. e 2 A 2 Q m m m 0.72; 22.5 13. 41.8 M 51. 21. 10.2 42.8; J 17. w Substitution y 80; M 0.71; X 56, 18.5; 17.9; m 54 m 2.7 m 2 2 18.6; 14.2 F 20.3 m 15. M 1 0 L 86.2; P 82, B B G N 16 C 9. r r quadrilateral thatisnot hpe Quadrilaterals Chapter 8 21. 17. 1. P 59. 47. 1. P 1. P 3. means thesumwillbedifferent from theformula. h perpendicular 27. 29. 23. 45. answer: 36,72,108,144 31. 33. 57. 39. 90, 57. egular, egular quadrilateral,360°; gs3236Catr7StudyGuideandReview Chapter7 ages 392–396 gs4749Lesson8-1 ages 407–409 GettingStarted Chapter8 age 403 6 A 18 true Sample answer: 130 c 1. 4. 2. 3. m B m m 26.9 105, 110, 120,130,135,140,160,170,180,190 m Given: 147.3, 32.7 ( Proof: and ttmnsReasons Statements Prove: Congruence. Then,bySASSimilarity, Since Proof: then bythedefinitionofsimilartriangles, x 5 3 180( E 16.1 JF concave polygonhasatleastoneobtuseangle,which F ; cons.int. 65. J K .,9.6) 1.6, 11. L 9.9 B 49. M M n 0.60; L 360° n 13 M JKL JKL MKL 3. BF 3. none G 25. LF 2 F K 55. 6 92.1 120 false; aright F JFM 2) 41, J J M K 2 37. K L . 5 60, 30, 4 ; L 43.0 JKL , 22 m y G LF J m F K K L z 59. 41. m m 9. F 5. 180 0.80; MLK MLK 51. M M . BytheTransitive Property ofEquality, 61. F JLK C 13 28.1 , L 6 1 27. 28 5.2) (2.8, 150, 30 n 5.9 N P EFB M , 2 b a 51.0 n n F , 4 3 57, MLK ; alt.ext. 2 75, by theReflexiveProperty of 360 6; perpendicular 120, and 120, 35. 39. 13. 5. f 22.6° 19. 0.75; c 53. true 36, 144 a 43. JK 25 m 63.7, m z 18 16.1 m LFM 5 4 n 0 108, 72 17 29. 3. 1. 2. 4. Q 63. n P 15. 18 9. m 5. 15. 11. m 25. 19. 7. h G . 0, 0.80; Reflexive Property Given Alt. int. ASA 35. m 1800 false; depression L 4 37. m 60, 60, 3060 20, 160 16 1080 3 P K 3 70.0 31.1 yd n 60 3 and 67, , m GFB J a 17 5 3 m m 40, 140 B 7. m m m JF 7. B M 180 m 17. 21. 36 , 3 4 Q R L 0.60; 43, E 4 JF , 16.5 N L H are 13. F 5, 61, 31. M 360(2 9 33. m 3 4 3 150 120 5400 3 4 ; 3 165 60, EFG m 23. M 2 and 21.3 yd n BF 60 Sample . C 30, y F 1.33 C 18 . 73 1) Pages 414–416 Lesson 8-2 45. Given: WXYZ WX 1. Opposite sides are congruent; opposite angles are Prove: WXZ YZX congruent; consecutive angles are supplementary; and if there is one right angle, there are four right angles. Z Y Proof: 3. 5. Statements Reasons Sample answer: VTQ, SSS; diag. bisect each other and opp. sides of are . 1. WXYZ 1. Given 2. WX ZY, WZ XY 2. Opp. sides of are . x 7. 100 9. 80 11. 7 3. ZWX XYZ 3. Opp. of are . 2x 4. WXZ YZX 4. SAS 47. Given: BCGH, HD FD H Prove: F GCB B 13. Given: VZRQ and WQST Q G R Prove: Z T S D F C W Proof: T Statements Reasons V 1. BCGH, HD FD 1. Given Z Proof: 2. F H 2. Isosceles Triangle Th. Statements Reasons 3. H GCB 3. Opp. of are . 1. VZRQ and WQST 1. Given 4. F GCB 4. Congruence of angles 2. Z Q, Q T 2. Opp. of a are . is transitive. 3. Z T 3. Transitive Prop. 49. The graphic uses the illustration of wedges shaped like 15. C 17. CDB, alt. int. are . 19. GD, diag. of parallelograms to display the data. Answers should include bisect each other. 21. BAC, alt. int. are . 23. 33 the following. 25. 109 27. 83 29. 6.45 31. 6.1 33. y 5, FH 9 • The opposite sides are parallel and congruent, the 35. a 6, b 5, DB 32 37. EQ 5, QG 5, HQ 13, opposite angles are congruent, and the consecutive QF 13 39. Slope of EH is undefined, slope of EF angles are supplementary. 1 • Sample answer: 100 ; no, the slopes of the sides are not negative reciprocals 79% 3 80 of each other. 60% 60

40 29% Percent That Percent 41. Given: PQRS PQ Use the Web 20 1 3 Prove: PQ RS 0 QR SP 4 2 1998 1999 2000 S R Proof: Year Statements Reasons 51. B 53. 3600 55. 6120 57. Sines; m C 69.9, 7 7 1. PQRS 1. Given mA 53.1, a 11.9 59. 30 61. side, 63. side, 3 3 2. Draw an auxiliary 2. Diagonal of PQRS segment PR and label Pages 420–423 Lesson 8-3 angles 1, 2, 3, and 4 as 1. Both pairs of opposite sides are congruent; both pairs of shown. opposite angles are congruent; diagonals bisect each other; 3. PQ SR, PS QR 3. Opp. sides of are . one pair of opposite sides is parallel and congruent. 4. 1 2, and 3 4 4. Alt. int. are . 3. Shaniqua; Carter’s description could result in a shape 5. PR PR 5. Reflexive Prop. that is not a parallelogram. 5. Yes; each pair of opp. 6. QPR SRP 6. ASA is . 7. x 41, y 16 9. yes 7. PQ RS and QR SP 7. CPCTC 11. Given: PT TR P Q TSP TQR T Prove: PQRS is a 43. Given: MNPQ MN parallelogram. SR M is a right angle. Proof: Selected Answers Prove: N, P and Q Q P Statements Reasons are right angles. 1. PT TR, 1. Given Proof: TSP TQR By definition of a parallelogram, MN QP. Since M 2. PTS RTQ 2. Vertical are . is a right angle, MQ MN. By the Perpendicular 3. PTS RTQ 3. AAS Transversal Theorem, MQ QP. Q is a right angle, 4. PS QR 4. CPCTC because perpendicular lines form a right angle. N 5. PS QR 5. If alt. int. are , Q and M P because opposite angles in a lines are . parallelogram are congruent. P and N are right 6. PQRS is a 6. If one pair of opp. sides angles, since all right angles are congruent. parallelogram. is and , then the quad. is a .

Selected Answers R57 Selected Answers R58 41. opposite sidesisparallelandcongruent. opposite anglesare congruent. 39. other. 13. 45. 43. 21. 12 29. (0, 37. Selected Answers 6. 3. 5. 2. 1. 4. 1. 1. 7. 5. 7. 5. 3. 2. 4. 2. 3. 4. 6. Y Given: Given: B Given: x no Parallelogram; ttmnsReasons Statements Reasons Statements Prove: Reasons Statements Prove: Prove: Proof: Proof: Proof: 2 9), or es; eachpairofoppositeanglesiscongruent. A D A hexagon. ABCDEF A A A ABCD FDCA ABCD Draw Draw A A 47. C B D C B D D 1, B 59. ABC E ABC 1 1 ABD 31. y R 12 16 B ABCD ABCD FDCA A A A ABCDEF yes A D D D D A A D B B to ( is aparallelogram. is a C is aparallelogram. B B D B E C C C C C B F B 2 , 2 2, B , , , 49. . is aregular A , 3 B A A D B , 16 F 33. D 23. C 7, 3). DEF CDA B B B D CDB is aparallelogram. is aparallelogram. is aparallelogram. C A K 3 14 units . C C C M D is aregular hexagon. Move x D C E D 61. and C C F C 35. D 4 34, 5, J M L ( y 51. are diagonalsthatbisecteach to ( 2 3 2, ; not 17. 8 5. 2. 4. 3. 1. 4. 2. 6. 4. 2. 1. 7. 6. 3. 1. 7. 5. 3. 5. 44 2), (4,10),or(10,0) 4, 1), CPCTC line. Tw SSS Angles Theorem Alternate Interior Given CPCTC Def. ofregular hexagon CPCTC Reflexive Property line. Tw Given parallelogram Definition of If alt.int. Reflexive Property Given sides are If bothpairsofopp. quad. is sides are If bothpairsofopp. quad. is SAS SAS lines are Y 53. es; onepairof 25. o pointsdeterminea o pointsdeterminea 30 D D N 63. yes 19. A A to ( 1 55. 3 2 F . x , 4 27. . . , thenthe , thenthe 1 AB 72 are 3, 4), 2 3 ED 2 6, ; yes 3 15. y 57. P , 2 C C Y to 45, es; B B 24 C A 39. same. are 43. diagonals are congruent. 9. opposite sidesisparallelandcongruent. 3. congruent, thentheparallelogramisarectangle. 1. P 1. P 35. 19. the diagonalstomakesure theyare congruent. 41. not arectangle D gs4740Lesson8-4 ages 427–430 Practice Quiz1 Chapter8 age 423 C H Make sure thattheanglesmeasure 90orthatthe McKenna; Consuelo’sdefinitioniscorrect ifonepairof If consecutivesidesare perpendicularordiagonalsare 11 4. 5. 1. 1. 8. 6. 3. 9. 7. 4. 2. 2. 6. 5. 7. 3. Sample answer: Given: Move 30 Given: Prove: ttmnsReasons Statements Reasons Statements Prove: Proof: Proof: and . B G aepae lines G same plane. ih nls 4rt. right angles. ar r DEAC G plane WXYZ are suppl. are supplementary. X G D GHJK W 3. ectangles. D , K H J Y 37. E e 21. 31. X WZY ZWX GKH ZWX ZWX WZX F WXYZ J and 66 ih nls suppl,each right angles. G , but L H W A H See students’work. K WXYZ GHJK are rectangles. DEAC G and 60 are notparallel. N W is aparallelogram. , 2 1 W C and J J Y J is arectangle. H 5. WXYZ K ABCD GKH and and and , Z and and X , K and x K 23. L plane intersect at FEAB 2 3 is aparallelogram. X are inthe until thelengthofdiagonalsis JHK and YXW XYW is arectangle. H Z F , 30 XYZ YXW 8, W is E YXW K and 2 7 Y M y intersect at FEAB , are JHK A 2 3 25. are B D 11. 6 X AB ; Measure theoppositesidesand 33. Z L 29. . 11 Y 3. 4. 8. 1. 6. 5. 2. 9. 3. 2. 1. 6. 7. 5. 4. 7. Y es; consec.sidesare L es; opp.sidesare 13. . W Def. ofparallelplanes Def. ofintersecting If Given If alt.int. Def. ofparallellines Def. ofparallelogram Def. ofrectangle Reflexive Property Opp. Given Consec. If 2 rt. CPCTC SSS are Def. ofparallelogram Z M 29 B . A . 3 1 has 1rt. C sides of of sides K 5. . are C 15. J 40 of 4 are and 27. 7. X is a N L , ithas 17. Y 52 or10 are No; F , diag. , lines 60 . E D G . H 45. No; there are no parallel lines in spherical geometry. 43. Given: QRST and QRTV QR 47. No; the sides are not parallel. 49. A 51. 31 53. 43 are rhombi. 55. 49 57. 5 59. 297 17.2 61. 5 63. 29 Prove: QRT is equilateral. V S T Pages 434–437 Lesson 8-5 Proof: 1. Sample answer: Statements Reasons Square 3. A square is a rectangle 1. QRST and QRTV are 1. Given (rectangle with all sides congruent. rhombi. with 4 5. 5 7. 96.8 9. None; sides) 2. QV VT TR QR, 2. Def. of rhombus the diagonals are QT TS RS QR Rectangle Rhombus not congruent or 3. QT TR QR 3. Substitution Property ( with ( with 4 perpendicular. 11. If 4. QRT is equilateral. 4. Def. of equilateral 1 right ) sides) the measure of each triangle angle is 90 or if the Parallelogram diagonals are congruent, || then the floor is a (opposite sides ) 45. Sample answer: You can ride a bicycle with square square. 13. 120 15. 30 wheels over a curved road. Answers should include the 17. 53 19. 5 21. Rhombus; the diagonals are following. perpendicular. 23. None; the diagonals are not congruent • Rhombi and squares both have all four sides congruent, or perpendicular. but the diagonals of a square are congruent. A square has 25. Sample answer: 27. always 29. sometimes four right angles and rhombi have each pair of opposite 31. always 33. 40 cm angles congruent, but not all angles are necessarily congruent. • Sample answer: Since the angles of a rhombus are not all congruent, riding over the same road would not be smooth. 5 cm 47. C 49. 140 51. x 2, y 3 53. yes 55. no 57. 13.5 59. 20 61. AJH AHJ 63. AK AB 65. 2.4 67. 5

35. Given: ABCD is a parallelogram. A B AC BD E Prove: ABCD is a rhombus. Pages 442–445 Lesson 8-6 Proof: We are given that ABCD is 1. Exactly one pair of opposite sides is parallel. a parallelogram. The diagonals of D C 3. Sample answer: a parallelogram bisect each other, The median of a so AE EC. BE BE because trapezoid is parallel congruence of segments is reflexive. We are also given to both bases. trapezoid isosceles trapezoid that AC BD. Thus, AEB and BEC are right angles by the definition of perpendicular lines. Then AEB 5. isosceles, QR 20, ST 20 7. 4 9a. AD BC, BEC because all right angles are congruent. Therefore, CD / AB 9b. not isosceles, AB 17 and CD 5 AEB CEB by SAS. AB CB by CPCTC. Opposite 11a. DC FE, DE / FC 11b. isosceles, DE 50, sides of parallelograms are congruent, so AB CD and CF 50 13. 8 15. 14, 110, 110 17. 62 19. 15 BC AD. Then since congruence of segments is 21. Sample answer: triangles, quadrilaterals, trapezoids, transitive, AB CD CB AD. All four sides of hexagons 23. trapezoid, exactly one pair opp. sides ABCD are congruent, so ABCD is a rhombus by 25. square, all sides , consecutive sides 27. A(2, 3.5), definition. B(4, 1) 29. DG EF, not isosceles, DE GF, DE / GF 37. No; it is about 11,662.9 mm. 39. The flag of Denmark 31. WV 6 contains four red rectangles. The flag of St. Vincent and the Grenadines contains a blue rectangle, a green rectangle, a 33. Given: TZX YXZ, WX / ZY W yellow rectangle, a blue and yellow rectangle, a yellow and T Z Prove: XYZW is a trapezoid. 1 green rectangle, and three green rhombi. The flag of Trinidad and Tobago contains two white parallelograms 2 Proof: X Y and one black parallelogram. Selected Answers TZX YXZ 41. P Given: TPX QPX Q Given QRX TRX X Prove: TPQR is a rhombus. 1 2 || TR WZ XY Proof: CPCTC If alt. int. Statements Reasons are , then the lines are ||. 1. TPX QPX 1. Given QRX TRX ||| 2. TP PQ QR TR 2. CPCTC WX ZY XYZW is a trapezoid. 3. TPQR is a rhombus. 3. Def. of rhombus Given Def. of trapezoid

Selected Answers R59 Selected Answers R60 37. .5. 7. 3. another vertexliesonthepositive 1. P 1. P 61. do notbisecteachother. 51. 35. not 43. •T •T following. well asotherbuildings. Answers shouldincludethe 41. gs4941Lesson8-7 ages 449–451 Practice Quiz2 Chapter8 age 445 Place onevertexattheoriginandpositionfigure so 12 parallel. D Selected Answers rapezoids canbeusedaswindowpanes. rapezoids haveexactlyonepairofoppositesides Sample answer: No; oppositesidesare notcongruent andthediagonals Given: B Sample answer:Trapezoids are usedinmonumentsas Slope of Proof: Proof: Prove: D Prove: Given: The slopeof Slope of b c ( ABCE Def. ofisos.trapezoid EC of E 0, B y , sotheyare perpendicular. and 3. a a 45. Given to thethirdside. of atriangleisparallel midpoints oftwosides A segmentjoiningthe AD 5. || r

AB b hombus, opp.sides O 18 10 is anisos.trapezoid. C ) A trapezoid. ABCE A E and ABCD D A aremidpoints D A C and C B ( 0, 0 47.

DB A D C C is anisosceles D ) is asquare. a 0 0 0 B . 70 are midpointsof B is thenegative reciprocal oftheslope of B C 0 a a a AB ( ( 49. a a or or 1 , 0 D , a a 53. ) RS x 1 b 1 ) 5 , diag. 7 C 7 x D 55. y -axis. Def. of ( A AE AE AD 2 1 0, 2 D , AD ( O 1 a Given c 2 TV Substitution Def. ofMidpt. 3 ) , , consec.sides and b A

A )

BC BC DB 39. ( 57. E 0, 0

D 2 1 D B DB 4 ) 0 2 ; 113 1 C B 59. ( a 3 ( a , 4 C , 0 a ) 2 a ) b x B y .11. 9. QR m 19. 21. QR QT ST RS 23. QRST 17. 29. • • theorems. Answers shouldincludethefollowing. Midpoint FormulaandSlopeare usedtoprove plane isusedincoordinate proofs. TheDistanceFormula, the towerare parallel. is notenoughinformationgiventoprove thatthesidesof parallelogram are congruent. Sample answer:Theorem 8.3Opposite sidesofa coordinates thatwillsimplifycalculations. Keep thefigure inthefirstquadrant if possibleanduse Place atleastonesideofthefigure on thepositive Place thefigure sooneofthevertices isattheorigin. A Given: Given: Sample answer: Given: Prove: Midpoint Proof: Midpoint BD BD AC BD Proof: Prove: AC Use theDistanceFormulatofind Proof: Midpoint the samelength,sotheyare congruent. Midpoint Prove: O VXZ RS ( is arhombus. A D a 31. ( ( a a 0, 0 b 2 a 2 a , AC c 39. b 55 ) ) QRST r midpoints oftheir ABCD isosceles trapezoid Q ABCD B A ABCD ST ) espective sides. ( a (( a 2 2 a a 2 D C a 0 , 2 2 and Q T S R 0 a m R 2 2 2 C B is 2 is 33. is is , c ( ( S b b b a a a a QT 2 A D B is arhombus. b ) , and XZY with is arectangle. is arectangle. 2 2 a a a 2 ) 0 . D C 160 B 2 b 0 2 b and C b 2 A

b 2 2 2 2 ( so , b C a a 0 0 , 0 c 0) 0 (0 , A ) , b 2 b , 27. T 0 A and , Q 2 b 2 ) b 35. 0 D b 2 b C 2 2 are R c x 2 2 2 2 m , 2 Sample answer:Thecoordinate c b B 0 ( ) 0 b ), 0 c 2 C B R ZXY or D C 60 15. 13. or or or S have 2 a 2 a 2 a (2 0) 2 a A D 2 2 2 a 2 y 2 B T G 2 a a 2 2 0, 2 ( y ( y a 0, 0 0, S , , 0 ( ( 7.7 ( , a T 2 2 b O O b 2 2 b , 0), O b Q c ) b 2 ) . 2 b , 0) . , a 2 b 2 b A 2 b or A D , c or D 2 37. ( Q ) 2 c ( a ( E 2 ( 2 0, 0 ( 0, ), 0, 0 T b ( or or a , 25. 2 . a W b R m , c ) ) , ) ) b b 2 b ( T b ) , 2 No, there c C . XVZ B C C ) 2 a 2 2 a B B ( ( a a a ( ( ( ( a 2 2 a , a a a c , 0 x , , 0 , , 0 2 S -axis. b b b ) ) ) ) c ) , ) x x 2 b x 2 b c ) 2 2 Pages 452–456 Chapter 8 Study Guide and Review 25. 27. y 1. true 3. false, rectangle 5. false, trapezoid 7. true Q N PM 9. 120 11. 90 13. mW 62, mX 108, mY 80, mZ 110 15. 52 17. 87.9 19. 6 21. no 23. yes 25. 52 27. 28 29. Yes, opp. sides are parallel and diag. are congruent 31. 7.5 33. 102 O x

35. Given: ABCD is a square. y C(a, a) P M Q N Prove: AC BD D(0, a) Proof: a 0 29. 31. Slope of AC or 1 a 0 y y a 0 O A(0, 0) B(a, 0) x R B Slope of BD or 1 0 a Q C The slope of AC is the negative reciprocal of the slope of BD. Therefore, AC BD. S 37. P(3a, c) T B O T x O x S D D C Chapter 9 Transformations Q R Page 461 Chapter 9 Getting Started 1. 3. 33. 35. 2; yes 37. 1; no y y y 39. same shape, but turned F F or rotated B(1, 3) A(1, 3) E(2, 1) m n O x G G O x ( ) F 1, 2 O x

H H

(x, y) → (x, y) 5.J(7, 10) y 7. 36.9 9. 41.8 11. 41.4 5 1 8 13. 41. A(4, 7), B(10, 3), and C(6, 8) 43. Consider point K(6, 7) 10 5 (a, b). Upon reflection in the origin, its image is (a, b). 4 2 5 1 Upon reflection in the x-axis and then the y-axis, its image 15. 3 4 5 is (a, b) and then (a, b). The images are the same. 12 8 4 O x 45. vertical line of symmetry 47. vertical, horizontal lines of symmetry; point of symmetry at the center 49. D

51. Given: Quadrilateral LMNP; X, Y, Z, and W are midpoints of their respective sides. Pages 463–469 Lesson 9-1 Prove: YW and XZ y bisect each other. 1. Sample Answer: The centroid of an equilateral triangle is M(2d, 2e) not a point of symmetry. 3. angle measure, betweenness Proof: Y N(2a, 2c) of points, collinearity, distance 5. 4; yes 7. 6; yes Midpoint Y of MN is X 2d 2a 2e 2c , or Z 9. 11. 2 2 (d a, e c). A y A y I L(0, 0) W P(2b, 0) x

Midpoint Z of NP is Selected Answers 2a 2b 2c 0 , 2 2 J H 0 2b 0 0 or (a b, c). Midpoint W of PL is , or (b, 0). G 2 2 O x H I 0 2d 0 2e Midpoint X of LM is , or (d, e). Midpoint B 2 2 B O J x d a b e c 0 a b d c e of WY is , or , . C C G 2 2 2 2 d a b e c a b d c e Midpoint of XZ is , or , . 2 2 2 2 13. 4, yes 15. YX 17. XZW 19. UV 21. T The midpoints of XZ and WY are the same, so XZ and 23. WTZ WY bisect each other.

Selected Answers R61 Selected Answers R62 7 29. 27. ( up 2unitsfrom 5to7or the leftto 1. P 61. 53. ( 9 21. 19. 17. 9. 15. 7. oriented differently thanquadrilateral be reversed aswell. r x eflections wouldbetoofartotheright.Theimage gs4045Lesson9-2 ages 470–475 , Sample answer: 2, 1)to(1, y Selected Answers P 40 ) R Q 4 y → 8 5 Q P L P 55. ( y 8 4 4 8 x M O O 4 Q 36 4, whichis7unitstotheleftor L y P 7, Q P 4 1) isright3anddown2totheimage.The O 8 4 4 8 Q y M 57. S R A y Q K 2). 8 f (3, 5)and 4 R 5. O R K 25.5, x x No; quadrilateral 3. 8 2. Thetranslationfrom Allie; countingfrom thepoint P x m A x B 8 ( H to thetwoparallellines. after anotherwithrespect 13. followed arotation. 11. to thetwoparallellines. after anotherwithrespect J A C S 4, 7);startat3,countto A (7, ( J Y – 4 24 25. 23. down 7squares Y No; itisareflection es; itisonereflection 4, 76, – es; itisonereflection NPQR 0), 4) 4 4 8 72 in.right, 48 in.right left 3squares and h 3 M y O WXYZ y B in. down M (6, 2), B . 7. Thencount 28.8 C O P is B C P A 8 59. to x x B is 2 an application. 43. 39. 37. an isometry. preserve bothcongruences. Therefore, aglidereflection is angles. Thecompositionofthetwotransformationswill r figure isdifferent ineachcategory. 53. 31. 5 57. 59. 55. 1. P ( .9. 7. 5. parallel linesandrotations reflect across intersectinglines. r 3. eflections preserve thecongruences ofsegmentsand eflections. Thedifference isthattranslationsreflect across gs4642Lesson9-3 ages 476–482 D A clockwise ( Both translationsandrotations are madeupoftwo y Q 3 more brains;more free time , A x B ( ) X C m a A' 2 ( B' y y B 5, 8 45 y b A O C' , Y A D' O c ) ( ), x C 0, 3 , 150 T A O y 49. ) (0, 0) ) C B → B The twolinesare notparallel. C ( B y Y , X 45. x ( ( x 3 C ); counterclockwise ( 3 23 ft x 2 2 2

2

,

3 , x

60 13 C 2 41. 33. 2 2 47.

) 2 A 35. ) B No; thepercent per y Yo O T u didnotfillout magnitude 120° order 3and magnitude 90°; order 4and magnitude 72°; 11. magnitude 60° ranslations and A C order 6; order 5and A x , y B ) m 51. → x 5 13. M Pages 483–488 Lesson 9-4 N' P' 1. Semi-regular tessellations contain two or more regular Q polygons, but uniform tessellations can be any combination of shapes. 3. The figure used in the tesselation appears to P N M' be a trapezoid, which is not a regular polygon. Thus, the tessellation cannot be regular. 5. no; measure of interior 15. 17. 72° y T angle 168 7. yes 9. yes; not uniform 11. no; measure P of interior angle 140 13. yes; measure of interior angle R 60 15. no; measure of interior angle 164.3 17. no 19. yes 21. yes; uniform 23. yes; not uniform 25. yes; T 27. 29. R S not uniform yes; uniform, regular semi-regular, O x uniform 31. Never; semi-regular tessellations have the same combination of shapes and angles at each vertex like S uniform tessellations. The shapes for semi-regular tessellations are just regular. 33. Always; the sum of the 19. 21. measures of the angles of a quadrilateral is 360°. So if each m J K t Y t angle of the quadrilateral is rotated at the vertex, then that N L M equals 360° and the tessellation is possible. 35. yes Z X' 37. uniform, regular 39. Sample answer: Tessellations X can be used in art to create abstract art. Answers should Z' N' M' include the following. Y' L' J' • The equilateral triangles are arranged to form hexagons, K' m which are arranged adjacent to one another. • Sample answers: kites, trapezoids, isosceles triangles 23. K(0, 5), L(4, 2), and 25. 3 , 1 27. Yes; it 41. A M (4, 2); 90° clockwise is a proper successive 43. y reflection with respect to F y D M L the two intersecting lines. 29. yes 31. no 33. 9 → M 35. (x, y) (y, x) O F x 37. any point on the line P (8, 1) of reflection 39. no O K x invariant points 41. B E

L E(2, 5) D(7, 5)

K 45. 47. x 4, y 1 y M ( 2, 9) 49. x 56, y 12 43. 8 51. no, no 53. yes, no 55. no, no 57. AB 7, angle betweenness distance L TTTransformation T T T orientation collinearity 4 N (6, 3) measure of points measure (5, 5) L BC 10, AC 9 P M 59. 1(1) 1 and reflection yes yes no yes yes O x 4 K 4 8 1(1) 1 61. square translation yes yes yes yes yes 4 (3, 1) 63. 15 65. 22.5 rotation yes yes yes yes yes K 8 45. direct 47. Yes; it is one reflection after another with N respect to the two parallel lines. 49. Yes; it is one reflection after another with respect to the two parallel lines. 51. C 53. AGF 55. TR; diagonals bisect each other 57. QRS; Pages 490–497 Lesson 9-5 opp. 59. no 61. yes 63. (0, 4), (1, 2), (2, 0) 65. (0, 1. Dilations only preserve length if the scale factor is 1 12), (1, 8), (2, 4), (3, 0) 67. (0, 12), (1, 6), (2, 0) or 1. So for any other scale factor, length is not preserved and the dilation is not an isometry. 3. Trey; Desiree found Page 482 Chapter 9 Practice Quiz 1 the image using a positive scale factor. 1. 3. Selected Answers y E y Q 5. 7. A'B' 12 9. y F D O Q O D x P x F Q C Q

E P O P P x 5. order 36; magnitude 10°

Selected Answers R63 Selected Answers R64 9 31. 29. 25. 23. 11. 41. 7 19. 21. 17. 15. Selected Answers S ST r Given: S so are proportional, so So, to r Multiplication Property ofEquality. Prove: dilation. by thedefinitionofa CE Proof: M eflexive. Therefore, bySASSimilarity, 12 T T ACB A ED C C 2; enlargement B ECD A E 4 L r ( 0.9 5 3 8 CA dilation withcenter ED r C C . Thecorresponding sidesofsimilar triangles C C A E by substitution. Therefore, ) and M D B y ECD 4 L by substitution. r ( r 20 16 12 12 y AB O 27. 8 4 4 8 CD , sincecongruence ofanglesis and C C ) K E O A 13. C C D B N 4 D X r B ( CB C X C C ) 8 Y r K C A E . . We knowthat and scalefactor Z C 12 N x A ED perimeter. original four timesthe perimeter is 39. 37. enlargement 35. 33. C ACB E B Y The 7.5 by10.5 3 1 2 1 r ( ; reduction ; reduction 2; AB C C is similar r A E ) bythe D r Z , x 63. 61. 55. 53. 51. 43. 2 A Given: Thus, midpoint of Proof: the MidpointTheorem. Prove: y V JBH J 45. E O 57. D G V P 2 no 1 y midpoint of It isknownthat 0 L JHB U . Since JHB J D O LBC 47. 59. J L T ,J 60% because vertical anglesare congruent. no LB B B U LCB is the LCB J F is the 49. L L x F B by ASA. . G T by C E • • include thefollowing. should original. Answers image congruent tothe cut andpasteproduces an x 12 are creating adilation. by thesamefactor, you horizontally andvertically transformation. pasting isasimilarity similar, socuttingand If youscaleboth Congruent figures are C Sample answer: Y J B 65. 8 76.0 D y 4 B 16 A 12 4 4 D A es; a 4 L H C B x Page 497 Chapter 9 Practice Quiz 2 41. 43. 1. yes; uniform; semi-regular 3. S y y O 12 C R 8 4 O 4 8 x A 8 P D A B Q S 4 O D C B 5. A ( 5, 1), y R 1 8 C 12 B, 3, B 8 4 O 4 x 2 C C (2, 2) 4 4 P Q A O x A 4 8 45. 47. 13, 67.4° C G y 16 49. B G 5, 306.9° 51. 25 4.5, 26.6° 12 F 53. about 44.8 mi; F about 38.7° south Pages 498–505 Lesson 9-6 8 of due east 1. Sample answer; 〈7, 7〉 3. Sample answer: Using a 55. 〈350, 450〉 mph vector to translate a figure 4 y 57. 52.1° north of due is the same as using an E H west 4, 4 3, 3 ordered pair because a 12 8 4 O 4 x H vector has horizontal and E vertical components, each of O x which can be represented 59. Sample answer: Quantities such as velocity are vectors. by one coordinate of an The velocity of the wind and the velocity of the plane ordered pair. together factor into the overall flight plan. Answers should include the following. 〈 〉 5. 4, 3 9. y •Awind from the west would add to the velocity L 16 7. 213 7.2, 213.7° contributed by the plane resulting in an overall velocity 12 with a larger magnitude. •When traveling east, the prevailing winds add to the L J velocity of the plane. When traveling west, they detract 8 K from it. 4 61. D 63. AB6 65. AB 48 67. yes; not uniform 69. 12 71. 30 O 3 12 4 8 4 4 x 73. 4 75. 27 15 3 77. J 10 4 27 3 15 4 12 K

11. y Z Pages 506–511 Lesson 9-7 O x 0 1 2 2 2 2 4 Z 4 812 1. 3. Sample answer: Y 1 0 1 1 1 1 4 1 1 Y W 5. D(1, 9), E(5, 9), F(3, 6), G(3, 6) 7. A, , W 4 2 8 X 3 3 3 5 1 X B, , C, , D, 1 9. H(5, 4), I(1, 1), 4 4 4 4 4 J(3, 6), K(7, 3) 11. P(3, 6), Q(7, 6), R(7, 2) 13. 613 21.6, 303.7° 15. 〈2, 6〉 17. 〈7, 4〉 13. (1.5, 0.5), (3.5, 1.5), (2.5, 3.5), (0.5, 2.5) 19. 〈3, 5〉 21. 5, 0° 23. 25 4.5, 296.6° 25. 75 15. E(6, 6), F(3, 8) 17. M(1, 1), N(5, 3), O(5, 1), P(1, 1) 19. A(12, 10), B(8, 10), C(6, 14) 21. G(2, 1), 15.7, 26.6° 27. 25, 73.7° 29. 541 32.0, 218.7° H(2, 3), I(3, 4), J(3, 5) 23. X(2, 2), Y(4, 1) 31. 6 2 8.5, 135.0° 33. 4 10 12.6, 198.4° 25. D(4, 5), E(2, 6), F(3, 1), G(3, 4) 35. 2122 22.1, 275.2° 2 4 27. 29. Selected Answers 37. 39. V ( 2, 2), W , 2 , X 2, V ( 3, 3), y A J y 3 3 4 W(3, 1), X(2, 3) 31. P(2, 3), Q(1, 1), R(1, 2), 4 I G S(3, 2), T(5, 1) 33. P(1, 1), Q(4, 1), R(2, 4), S(0, 4), O 35. 37. C A 12 8 4 x T ( 2, 1) M ( 1, 12), N ( 10, 3) S ( 1, 2), J 1 2 4 8 4 O 4 8 x T(1, 6), U(3, 5), V(3, 1) 39. A1, , B, , H 4 3 3 3 4 2 4 1 2 2 2 2 C C , , D 1, , E , , F , 41. A (2, 1), I 8 3 3 3 3 3 3 3 B G 8 B(5, 2), C(5, 6), D(2, 7), E(1, 6), F(1, 2) 43. Each foot- H print is reflected in the y-axis, then translated up two units. 12 B 45. 1 0 47. 0 1 49. 0 1 0 1 1 0 1 0

Selected Answers R65 Selected Answers R66 3 15. 11. 13. 9. of rotation 11. to begreater thanthe measure ofanychord thatisnota So,2 sum oftwosides hastobegreater thanthethird. TheTriangle InequalityTheorem states thatthe triangle. a diameter, tworadiiandachord ofacircle canforma the circumference ofacircle bythe diameter. 1. P 1. P Circles Chapter 10 45. 1. P 53. 51. 17. ( 6), 31. 35. N gs5258Lesson10-1 ages 522–528 GettingStarted Chapter10 ages 521 StudyGuideandReview Chapter9 ages 512–516 Sample answer:Thevalueof 162 false, center 213.3° ,3;9°counterclockwise 6, 3);90° y Y Selected Answers 1.5, D L C 〈 3, 4 ( O A ( 8 D ( A S 〉 2, 0), 3. 2, 2), 2, 3), M C A T 0.9 y C 4 2.4 43. 37. A y 10 L 7. Z 8 4 4 8 M O y E S D 〈 13. O false, scalefactor C 0, 8 ( 5. 33. (5, 0), 3. ( O C B B 12, r 2.5, L B 4 〉 false, componentform B 3, 5), T 1 P 5 2 x 39. (2, N , 2 F C 8 x p 6) x 3 ( N 5 8 M 6), 7. D 4, , D 14.8, 55. E x 15 (5, Q (0, 4), B 2) E 60, 120 ( H (3, 19. 29. 27. is afactorof360. interior angleis60,which uniform 25. 9. B is calculatedbydividing 2 1 3); 180° 4, 47. ; reduction C 208.3° y 17.0 Ye 200° CD C F 5), y D O s; themeasure ofan W 4), 5 8 F O C , 57. G ( R (3, 4 21. 23. 5. 41. 1 ( 24 16, 2), 36, 144 5 C x 6

G F false, center B 3. yes; not yes; uniform 2, 2) 1 3), Except for 72.9, X D x ( r H E x has 4, 13. 47. 39. 22% 23. 37. units 7. it mustbeadiameterand perpendicular. 29. same radiusmeasure. congruent circles usuallyhavedifferent centersbutthe center, butdifferent radiusmeasures; answer: Concentriccircles havethesame 75. three categoriesare minorarcs. 120; 21. vertices lieonthecircle. around sothatthepolygonliesinitsinterior andall circumscribed circle meansthecircle isdrawn A the circle. 1. P AB 1. P ABC 5. diameter hastobelongerthananyotherchord ofthecircle. diameter, but2 53. 73. mEF 35. mBCA the arcs wouldnotbecongruent. Thus, of thecircumferences wouldnotbethesame. 55. 65. 51. 13. 17. mBC 61. the diameter, whichcontainsthecenter. 69. 61. gs5653Lesson10-3 ages 536–543 Lesson10-2 ages 529–535 5 Sample answer: An inscribedpolygonhas allverticeson Sample answer: E , 6. 5. 1. 3. 2. 4. Sample answer:25% 24.32 m,12.16m 20 65 256 4i.or5ft4in. 64 in. 60 5 Given: No; theradiiare notequal,sotheproportional part 36.68 m 10 m,31.42m 80 24 44.7; 27° 28; 14 Prove: ttmnsReasons Statements Proof: A BC mNR is atriangle. , R m m m R m m 3 2 E ft 79°, 2% 45. , mFG 1 Q 160, B units 41. 25. 77. 19. SRQ AC 39. , 250, 23. 9. bisects SQR QRT SQR SRQ SQR E 1.41 55. The firstcategoryisamajorarc, andtheother 67. 63. , R m 80, C 5 90 30 4 10 mAC mPQ ABC 308 Q Z , or mCAB 71. 8 V r m 43. 21. 63. 49nwos 2 northofdueeast 84.9 newtons,32° mGH bisects SQR 27. 79. stemaueo h imtr Sothe is themeasure ofthediameter. , 5 E , 24.5 7° cm T m m m m 24 QRT BCA 41. D X 2.5 5 27 2 SRT 71. 15. 90 30 90, , or 60 QRT T QRT SRQ 15. 22.36 24 23. 49. 7. 31. 57. , m 42 B 200 120, . 65. 69. mHA mABC 5. 45. CAB W 29. SRT A 27. 60 3. mAB 13 137 0.6 m 0; Thelongestchord ofacircle is Z C 17. SRQ 10 90°, 23% If 73. 13.4 cm,84.19cm m 135 T 2 1 or . ; 30 3. 75.40 units 11. ABC okei; tobisectthechord, 17. mAB 25. in., 42.41in. F , 20 100 Sample B 3 A 7. 45 D 15 5. 4. 3. 6. 2. 1. 47. 270, 29. , 30 31. mBC R 33. 103 has three sides,then F 57. U , 30 B e.ofInequality Def. Exterior Angle of Def. Substitution Theorem e.of Def. Given 160 25. always 9. 75. 15 , or ST , 13. 270 3 1 19. R B 83°, 28% 10, 10.4 in. V 9. mNP 36 15 35. F 59. 31. 53. mCD 120 43. 59. E 180 67. 33. 27. R 51. 12 500–600 ft 16 56.5 ft 15. 5. 4 20; 62.83 19. 49. bisector Q C 9.8; 66° 76 2.5 ft 21. 11. 30 mRQ 0.33 11. 40 37. 90 B never 33. mDE 101°, 160 E 115 35. 12.57 A 138 a 6 110 34 , 1.05 6 52 B a 37. Given: O, OS RT, OV UW, OS OV other chords are 5.29 in. and 7.75 in., but exactly 8 in. for the diameter. Prove: RT UW S R T 53. 14,400 55. 180 57. SU 59. RM, AM, DM, IM 61. 50 63. 10 65. 20 O W V Page 543 Chapter 10 Practice Quiz 1 U 1. BC, BD, BA 3. 95 5. 9 7. 28 9. 21 Proof: Statements Reasons Page 544–551 Lesson 10-4 1. OT OW 1. All radii of a are . 1. Sample answer: 3. m1 30, m2 60, m3 60, 2. OS RT, OV VW, 2. Given C m 4 30, m 5 30, m 6 60, OS OV m7 60, m8 30 5. m1 35, 3. OST, OVW are 3. Definition of lines D m2 55, m3 39, m4 39 right angles. 7. 1 9. m1 m2 30, m3 25 4. STO VWO 4. HL B 5. ST VW 5. CPCTC A 6. ST VW 6. Definition of segments 11. Given: AB DE, AC CE BD 7. 2(ST) 2(VW) 7. Multiplication Prove: ABC EDC 12 Property A E 8. OS bisects RT; 8. Radius to a chord OV bisects UW. bisects the chord. 9. RT 2(ST), UW 9. Definition of segment Proof: C 2(VW) bisector Statements Reasons 10. RT UW 10. Substitution 1. AB DE, AC CE 1. Given 11. RT UW 11. Definition of 2. mAB mDE, 2. Def. of arcs segments mAC mCE 1 1 3. mAB mDE 3. Mult. Prop. 39. 2.82 in. 2 2 41. 18 inches 43. 2135 23.24 yd 1 1 mAC mCE B 2 2 1 24 in. 4. mACB mAB, 4. Inscribed Angle 30 in. 2 1 Theorem 16 yd 11 yd m ECD mDE, x yd 2 A x C 1 E F m1 mAC, 2 30 in. 1 m2 mCE 2 D 5. mACB mECD, 5. Substitution m1 m2 45. Let r be the radius of P. Draw radii to points D and E 6. ACB ECD, 6. Def. of to create triangles. The length DE is r3 and AB 2r; 1 2 1 r3 . 47. Inscribed equilateral triangle; the six 7. AB DE 7. arcs have chords. 2(2r) arcs making up the circle are congruent because the chords 8. ABC EDC 8. AAS intercepting them were congruent by construction. Each of 13. m 1 m 2 13 15. m 1 51, m 2 90, m 3 the three chords drawn intercept two of the congruent 39 17. 45, 30, 120 19. m B 120, m C 120, m D chords. Thus, the three larger arcs are congruent. So, the three 60 21. Sample answer: EF is a diameter of the circle and a chords are congruent, making this an equilateral triangle. diagonal and angle bisector of EDFG. 23. 72 25. 144 8 49. No; congruent arcs are must be in the same circle, but 27. 162 29. 9 31. 33. 1 9 these are in concentric circles. 51. Sample answer: The 35. Given: T lies inside PRQ. RK K grooves of a waffle iron are chords of the circle. The ones that is a diameter of T. P Q pass horizontally and vertically through the center are Prove: mPRQ 1mPKQ diameters. Answers should include the following. 2 • If you know the measure of the T Selected Answers radius and the distance the chord A F B is from the center, you can use the Pythagorean Theorem to find R the length of half of the chord E Proof: and then multiply by 2. CDStatements Reasons 1. mPRQ mPRK 1. Angle Addition • There are four grooves on either side of the diameter, so each mKRQ Theorem groove is about 1 in. from the center. In the figure, 2. mPKQ mPK mKQ 2. Arc Addition Theorem 1 1 EF 2 and EB 4 because the radius is half the 3. mPKQ mPK 3. Multiplication Property 2 2 diameter. Using the Pythagorean Theorem, you find that 1mKQ FB 3.464 in. so AB 6.93 in. Approximate lengths for 2

Selected Answers R67 Selected Answers R68 37. making itisoscelesand 41. • • Answersshouldinclude thefollowing. socket. on acircle thatisconcentricwiththe outercircle ofthe polygon becausethevertices of thehexagoncanbeplaced 45. congruent. side isachord ofcongruent arcs, sothechords are Each angles. angle intercepts asemicircle, makingthem90° of 90°,makingitarightangle. 39. The sideofthe regular hexagoninscribed inacircle are pointsonacircle. An inscribedpolygonisonein whichallofitsvertices 4 3 Selected Answers 1. 6. 4. 5. 6. 3. 2. 5. 4. Given: Isosceles righttrianglebecausesidesare congruent radii Sample answer:Thesocketis similartoaninscribed and thesumofcentralangles, Proof: m 360, the measures oftheinterioranglesaquadrilateralis makes them supplementaryalso. m ttmnsReasons Statements Proof: Prove: mDAB 6.Since 360. Prove: Given: inch wideis m m m mCD inscribed; m m m 2 1 2 1 2 1 C A mPKQ mPKQ mCD m CED MLN CED CED MLN KRQ KRQ PRK supplementary. By arc addition andthedefinitionsofarc measure supplementary. inscribed inscribed in quadrilateral m A m C 360, so B A CED CED m and mMN and and A and 2 1 C CD m m m mMN 8 3 m 2 1 C 2 1 2 1 MLN , 2 1 mPK inch. mCD mKQ CD mMN m CED C 180, so 2 1 PRQ PRK A MLN ( D C 2 1 mDCB MN MLN MLN upeetr.Becausethesum of supplementary. , mDAB C are are ABCD m ; AOC are MN O m m B and and is acentralangleforanarc B mDAB 43. A 2. 6. 5. 4. 3. 5. 1. 6. 4. m mDCB m Measure ofaninscribed Substitution (Steps5,1) Substitution (Steps3,4) ut Prop. Mult. of Def. Substitution of intercepted arc. Given intercepted arc (Case1). half themeasure ofthe side isadiameter inscribed anglewhose e.of Def. The measure ofan Square becauseeach D m M D A 2 1 ), but 30 r8.This (360) or80. A D half measure O 6.But 360. N L mDAB 2 1 mDCB 180, making O mDCB E arcs C B D , C units 47. 29. QL 5. 3. is onetangentcontainingapointonthecircle. since atangentintersectscircle inexactlyonepoint,there tangent canonlyintersectacircle inonepoint. A inside thecircle wouldintersectthecircle intwopoints. only twotangents. 1a. P 31. ( the Pythagorean Theorem, NL LQGN 15. Construct 27. GN 33. 41. 55. QP g 5–5 Lesson10-5 age 552–558 Y bu ice inacircle about acircle polygon cir Sample answer: 8. 1. 4. 7. 6. 5. 2. 3. 234 A T 12; Draw 12 27 Sample answer: sometimes es; 5 Prove: Prove: ttmnsReasons Statements Proof: Given: Given: ) 2 wo; from anypointoutsidethecircle, youcandraw E 4, so 2 Since 12. A A A A B right angles. Draw 12. A is arectangle. 25. and X B X C B 2 17. 35. ABX ABX B ( QL 49. is tangentto is tangentto L 15 QP Q to A isosceles. of the midpoint A re ABCD B 3 AD 12 ) A C B A 2 P F B B B ctangle. and usrbdpolygoninscribed cumscribed X X CED C X 2 A G X 19. is tangentto GN , 135 and 3 B , , X 57. C G ( 5. Using N . A units PL A X 13 ACX P is a at L C 30 is C , , thus BC ACX , and no ) 2 1b. and 2 GQ C QL . E 1 . So, 1.62 7. 21. is C X X None; alinecontainingpoint , A 37. X y are P 576 ft at at X L See students’work. . . B C 45, 45 51. . X . at D 4 A 1. 4. 7. 8. 6. 5. 3. 2. 9. ( ( B 0, 0, 0 X Given e.of Def. HL h t oftangency. the pt. CPCTC Reflexive Prop. are All radiiofacircle is Line tangenttoacircle line. points, there isone Through anytwo no units . B 39. b A ) ) C C GN 4 11. to theradiusat 5 44 is tangent E . P Q ( 53. a , 0 yes lines ) always 13 1c. 23. B 13. C ( L 2 ( One; 2 60 a a , 0 16 , b ) ) x A 1 Proof: Let the coordinates of E be (a, 0). Since E is 45. 3, 1, 2; m3 mRQ, m1 mRQ so m3 2 the midpoint and is halfway between A and B, the 1 1 1 m1, m2 (mRQ mTP) mRQ mTP, which coordinates of B will be (2a, 0). Let the coordinates of 2 2 2 1 D be (0, b). The coordinates of C will be (2a, b) because is less than mRQ, so m2 m1. 47. A 49. 16 it is on the same horizontal as D and the same vertical 2 51. 33 53. 44.5 55. 30 in. 57. 4, 10 59. 3, 5 as B. ED (a 0)2 (0 b)2 EC (a 2a)2 (0 b)2 Page 568 Chapter 10 Practice Quiz 2 a2 b2 a2 b2 1. 67.5 3. 12 5. 115.5 Since ED EC, ED EC. DEC has two congruent Page 569–574 Lesson 10-7 sides, so it is isosceles. 1. Sample answer: The product equation for secant 43. 6 45. 20.5 segments equates the product of exterior segment measure and the whole segment measure for each secant. In the Page 561–568 Lesson 10-6 case of secant-tangent, the product involving the tangent 1. Sample answer: A tangent intersects the circle in only segment becomes (measure of tangent segment)2 because one point and no part of the tangent is in the interior of the the exterior segment and the whole segment are the same circle. A secant intersects the circle in two points and some segment. 3. of its points do lie in the interior of the circle. 138 3. Sample answer: 5. 28.1 7. 7:3.54 9. 4 5. 20 7. 235 9. 55 11. 110 13. 60 15. 110 17. 90 B 11. 2 13. 6 15. 3.2 19. 21. 23. 25. 27. 29. 31. 50 30 8 4 25 130 10 A 17. 4 19. 5.6 33. 141 35. 44 37. 118 39. about 103 ft 41. 4.6 cm 43a. Given: AB is a tangent to O. AC is a secant to C O. CAB is acute. 1 D D Prove: mCAB = mCA 2 21. Given: WY and ZX intersect at T. W O C Prove: WT TY ZT TX X

A B T Y Z Proof: DAB is a right with measure 90, and DCA Proof: is a semicircle with measure 180, since if a line is Statements Reasons tangent to a , it is to the radius at the point of a. W Z, X Y a. Inscribed angles that tangency. Since CAB is acute, C is in the interior of intercept the same arc DAB, so by the Angle and Arc Addition Postulates, are congruent. mDAB mDAC mCAB and mDCA mDC b. WXT ZYT b. AA Similarity WT TX c. c. Definition of similar mCA. By substitution, 90 m DAC m CAB and ZT TY 1 1 180 mDC mCA. So, 90 mDC mCA by triangles 2 2 1 d. WT TY ZT TX d. Cross products Division Prop., and mDAC mCAB mDC 2 1 1 23. 4 25. 11 27. 14.3 29. 113.3 cm mCA by substitution. mDAC mDC since 2 2 1 31. Given: tangent RS and secant US DAC is inscribed, so substitution yields mDC 2 2 Prove: (RS) US TS 1 1 R mCAB mDC mCA. By Subtraction Prop., 2 2 1 S mCAB mCA. 2 43b. Given: AB is a tangent to O. AC is a secant to O. T CAB is obtuse. 1 U Prove: mCAB mCDA D 2 Proof: C O Statements Reasons 1. tangent RS and 1. Given secant US E AB Selected Answers 1 2. mRUT mRT 2. The measure of an Proof: CAB and CAE form a linear pair, so 2 mCAB mCAE 180. Since CAB is obtuse, inscribed angle equals half the measure of its CAE is acute and Case 1 applies, so mCAE intercepted arc. 1 1 1 mCA. mCA mCDA 360, so mCA mCDA 1 2 2 2 3. mSRT mRT 3. The measure of an 1 2 180 by Divison Prop., and mCAE mCDA 180 by angle formed by a 2 substitution. By the Transitive Prop., mCAB secant and a tangent 1 equals half the measure mCAE mCAE mCDA, so by Subtraction 2 of its intercepted arc. 1 Prop., mCAB mCDA. 4. mRUT mSRT 4. Substitution 2

Selected Answers R69 Selected Answers R70 13. 19. 29. 27. 25. 13. 1. P 21. 17. 1. P • • Answersshouldincludethefollowing. other chord. intersecting chord equalstheproduct oftheparts 33. 9. 35. 45. gs5156Catr1 StudyGuide andReview Chapter10 ages 581–586 Lesson10-8 ages 575–580 a Sample answer: x AF A 8 Selected Answers 2 9. 7. 8. 6. 5. 10.82 yd;21.65 yd 117 x ( ( Sample answer:Theproduct ofthepartsone C equilateral, acuteorequiangular F x x 2 , F 3. ( U RS RS FD y 37. 4 D ( S h SUR S RUT y y 2 2) 3) y 21. , ) 2 1, 2 2 2 E 157.5 O O 8 4 4 8 5. F 1600 TS y R 30 , EF 2 O 36 S b US F ( ( ) S y y B 23. 4 7. FB SRT SRT 15. 39. TS 11. 1) 10) d 30 2 x 7 2 8 ( 2 15. x 9. x x x 10 25. 41. 144 c ( 21.96 ft;43.93ft y 2) 5. 3. 7. 150 36 11. 23. 2 31. 35. 39. students’ work 43c. or ( 49. 55. y ( ( 5) 19. x x 6. 8. 7. 9. 5. 7.5 in.;47.12in. 8 ( ( 2 43. x ( x (2, 18 (3, 2),( y 27. x O 2 (0, 0) x 3, 0) Reflexive Prop. Definition of ASimilarity AA Cross products Definition of 2) 3) 2 scalene, obtuse 100 47. 2 2 7) 4 51. 8) 2 4); 3) y y 5 2 2 2 2 y 2 2 O r 43b. ( ( 8 4 4 8 59 45. 4, y y ( 49 ( 25 13 29. y y 3, 0 17. 8 2 6 B 53. 1), (0, 11) 5) none 43a. ) 4 10 60 37. 2 8) 41. 47. 9 2 20 2 13 (0, 3) 100 31. See 8 33. 24 s 32 x x 25 4) 10 2 y area is20m 31. 40 units, 61. 170 units altitude. the lengthofonesideismultipliedby base andtheheight.Forbothquadrilaterals,measure of the width.Thearea ofaparallelogramistheproduct ofthe 1. P 15. 1. P AreasofPolygons Chapter 11 perimeter oftheotheris8 one triangleis12andthe units. Theperimeterof right trianglesis6square the area foreachofthese 47. 31. 23. 15. 7. 1. P 27. 23. 13. 39. ( 41. 49. 33. 41. perimeter G factor. 55. 5 57. 55. 51. 2 1 y gs5860Lesson11-1 ages 598–600 GettingStarted Chapter11 age 593 gs6569Lesson11-2 ages 605–609 , ratioofareas The area ofarectangle istheproduct ofthelengthand 10 21 units Sample answer: ( O area False; sampleanswer: 55 units 95 km 150 units parallelogram, 64units 44 m;103.9m 13,326 ft 9 area 45 4: 16 15 8) 2, 2 13.9 ft 26.8 ft 1; Theratiooftheareas isthesquare ofthescale 3 2 2 3. 43. , 2 35. 53. 57. 2 b 9 1 2), 4.6 2 2 3. 9 18 12, area 2 1 , 2 2 48 20 2 9. 45 ft H r ab . 2 4 17. 28 ft;39.0ft 50 units 9. 53. 33. 33. 5. 80 in.;259.8in 25. 41. ( sin 45. 35. 29. 4 units 2 37. 3 2 13 1200 ft 18 h ai fteaesi 9. : ; Theratioofthe areas is5 2 ,2;9°counterclockwise 2, 2);90° ( The perimeteris19m,halfof38m. 22.5 units x C and Circles 120 in ; scalefactorandratioofperimeters 37 5 in.,7in. Y 2 1 43. 32 15. 22.6 m es; thedimensionsare 32in.by18 7. 3; perimeter 2 63. 2 1) 2 21. 47. 32 54 x 45.7 mm 2 39. 2 2 51. same area. diagonal lengthsandhavethe have different corresponding 3. 11. 6.02 cm 19. 2 y parallelogram, 56units 17.1 5. 43. Sometimes; tworhombi can 2 45. m 9. ( 2 1 2 3 y 45 m 35. 25. 50 m 12.8 m;10.2m 13 27. 37. 40 11. 21 53. 1 O 2 20 cm 49. square, 13 units 4) 2 10.8 in or about14.3. 20 units 4 A 2 C 11. 2 21.6 cm;29.2cm 5 17. The ratioisthesame. ( 47. 13. m 65. 8 0, 6 7.2 21. 5. 39. 9 4 108.5 m ) F 12.4 cm 3 374 cm 2 13 499.5 in 37. 129.9 mm 51. (5, 2), ( 2 13. 2 , C B 45. 2 30, ( about 8.7ft 4, 0), ( 6, 6 29. 6, 0 ( 6 49. x 21 ft 7. 2 2 m 2 r ) ) 59. 2 16 units 19. r 2 13 ft 3 6 ectangle, 4) 2 40 2 2 2 7 B 2 x h 60 2 1 2 121 31b. 67. 231 ft2 69. (x 4)2 y 71. 275 in. has a different central angle. No; there is not an 2 4 equal chance of landing on each color. 33. C 35. 1050 73. 〈172.4, 220.6〉 75. 20.1 units2 37. 110.9 ft2 39. 221.7 in2 41. 123 43. 165 45. g 21.5 Page 609 Practice Quiz 1 1. square 3. 54 units2 5. 42 yd Pages 628–630 Chapter 11 Study Guide and Review 1. c 3. a 5. b 7. 78 ft, 318.7 ft2 9. square; 49 units2 Pages 613–616 Lesson 11-3 11. parallelogram; 20 units2 13. 28 in. 15. 688.2 in2 1. Sample answer: Separate a hexagon inscribed in a circle 17. 31.1 units2 19. 0.3 into six congruent nonoverlapping isosceles triangles. The area of one triangle is one-half the product of one side of the hexagon and the apothem of the hexagon. The area of Chapter 12 Surface Area 1 the hexagon is 6sa. The perimeter of the hexagon is 6s, so 2 Page 635 Chapter 12 Getting Started 1 2 2 the formula is Pa. 3. 127.3 yd2 5. 10.6 cm2 7. about 1. true 3. cannot be determined 5. 384 ft 7. 1.8 m 2 9. 7.1 yd2 3.6 yd2 9. 882 m2 11. 1995.3 in2 13. 482.8 km2 15. 30.4 units2 17. 26.6 units2 19. 4.1 units2 21. 271.2 Pages 639–642 Lesson 12-1 units2 23. 2:1 25. One 16-inch pizza; the area of the 1. The Platonic solids are the five regular polyhedra. All of 16-inch pizza is greater than the area of two 8-inch pizzas, the faces are congruent, regular polygons. In other so you get more pizza for the same price. 27. 83.1 units2 polyhedra, the bases are congruent parallel polygons, but 29. 48.2 units2 31. 227.0 units2 33. 664.8 units2 the faces are not necessarily congruent. 35. triangles; 629 tiles 37. 380.1 in2 39. 34.6 units2 3. Sample answer: 41. 157.1 units2 43. 471.2 units2 45. 54,677.8 ft2; 899.8 ft 47. 225 706.9 ft2 49. 2:3 51. The ratio is the same. 53. The ratio of the areas is the square of the scale factor. 55. 3 to 4 57. B 59. 260 cm2 61. 2829.0 yd2 63. square; 36 units2 65. rectangle; 30 units2 67. 42 5. Hexagonal pyramid; base: ABCDEF; faces: ABCDEF, 69. 6 71. 42 AGF, FGE, EGD, DGC, CGB, BGA; edges: AF, FE, ED, DC, CB, BA, AG, FG, EG, DG, CG, and BG; vertices: Pages 619–621 Lesson 11-4 A, B, C, D, E, F, and G 7. cylinder; bases: circles P and Q 1. Sample answer: 18.3 units2 3. 53.4 units2 5. 24 units2 2 2 y 7. 1247.4 in 9. 70.9 units 9. 11. 11. 4185 units2 13. 154.1 units2 15. 2236.9 in2 back view 17. 23.1 units2 19. 21 units2 back view 21. 33 units2 23. Sample corner view answer: 57,500 mi2 25. 462 27. Sample answer: Reduce corner view O x the width of each rectangle. 13.

29. Sample answer: Windsurfers use the area of the sail to top view left view front view right view catch the wind and stay afloat on the water. Answers 15. should include the following. •To find the area of the sail, separate it into shapes. Then find the area of each shape. The sum of areas is the area top view left view front view right view of the sail. • Sample answer: Surfboards and sailboards are also 17. rectangular pyramid; base: DEFG; faces: DEFG, irregular figures. DHG, GHF, FHE, DHE; edges: DG, GF, FE, ED, 31. C 33. 154.2 units2 35. 156.3 ft2 37. 384.0 m2 DH, EH, FH, and GH; vertices: D, E, F, G, and H 39. 0.63 41. 0.19 19. cylinder: bases: circles S and T 21. cone; base: circle B; Page 621 Practice Quiz 2 vertex A 23. No, not enough information is provided by 1. 679.0 mm2 3. 1208.1 units2 5. 44.5 units2 the top and front views to determine the shape. Selected Answers Pages 625–627 Lesson 11-5 25. parabola 27. circle 29. rectangle 1. Multiply the measure of the central angle of the sector by 31. intersecting three faces 33. intersecting all four faces, the area of the circle and then divide the product by 360°. and parallel to base; not parallel to any face; 62 3. Rachel; Taimi did not multiply by the area of the 360 2 circle. 5._ 114.2 units , 0.36 _7. 0.60 9. 0.54 11. 58.9 units2, 0.3 13. 19.6 units2, 0.1 15. 74.6 units2, 0.42 17. 3.3 units2, 0.03 19. 25.8 units2, 0.15 21. 0.68 23. 0.68 25. 0.19 27. 0.29 29. The chances of landing on a black or white sector are the same, so they should have the same point value. 31a. No; each colored sector 35. cylinder 37. rectangles, triangles, quadrilaterals

Selected Answers R71 Selected Answers R72 7. 5. 1. P 55. 39c. 39a. .11. 9. 45. •T •V following. differences. Answers shouldincludethe drawings ofthepyramidsandnotesimilaritiesany are studying.Egyptologistscancompare two-dimensional dimensional drawingstolearnmore aboutthestructure they they are studying.Egyptologistscancompare two- two dimensionaldrawingstolearnmore aboutthestructure polyhedron. pyramid. More informationisneededtoclassifya cube, rectangular prism,hexahedron, orapentagonal polyhedron with6facescouldbea classify apolyhedron. A 41. gs6568Lesson12-2 ages 645–648 64 cm 188 in Sample answer: know theviewsfrom thefront, top,andeachside. two-dimensional drawingsthatshowthree dimensions. Selected Answers 11 11 No; thenumberoffacesisnotenoughinformationto o showthree dimensionsinadrawing,youneedto D iewpoint drawingsandcornerviewsare typesof pentagonal triangular units 47. 2 2 ; ; 2 infinite 4 43. 57. 39b. 77 Sample answer: Archaeologists use 90 ft,433.0ft 39d. 49. cube, rectangular, orhexahedron hexagonal 0.242 6 3. 2 51. 59. 0.611 39e. 300 cm hexagonal 53. 2 61. 21 units 4320 in 2 2 3 15. 13. 19. 17. 21. 121.5 units 56 units 1 16.3 units 2 ; 2 2 ; ; 66 units 2 ; 23. 108.2 units2; or 6 square units. When the dimensions are doubled the surface area is 6(22) or 24 square units. 39. No; 5 and 3 are opposite faces; the sum is 8. 41. C 43. rectangle 45. rectangle 47. 90 49. 120 51. 63 cm2 53. 110 cm2

Pages 651–654 Lesson 12-3 1. In a right prism a lateral edge is also an altitude. In an oblique prism, the lateral edges are not perpendicular to the bases. 3. 840 units2, 960 units2 5. 1140 ft2 7. 128 units2 9. 162 units2 11. 160 units2 (square base), 126 units2 (rectangular base) 13. 16 cm 15. The perimeter of the base must be 24 meters. There are six rectangles with integer values for the dimensions that have a perimeter of 24. The dimensions of the base could be 1 11, 2 10, 3 9, 4 8, 5 7, or 6 6. 17. 114 units2 19. 522 units2 21. 454.0 units2 23. 3 gallons for 2 coats 25. 44,550 ft2 27. The actual amount needed will be higher because the area of the curved architectural element appears to be greater than the area of the doors. 29. base of A base of C; base of A base of B; base of C base of B 31. A:B 1:4, B:C 4:1, A : C 1:1 33. A:B, because the heights of 25. 27. A and B are in the same ratio as perimeters of bases 35. No, the surface area of the finished product will be the sum of the lateral areas of each prism plus the area of the bases of the TV and DVD prisms. It will also include the area of the overhang between each prism, but not the area of the overlapping prisms. 37. 198 cm2 39. B 41. L 1416 cm2, T 2056 cm2 43. See students’ work. 29.N 31. WX Z 45. 108 units2;

U M R R T T R P V N N W YZ XZ

33.

47. 49. 43 1 51. 35 53. 3 72 35. A 6 units2; B 9 9.87 units2; 2 back view 55. 1963.50 in2 57. 21,124.07 mm2 corner view Selected Answers Pages 657–659 Lesson 12-4 C 76 units2; 1. Multiply the circumference of the base by the height and add the area of each base. 3. Jamie; since the cylinder has one base removed, the surface area will be the sum of the lateral area and one base. 5. 1520.5 m2 7. 5 ft 9. 2352.4 m2 11. 517.5 in2 13. 251.3 ft2 15. 30.0 cm2 17. 3 cm 19. 8 m 21. The lateral areas will be in the ratio 3:2:1; 45 in2, 30 in2, 15 in2. 23. The lateral area is tripled. The surface area is increased, but not tripled. 25. 1.25 m 27. Sample 37. The surface area quadruples when the dimensions are answer: Extreme sports participants use a semicylinder for doubled. For example, the surface area of the cube is 6(12) a ramp. Answers should include the following.

Selected Answers R73 Selected Answers R74 • radius ofthebase andtheslantheightof cone. •W following. covered incircular canvas. Answers shouldincludethe Lateral area isusedbecausetheground may notalwaysbe 33. the heightsare inthesameratioasradiiofbases. computation ofthelateralarea. changes thevalueofslantheight,whichaffects thefinal slant heighttoeitherthetenthsplaceorhundredths place accurate techniquetofindthelateralarea. Roundingthe 29. 1. P 21. 15. 1. P 3. 1. P 29. •A •T 285.8 ft each pairofoppositesidesare congruent. 5 37. 35. square inch. corner ofthecubereduces thesurfacearea byabout0.63 cube isapproximately 5.37square inches.Truncating the cube is6square inches.Thesurfacearea ofthetruncated 33. gs6860Lesson12-6 ages 668–670 Lesson12-5 ages 663–665 Practice Quiz1 age 659 Sample answer: Sample answer: cover thesides. To findtheactuallateral area, subtract The opentopreduces the lateralarea ofcanvasneededto equal distancefrom theaxisofcylinder. base andthendivideby2. pipe, multiplytheheightbycircumference ofthe Selected Answers corner view 20 ft Sample answer:Tepees are conical shapedstructures. o findthelateralarea ofasemicylinderlikethehalf- 27.7 ft Using thestore feature onthecalculatorismost C 300 units e needtoknowthecircumference ofthebase orthe half-pipe rampishalfofacylinderifthean square base (regular) 2 2 23. 35. vertex 2 17. 960 ft center ofbase 27. 98 m,366m D 2 231.5 m 2.3 31. rectangular base 29. 25. (not regular) inches oneachside of thefacecube the oppositevertices to thelinecontaining a planeperpendicular 41. 43. 967.6 m The surfacearea ofthe 2 2 11. 7. 23. 15. 19. 27. 3. m 27 54 cm 37. 5. 282.7 cm 848.2 cm 5.4 ft 628.8 m 1613.7 in 7.9 m 475.2 in 8.1 in.;101.7876in 2 A G 39. 31. 2 F 31. 8 Sometimes; onlywhen 64, 39. 1809.6 yd 13. 11. 9. 7. 5. 2 2 3. 2 17. 2 b 2 J 147.7 ft 1 340 cm 74.2 ft 19. M 9. 5. 21. 13. 19 cm 326.9 in 173.2 yd 5.6 ft 43. 25. 12.2, 614.3 in 485.4 in 1509.8 m 679.9 in 41. original 21.3 m 2 2 615,335.3 ft 2 2 2 c 12 ft 2 Tr 2 33. 2 ue; 2 2 15.6 2 74 ft, 2 2 47. to about1256.6mi 15. C P 51. 35. 1. 1. P 1. intersect. that passesthrough the sideofcube. 35. of thestructure. the lateralarea oftheconicalopeningfrom thelateralarea 17. P gs6862Catr1 StudyGuideandReview Chapter12 ages 678–682 Practice Quiz2 age 670 D gs6466Lesson12-7 ages 674–676 ABD G Sample answer: 423.9 cm d ; vertices: 21 340 units 649 cm D 2 13. 3. 133.7 units 2 , 37. :1 49. b triangular prism;base: 2 ACD 43. 2 5.8 ft 8 5. 2 37. A ; 53. 3. A a , , and 11 B 144.9 ft The surfacearea canrangefrom about452.4 2 , ( ; 2 7. x 45. . C 39. 33. 23. 19. 11. 29. 3. , and X e 13 26.5 39. 1430.3 in 41. 6.0 yd 15 will alsointersect 2) 3257.2 m 636,172.5 m BCD 398.2 ft true 9. 2 2 The radiusofthesphere ishalf D 8.5 None; everyline(great circle) c 5. 5. 51. ; edges: ( 18 y 3.9 in. 11. 31. 41. 13. 2 25.1 2 2 7) cylinder; bases: 7. 47. 8 48 2 12 25. 2 BCD 150.8 cm 10 A 206,788,161.4 mi 53. 254.7 cm 15. 21. B true 43. 50 , ; faces: B g 51.5 12.8 397.4 in . All great circles great . All C 24 , 2 A 27. 2 C 9. 55. 45. 17. , true 2 A 49. ABC D 7854.0 in 25.8 45 F 283.5 in , 969 yd and 2 B , D , 2 2 2 19. 228 units2; Pages 704–706 Lesson 13-3 1. The volume of a sphere was generated by adding the volumes of an infinite number of small pyramids. Each pyramid has its base on the surface of the sphere and its height from the base to the center of the sphere. 3. 9202.8 in3 5. 268.1 in3 7. 155.2 m3 9. 1853.3 m3 11. 3261.8 ft3 13. 233.4 in3 15. 68.6 m3 17. 7238.2 in3 19. 21,990,642,871 km3 21. No, the volume of the cone is 41.9 cm3; the volume of the ice cream is about 33.5 cm3. 2 23. 20,579.5 mm3 25. 1162.1 mm2 27. 3 29. 587.7 in3 31. 32.7 m3 33. about 184 mm3 35. See students’ work. 37. A 39. 412.3 m3 41. (x 2)2 (y 1)2 64 43. (x 2)2 (y 1)2 34 8k3 45. 27x3 47. 125 Pages 710–713 Lesson 13-4 4 1. Sample answer: 3. congruent 5. 3 64 7 in. 7. 9. 1:64 5 in. 27 5 in. 7 in. 11. neither 13. congruent 15. neither 21. 72 units2 23. 175.9 in2 25. 1558.2 mm2 27. 304 17. 7 in. 130 m high, 245 m units2 29. 33.3 units2 31. 75.4 yd2 33. 1040.6 ft2 wide, and 465 m long 2 2 2 6 in. 35. 363 mm 37. 2412.7 ft 39. 880 ft 12 in. 8 in. 19. Always; congruent solids have equal Chapter 13 Volume dimensions. 21. Never; different types of solids cannot be similar. Page 687 Chapter 13 Getting Started 23. Sometimes; solids that are not similar can have the 2 2 1. 5 3. 3 5. 305 7. 134.7 cm 9. 867.0 mm 2 2 8 2 same surface area. 25. 1,000,000x cm 27. 29. 2 9x 5 125 11. 25b 13. 2 15. W( 2.5, 1.5) 17. B(19, 21) 29 24,389 16y 31. 18 cm 33. 35. 37. 0.004 in3 39. 3:4; 3:1 30 27,000 Pages 691–694 Lesson 13-1 41. The volume of the cone on the right is equal to the sum of 1. Sample answers: cans, roll of paper towels, and chalk; the volumes of the cones inside the cylinder. Justification: boxes, crystals, and buildings 3. 288 cm3 5. 3180.9 mm3 Call h the height of both solids. The volume of the cone on 3 3 3 3 1 7. 763.4 cm 9. 267.0 cm 11. 750 in 13. 28 ft the right is r2h. If the height of one cone inside the cylinder 3 3 3 15. 15,108.0 mm 17. 14 m 19. 24 units 21. 48.5 3 23. 3 25. 3 27. is c, then the height of the other one is h c. Therefore, the mm 173.6 ft 304.1 cm about 19.2 ft 1 1 3 3 2 sum of the volumes of the two cones is: r2c r2(h c) 29. 104,411.5 mm 31. 137.6 ft 33. A 35. 452.4 ft 3 3 1 1 37. 1017.9 m2 39. 320.4 m2 41. 282.7 in2 43. 0.42 or r2(c h c) or r2h. 43. C 45. 268.1 ft3 45. 2 47. 49. 2 51. 2 3 3 186 m 8.8 21.22 in 61.94 m 47. 14,421.8 cm3 49. 323.3 in3 51. 2741.8 ft3 53. 2.8 yd 55. 36 ft2 57. yes 59. no Pages 698–701 Lesson 13-2 1. Each volume is 8 times as large as the original. Page 713 Practice Quiz 2 3. Sample answer: 3 7 343 1 1. 67,834.4 ft 3. 5. V (32)(16) 5 125 3 48 Pages 717–719 Lesson 13-5 1 V (42)(9) 1. The coordinate plane has 4 regions or quadrants with 4 3 possible combinations of signs for the ordered pairs. Three- 48 16 dimensional space is the intersection of 3 planes that create 9 8 regions with 8 possible combinations of signs for the 3 4 ordered triples. 3. A dilation of a rectangular prism will Selected Answers provide a similar figure, but not a congruent one unless r 1 or r 1. 5. 3 7. 3 9. 3 603.2 mm 975,333.3 ft 1561.2 ft 5. z 11. 8143.0 mm3 13. 2567.8 m3 15. 188.5 cm3 Q(1, 0, 2) 17. 1982.0 mm3 19. 7640.4 cm3 21. 2247.5 km3 ( ) 3 3 3 P 1, 4, 2 23. 158.8 km 25. 91,394,008.3 ft 27. 6,080,266.7 ft R(0, 0, 2) 3 3 3 S(0, 4, 2) 29. 522.3 units 31. 203.6 in 33. B 35. 1008 in U(1, 0, 0) 3 2 (1, 4, 0) 37. 1140 ft 39. 258 yd 41. 145.27 43. 1809.56 V(0, 0, 0) T O ( ) y Page 701 Practice Quiz 1 W 0, 4, 0 x 1. 125.7 in3 3. 935.3 cm3 5. 42.3 in3

Selected Answers R75 Selected Answers R76 ( (12, 8,0),0,(0,and0);( 3 25. 23. 21. 7. ( 17. 15. 13. 11. 36, 0,24),( 36, 0,16),( x Selected Answers BC PQ W H x R 186 W ( ( L 4, 0, ( 3, A ( E ( I 4, 0,0 1, ; 0, ( S 0, V 1, ( 4, 1 z 0, ( 3, 4, 0 3, 3 39 T 115 x ) ) ) ( B ; 0, M 3, 0 2 7 3 ) 48, 0,24),( 48, 0,16),and( S 6 R 4, 0 ; ) , ) ( ( ( G H 0, 0 O 0, 2 1 ) 2 1 3, ) z ( , 1, U 4, 1, 2 D F 4, 1 3 ( 6 9. x 3, 0,1 , 3,3 ) 2 7 ) 3, 0 O , ) (12, 8,8),0,(0, K 3 2 7 ) ) O B ( T ) G J C N 4, 1,0 48, 8,24)( ( ( ( U z 0, 1 3, 0,0 ( 0, 1, 19. 2 0, 1,0 ( V 0, Q z y 48, 8,16) ) ( O GH ( P ) ) 3 ) 1, 0, 1, 0,0 A ( ) 0, 0 6 y Q R S T U V and ( ) 0, 0 (2, 2, (0, 5, (2, 5, (2, 2, y (0, 5, (0, 2, 36, 8,16), 6 ) P ) W ) 17 (0, 2, 36, 8,24), ) (2, 5, y ; 2) 5), 2), 5), 2), 5), 5 3 , 2), 5) 1 7 0 , 4 11. Distance FormulainSpace 31. 19. 1. P that satisfytheequationof •O following. them tomoverealistically. Answers shouldincludethe used incomputeranimationtorender imagesandallow z 39. • xz 43. perpendicular tothe 5, C 27. 2, B 29. E D C F V and G gs7072Catr1 StudyGuideandReview Chapter13 ages 720–722 (6, 0,0), -plane isthegraphof (6, 0,6), (6, 6,0), (1, 2,1), (0, 0,6), (0, 0,0), pyramid an animatortocreate realistic movementinanimation. Applying transformationstopointsinspacewouldallow points inspace. An ordered tripleisuniquetoonepoint. (0, 6,6), 58 504 in 8.2 mi 4637.6 mm B similar A A 5) 2), 2), and r H 216 units ; ( dered triples are amethodoflocatingandnaming (4, 5,1), (6, 6,6), (0, 6,0); F 41. 37. 9, 5.5,5.5) (4, 2, 3 D The locusofpointsinspacewithcoordinates H Sample answer:Three-dimensional graphingis 33. 13. 45. (1, 5,1) 3. 3 (1, 5, 3 B ; (0, an ordered triple 2), 749.5 ft 1 (4, 2,1), 21. 150.3 yd G B 14, 14) 27. E xz x 523.6 units 2); (1, F (4, -plane whoseintersectionwiththe 3 z FG 3 15. x B 9. 47. x 35. x 1466.4 ft B G Cavalieri’s Principle z F 3 z 12,770.1 ft ( 422 C x 4 inthe G 5. , C 23. B 4 isaplane F y F ; O , z similar 1.5 z G C 3 similar ) A O E → E A 17. xz 3 A ( G C 2 E x -plane. , 3 D 33.5 ft E 7. H A 25. 2, the D 7 H y , 3 CD D 3 H 3, H D y y Photo Credits

About the Cover: This photo of the financial district of Hong Kong illustrates a variety of

geometrical shapes. The building on the right is called Jardine House. Because the circular Photo Credits windows resemble holes in the rectangular blocks, this building was given the nickname “House of a Thousand Orifices.” The other building is one of the three towers that comprise the Exchange Square complex, home to the Hong Kong Stock Exchange. These towers appear to be a combination of large rectangular prisms and cylinders.

Cover Wilhelm Scholz/Photonica; vii Jason Hawkes/ Photo Researchers; 209 (l)Dennis MacDonald/PhotoEdit, CORBIS; viii Galen Rowell/CORBIS; ix Lonnie (r)Michael Newman/PhotoEdit; 212 Courtesy Peter Lynn Duka/Index Stock Imagery/PictureQuest; x Elaine Kites; 216 Marvin T. Jones; 220 Dallas & John Heaton/ Thompson/AP/Wide World Photos; xi Jeremy Stock Boston; 223 Francois Gohier/Photo Researchers; Walker/Getty Images; xii Lawrence Migdale/Stock 224 John Elk III/Stock Boston; 225 Christopher Morrow/ Boston; xiii Alexandra Michaels/Getty Images; xiv Izzet Stock Boston; 234–235 Mike Powell/Getty Images; Keribar/Lonely Planet Images; xv Phillip Wallick/ 238 Michael S. Yamashita/CORBIS; 244 Getty Images; CORBIS; xvi Aaron Haupt; xvii Paul Barron/CORBIS; 250 Tony Freeman/PhotoEdit; 253 Jeff Greenberg/ xviii First Image; xix CORBIS; xx Brandon D. Cole; PhotoEdit; 255 Joshua Ets-Hokin/PhotoDisc; 256 James 2 Wayne R. Bilenduke/Getty Images; 2–3 Grant V. Marshall/CORBIS; 265 British Museum, London/Art Faint/Getty Images; 4–5 Roy Morsch/CORBIS; Resource, NY; 267 Jeremy Walker/Getty Images; 6 C Squared Studios/PhotoDisc; 9 Ad Image; 10 (l)Daniel 270 Bob Daemmrich/The Image Works; 271 C Squared Aubry/CORBIS, (cl)Aaron Haupt, (cr)Donovan Reese/ Studios/PhotoDisc; 272 Rachel Epstein/PhotoEdit; PhotoDisc, (r)Laura Sifferlin; 16 (t)Rich Brommer, 280–281 David Weintraub/Stock Boston; 282 Christie’s (b)C.W. McKeen/Syracuse Newspapers/The Image Images; 285 Courtesy University of Louisville; 286 Walt Works; 17 (l)PhotoLink/PhotoDisc, (r)Amanita Pictures; Disney Co.; 289 Art Resource, NY; 294 Joe Giblin/ 18 (l)Getty Images, (r)courtesy Kroy Building Products, Columbus Crew/MLS; 298 Jeremy Walker/Getty Images; Inc.; 32 Red Habegger/Grant Heilman Photography; 304 Macduff Everton/CORBIS; 305 Lawrence Migdale/ 35 (l)Erich Schrempp/Photo Researchers, (r)Aaron Stock Boston; 310 JPL/NIMA/NASA; 316 (l)Kelly- Haupt; 37 Jason Hawkes/CORBIS; 41 Reuters New Mooney Photography/CORBIS, (r)Pierre Burnaugh/ Media/CORBIS; 45 Copyright K’NEX Industries, Inc. PhotoEdit; 318 Beth A. Keiser/AP/Wide World Photos; Used with permission.; 49 Getty Images; 60–61 B. Busco/ 325 (t)C Squared Studios/PhotoDisc, (bl)CNRI/ Getty Images; 62 Bob Daemmrich/Stock Boston; PhotoTake, (br)CORBIS; 329 Reunion des Musees 65 Mary Kate Denny/PhotoEdit; 73 Bill Bachmann/ Nationaux/Art Resource, NY; 330 (t)Courtesy Jean-Paul PhotoEdit; 79 Galen Rowell/CORBIS; 86 AP/Wide Agosti, (bl)Stephen Johnson/Getty Images, (bcl)Gregory World Photos; 89 Jeff Hunter/Getty Images; 92 Spencer Sams/Science Photo Library/Photo Researchers, Grant/PhotoEdit; 94 Bob Daemmrich/The Image Works; (bcr)CORBIS, (br)Gail Meese; 340–341 Bob Daemmrich/ 96 Aaron Haupt; 98 Duomo/CORBIS; 105 (t)David The Image Works; 342 Robert Brenner/PhotoEdit; Madison/Getty Images, (b)Dan Sears; 107 (t)C Squared 350 Alexandra Michaels/Getty Images; 351 StockTrek/ Studios/PhotoDisc, (b)file photo; 113 (l)Richard Pasley/ PhotoDisc; 354 Aaron Haupt; 355 Phil Mislinski/ Stock Boston, (r)Sam Abell/National Geographic Image Getty Images; 361 John Gollings, courtesy Federation Collection; 124–125 Richard Cummins/CORBIS; Square; 364 Arthur Thevenart/CORBIS; 368 David R. 126 Robert Holmes/CORBIS; 129 Angelo Hornak/ Frazier/Photo Researchers; 369 StockTrek/CORBIS; CORBIS; 133 Carey Kingsbury/Art Avalon; 137 Keith 374 R. Krubner/H. Armstrong Roberts; 375 John Wood/CORBIS; 151 David Sailors/CORBIS; 156 Brown Mead/Science Photo Library/Photo Researchers; Brothers; 159 Aaron Haupt; 163 (l)Lonnie Duka/Index 377 Roger Ressmeyer/CORBIS; 382 Rex USA Ltd.; Stock Imagery/PictureQuest, (r)Steve Chenn/CORBIS; 385 Phil Martin/PhotoEdit; 389 Pierre Burnaugh/ 174 A. Ramey/Woodfin Camp & Associates; PhotoEdit; 400 Matt Meadows; 400–401 James Westwater; 174–175 Dennis MacDonald/PhotoEdit; 176–177 Daniel 402–403 Michael Newman/PhotoEdit; 404 Glencoe J. Cox/Getty Images; 178 (t)Martin Jones/CORBIS, photo; 408 (l)Monticello/Thomas Jefferson Foundation, (b)David Scott/Index Stock; 181 Joseph Sohm/Stock Inc., (r)SpaceImaging.com/Getty Images; 415 (l)Pictures Boston; 185 Courtesy The Drachen Foundation; Unlimited, (r)Museum of Modern Art/Licensed by 188 Adam Pretty/Getty Images; 189 Doug Pensinger/ SCALA/Art Resource, NY; 417 Neil Rabinowitz/ Getty Images; 190 Jed Jacobsohn/Getty Images; CORBIS; 418 Richard Schulman/CORBIS; 418 Museum 192 Aaron Haupt; 193 Private Collection/Bridgeman Art of Modern Art/Licensed by SCALA/Art Resource, NY; Library; 196 North Carolina Museum of Art, Raleigh. 422 (l)Aaron Haupt, (r)AFP/CORBIS; 424 Simon Bruty/ Gift of Mr. & Mrs. Gordon Hanes; 200 Paul Conklin/ Getty Images; 426 Emma Lee/Life File/PhotoDisc; PhotoEdit; 201 Jeffrey Rich/Pictor International/ 428 Zenith Electronics Corp./AP/Wide World Photos; PictureQuest; 204 Elaine Thompson/AP/Wide World 429 Izzet Keribar/Lonely Planet Images; 431 Courtesy Photos; 205 (tl)G.K. & Vikki Hart/PhotoDisc, (tr)Chase Professor Stan Wagon/Photo by Deanna Haunsperger; Swift/CORBIS, (b)Index Stock; 207 Sylvain Grandadam/ 435 (l)Metropolitan Museum of Art. Purchase, Lila

Photo Credits R77 Photo Credits R78 Getty Images; 573 566 557 Aurness/CORBIS; Haupt; (b)Matt Meadows; Library; (r)Bridgeman Art Hermitage Museum,St.Petersburg, Russia/CORBIS, 595 Stock Boston; National World War IIMemorial; 550 544 Getty Images,(r)Aaron Haupt; The HouseonRock,SpringGreen WI; 520–521 Motion Picture &Television Photo Archive; Researchers; CORBIS; (tr)Aaron Haupt,(b)DigitalVision; Art Resource, NY; 486 Cordon Art, Baarn,Holland. All rightsreserved; Haupt; Stock Boston,(bc)SpencerGrant/PhotoEdit,(br)Aaron Electronics/AP/Wide World Photos,(bl)JimCorwin/ CORBIS, (r)MassimoListri/CORBIS; (c)Nick Carter/ElizabethWhiting& Associates/ 478 Amos/CORBIS; 469 Mafford/PhotoDisc, (r)Lynn Stone; Glusic/PhotoDisc; Images; 442 Newman Foundation/ArtistsRightsSociety, New York; Art. Purchased through agiftofPhyllisWattis/©Barnett Camp & Associates, (r)SanFranciscoMuseumofModern Bachmann/PhotoEdit; Dorothea Rockburneand Artists RightsSociety; Acheson Wallace Gift,1993(1993.303a–f),(r)courtesy Photo Credits Smithsonian American Art Museum,Washington DC/ David Young-Wolff/PhotoEdit; file photo; Ray Massey/GettyImages; Michael S. Yamashita/CORBIS; Courtesy JudyMathieson; Phillip Hayson/PhotoResearchers; Ti m Hall/PhotoDisc; 529 483 Aaron Haupt; 460–461 Michael Dunning/GettyImages; 501 Carl Purcell/Photo Researchers; Symmetry DrawingE103 506 CORBIS; 592–593 578 569 476 W Rob McEwan/TriStar/Columbia/ NOAA; 536 487 Matt Meadows; 467 illiam A. Bake/CORBIS; illiam A. 607 Sellner ManufacturingCompany; KS Studios; 552 Ken Fisher/GettyImages; (tl)Sue Klemens/StockBoston, 504 (l)Siede Pries/PhotoDisc,(c)Spike Chuck Savage/CORBIS; 440 451 Andy Lyons/Getty Images; 601 Georg Gerster/Photo 579 (l)Bernard Gotfryd/Woodfin Paul Trummer/Getty (t)Paul Baron/CORBIS, 479 NASA; 543 558 590–591 541 (l)Matt Meadows, 572 495 Profolio/Index Stock; . M.C.Escher. ©2002 Aaron Haupt; 468 575 599 480 (l)Hulton Archive/ 590 Doug Martin; Phillip Wallick/ Hulton Archive; 470 Pete Turner/ (l)State (t)Sony 522 Rob Crandall/ Courtesy 524 463 James L. 534 Courtesy Aaron Robert Craig 439 Bill Researchers. W 712 (b)Aaron Haupt; CORBIS, (c)courtesyM-KDistributors, Bruce Coleman,Inc.; Martin; T 782 Dagli Orti/CORBIS; (b)CORBIS; Alan Wilton/Getty Images; Researchers; CORBIS, (c)Zefa/IndexStock,(r)V. Fleming/Photo 638 Foundation. PhotobyJamesIsberner; Museum ofContemporary Art, Chicago,giftofLannan Images; S. Beyer/GettyImages; 690 Reprinted withpermission.,(b)MattMeadows; 688 (r)StockTrek/PhotoDisc; PhotoDisc; 671 T CORBIS; Callister/Online USA/GettyImages,(b)MassimoListri/ 664 658 Meadows; Liu/CORBIS; Ressmeyer/CORBIS; 699 (r)CORBIS; Images; 610 PhotoDisc; W CORBIS; ourism MedicineHat.PhotobyRoyceHopkins; imePix; orld Photos; oodfin Camp& Associates; (t)Walter Bibikow/StockBoston, Aaron Haupt; (l)Doug Pensinger/GettyImages,(r)AP/Wide Scala/Art Resource, NY; (t)Tribune MediaServices,Inc. All RightsReserved. StudiOhio; (tl)Elaine Rebman/PhotoResearchers, (tr)Dan Michael Newman/PhotoEdit; Courtesy American HeritageCenter; R. Gilbert/H. Armstrong Roberts; 789 636 615 617 784 666 709 675 626 696 John D.Norman/CORBIS; 649 (t)Steven Studd/GettyImages,(b)Collection Sakamoto PhotoResearch Laboratory/ 643 Peter Stirling/CORBIS; (l)Carl & Ann Purcell/CORBIS, (r)Doug EyeWire; 706 714 Aaron Haupt; (l)David Rosenberg/Getty Images, Stu Forster/GettyImages; 672 (t)Lightwave Photo,(b)MattMeadows; Lon C.Diehl/PhotoEdit; 794 (t)Image Port/IndexStock,(b)Chris Brian Lawrence/SuperStock; 693 Rein/CORBIS SYGMA; Aaron Haupt; 727 F. 793 702

(l)Peter Vadnai/CORBIS, Stuart Westmorland/ 668 655 Grant V. Faint/GettyImages; 686–687 (t)Yann Arthus-Bertrand/ Dominic Oldershaw; CORBIS; Paul A. Souders/CORBIS; Paul A. 647 641 711 622 (t)Doug Martin, (l)Charles O’Rear/ Courtesy DensoCorp.; Ron Watts/CORBIS; C Squared Studios/ 673 660 (b)Serge Attal/ 669 620 Don Tremain/ 637 First Image; 613 790 Conrad MT, Mark S.Wexler/ Courtesy 700 652 634–635 Aaron Haupt; Christie’s 717 Stella Snead/ Photo Roger G. Ryan & 705 Gianni 707 Getty Y Matt ang Index

A Angle of depression, 372 alphabet, 480 amusement parks, 374 AA. See Angle-Angle Angle of elevation, 371 amusement rides, 476, 480, 568 AAS. See Angle-Angle-Side Angle of rotation, 476 animation, 471, 472 aquariums, 407, 693 Absolute error, 19 Angle relationships, 120 archeology, 636 Acute angles, 30 Angles architecture, 178, 181, 209, 294, acute, 30 315, 344, 383, 408, 451, 615, 660, Acute triangle, 178 adjacent, 37 670, 693, 700, 706, 711 Adjacent angles, 37 alternate exterior, 128 arrowheads, 223 alternate interior, 128 art, 330, 435, 440, 483, 599, 654 Algebra, 7, 10, 11, 19, 23, 27, 32, 33, base, 216 artisans, 220 34, 39, 40, 42, 48, 49, 50, 66, 74, bisector, 32, 239 astronomy, 181, 260, 369, 557, Index 80, 87, 93, 95, 97, 99, 112, 114, 138, central, 529 675, 705 144, 149, 157, 163, 164, 180, 181, classifying triangles by, 178 aviation, 130, 284, 369, 371, 373, 183, 191, 197, 198, 213, 219, 220, complementary, 39, 107 374, 381, 382, 501, 504, 614, 717 221, 226, 242, 243, 244, 245, 253, congruent, 31, 108 banking, 327, 328 254, 265, 266, 273, 287, 297, 301, consecutive interior, 128 baseball, 205, 251, 285, 433, 673, 302, 305, 313, 322, 331, 348, 370, corresponding, 128 701 376, 382, 390, 405, 407, 408, 415, degree measure, 30 basketball, 674, 712 416, 423, 430, 432, 434, 437, 442, of depression, 372 beach umbrellas, 601 443, 445, 451, 469, 475, 481, 487, of elevation, 371 bees, 487 488, 496, 505, 511, 528, 532, 533, exterior, 128, 186, 187 bicycling, 258 535, 549, 553, 554, 558, 567, 574, exterior of an angle, 29 bikes, 600 580, 600, 605, 606, 614, 616, 621, of incidence, 35 billiards, 468 627, 648, 653, 658, 670, 676, 694, included, 201 biology, 93, 272, 347 701, 706, 713, 719 interior, 128, 186, 187 birdhouses, 661 and angle measures, 135 interior of an angle, 29 birdwatching, 374 definition of, 94 linear pair, 37 boating, 374, 503 indirect proof with, 255 obtuse, 30 brickwork, 487 Algebraic proof. See Proof properties of congruence, 108 broccoli, 325 relationships of sides to, 248, 250 buildings, 389 Alternate exterior angles, 128 remote interior, 186, 187 design of, 385 right, 30, 108, 111 Alternate Exterior Angles business, 149 of rotation, 476 Theorem, 134 cakes, 614 sides, 29 calculus, 620 Alternate interior angles, 128 straight, 29 camping, 658 supplementary, 39, 107 careers, 92 Alternate Interior Angles Theorem, of triangles, 185–188 134 car manufacturing, 643 vertex, 29, 216 carousels, 522 Altitude, 241 vertical, 37 carpentry, 137, 157, 541, 639 of cone, 666 congruent, 110 cartoons, 688 of cylinder, 655 Angle-Side-Angle Congruence cats, 205 of parallelogram, 595 (ASA) Postulate, 207 cell phones, 577 of trapezoid, 602 cost of, 147 of triangle, 241, 242 Angle Sum Theorem, 185 chemistry, 65, 83, 405 chess, 473 Ambiguous case, 384 Apothem, 610 circus, 564 Angle Addition Postulate, 107 Applications. See also Cross- civil engineering, 374, 397 Curriculum Connections; More clocks, 529, 534 Angle-Angle-Side (AAS) About commercial aviation, 498 Congruence Theorem, 188, 208 advertising, 87, 273 computer animation, 714 Angle-Angle (AA) Similarity aerodynamics, 579 computers, 143, 354, 490, 508, Postulate, 298 aerospace, 375 542, 714 agriculture, 72, 555, 658 construction, 100, 137, 163, 296, Angle bisectors, 32, 239 airports, 127 314, 347, 497, 511, 573, 663 algebra, 112 Angle Bisector Theorem, 319 contests, 657

Index R79 Index R80 houses, 64 home improvement, 156 hockey, 285 history, 107,265,303,526,572, historic landmarks,48 hiking, 225 highways, 113 health, 270 grills, 674 golf, 374,466 golden rectangles, 429 geysers, 374 geology, 647 geography, 27,104,355,620 gemology, 641 geese, 205 gazebos, 407,610 gates, 619 gardens, 607,615 gardening, 47,99,195,212,381, games, 718 furniture, 618,651 framing, 428 forestry, 78,305,511 football, 156 food, 531,646,665,705 folklore, 331 flags, 436 fireworks, 527 festivals, 712 fencing, 74 fans, 479 family, 705 extreme sports,655 formula,640 Euler’s entertainment, 286,709 enlargement, 27 engineering, 693 elevators, 715 education, 259,285 Earth, 675 drawing, 415 doors, 18,253,272 dog tracking,34 digital photography, 496 digital camera,691 diamonds, 469 desktop publishing,496 design, 104,220,415,435,443,598 delicatessen, 639 darts, 622,627 dancing, 91 dances, 346 currency, 285 crystal, 182 crosswalks, 599 crafts, 18,265,388,580,609 Index 664, 700 designing, 595 407, 652,683 design, 193 physical fitness, 320 photography, 114, 286,294,318, photocopying, 294,495 perimeter, 18,27,435 perfume bottles,663 pattern blocks,35 patios, 428 party hats,669 parachutes, 210,626 paleontology, 510 painting, 355,651,652 origami, 33 orienteering, 244 online music,534 one-point perspective,10 nutrition, 697 nets, 50 navigation, 225,355,654,713 nature, 330,467 music, 18,65,73,480 movies, 506 mountain biking,142 mosaics, 196,473 monuments, 129 models, 92,495 mirrors, 382 miniatures, 707,711 miniature gold,429 meteorology, 375,422 medicine, 375 mechanical engineering,699 masonry, 649 marine biology, 201 marching bands,470 maps, 9,149,182,287,292,293, manufacturing, 655,693 logos, 558 literature, 286 lighting, 104 lawn ornaments,710 law, 259 latitude andlongitude,351 language, 34,42 landscaping, 272,323,355,475, landmarks, 566 lamps, 667 knobs, 572 kitchen, 16,212 jobs, 149 irrigation, 534 Internet, 148 interior design,163,245,487,596, insurance, 84 indirect measurement, 301,574 igloos, 675 ice cream, 286 housing, 190,694 319, 321,442,557 306, 310,312 509 599, 605 surveying, 304, 364,366,368,381, students, 474 structures, 130 storage, 422,717 steeplechase, 225 statistics, 245,296,653 states, 450 stained glass,43,551 stadiums, 664 squash, 435 spreadsheets, 27 spotlights, 669 sports, 86,294,705,723 speed skating,190 speakers, 640 space travel,579 softball, 360 sockets, 544 soccer, 347,389,535 snow, 689 sledding, 374 ski jumping,188,189 skiing, 41,374 shopping, 256 shipping, 504 shadows, 373 sewing, 443,612 seasons, 79 sculpture, 285 school rings,550 school, 73,76,285 scallop shells,404 scale drawings,493 sayings, 541 satellites, 563,567 sailing, 355 safety, 369 r rivers, 113, 504 ripples, 575 r r r r r r ramps, 568 rainbows, 561 railroads, 374 radio astronomy, 377 quilts, 478 quilting, 181,195,486,604 programming, 356 probability, 204,265,537,549, precision flight,204 population, 143 physics, 98,154,614 oller coasters,306 emodeling, 434,488 efrigerators, 706 eflections innature, 463 ecycling, 70 ecreation, 18,140,479,607,676, eal estate,314,382,387,607 390, 482 equipment, 671 718 550, 648,700,705 surveys, 533, 626, 642 183, 191, 198, 206, 213, 235, 260, Test-Taking Tips, 23, 59, 96, 123, suspension bridges, 350 266, 281, 287, 297, 306, 315, 323, 135, 173, 217, 233, 262, 279, 283, swimming, 693 341, 348, 356, 370, 376, 383, 403, 339, 413, 459, 493, 519, 525, 589, swimming pools, 614 409, 416, 423, 430, 437, 445, 461, 633, 644, 685, 725 symmetry and solids, 642 469, 475, 482, 488, 497, 505, 521, tangrams, 421 528, 535, 543, 551, 558, 568, 574, Astronomical units, 369 television, 370, 428 593, 600, 609, 616, 635, 642, 648, Axis tennis, 626, 705 654, 665, 670, 687, 694, 701, 706, of circular cone, 666 tepees, 224, 666, 669 713 of cylinder, 655 textile arts, 430 Standardized Test Practice, 11, 19, in three-dimensions, 716 tools, 271 23, 25, 27, 35, 43, 50, 58, 59, 66, tourism, 652, 712 74, 80, 86, 93, 96, 97, 99, 106, towers, 668 114, 122, 123, 131, 135, 136, 144, Axiom, 89 track and field, 552 149, 157, 164, 171, 172, 183, 191, Transamerica Pyramid, 696 206, 213, 217, 219, 221, 232, 233, B travel, 253, 259, 304, 375, 397 245, 264, 265, 273, 278, 279, 282, Base Index treehouses, 250 285, 287, 297, 305, 314, 322, 331, cone, 666 tunnels, 570 338, 339, 348, 356, 362, 370, 372, cylinder, 655 two-point perspective, 11 373, 376, 382, 390, 397, 398–399, parallelogram, 595 umbrellas, 196 409, 413, 414, 416, 423, 430, 437, prism, 637 upholstery, 613 445, 451, 458, 459, 469, 475, 481, pyramid, 660 utilities, 162 493, 494, 496, 505, 511, 517, 518, trapezoid, 602 visualization, 9 519, 525, 535, 543, 551, 558, 567, triangle, 602 volcanoes, 700 574, 580, 588, 589, 600, 608, 616, volume of Earth, 702 621, 622, 625, 627, 632, 633, 642, Base angles, 216 waffles, 536 646, 648, 653, 658, 664, 670, 676, Between, 14 weather, 330, 578, 579 683, 684, 685, 694, 701, 703, 706, weaving, 565 713, 719, 724, 725 Betweenness, 14 windows, 436 Extended Response, 59, 123, 173, Biconditional statement, 81 windsurfing, 617 233, 279, 339, 399, 459, 519, 589, winter storms, 669 633, 685, 725, 806–810 C Multiple Choice, 11, 19, 23, 25, Arcs, 530 27, 35, 43, 50, 58, 66, 74, 80, 86, Career Choices chords and, 536 87, 93, 96, 99, 106, 114, 121, 122, agricultural engineer, 658 probability and, 622 131, 138, 144, 149, 157, 164, 172, architect, 209 Area. See also Surface area 183, 191, 198, 206, 213, 217, 221, atmospheric scientist, 422 circles, 612 226, 231, 232, 245, 253, 260, 262, bricklayer, 487 congruent figures, 604 264, 265, 273, 277, 278, 287, 297, construction worker, 573 inscribed polygon, 612 305, 314, 322, 338, 348, 356, 362, detective, 92 irregular figures, 617, 618 370, 376, 382, 390, 397, 398, 409, engineering technician, 10 lateral, 649, 655, 660, 666 413, 414, 416, 423, 430, 437, 445, forester, 305 parallelograms, 595–598 451, 457, 458, 469, 475, 481, 487, interior designer, 163 rectangles, 732–733 493, 494, 496, 505, 511, 517, 518, landscape architect, 272 regular polygons, 610, 611 525, 528, 535, 543, 551, 558, 567, machinist, 693 rhombi, 602, 603 574, 588, 600, 608, 616, 621, 627, military, 353 squares, 732–733 631, 632, 642, 644, 645, 648, 653, real estate agent, 607 trapezoids, 602, 603 658, 664, 665, 670, 676, 683, 684, Cavalieri’s Principle, 691 triangles, 601, 602 694, 701, 706, 713, 719, 723, 724, 796–797 Center ASA. See Angle-Side-Angle Open Ended, See Extended of circle, 522 Assessment Response; See also Preparing of dilation, 490 Practice Chapter Test, 57, 121, for Standardized Tests, of rotation, 476 795–810 171, 231, 277, 337, 397, 457, 517, Central angle, 529 587, 631, 683, 723 Short Response/Grid In, 43, 50, Practice Quiz, 19, 36, 80, 100, 138, 59, 106, 123, 131, 135, 136, 164, Centroid, 240 173, 233, 279, 283, 285, 287, 314, 150, 198, 221, 254, 266, 306, 323, Challenge. See Critical Thinking; 322, 331, 339, 362, 372, 373, 382, 363, 383, 423, 445, 482, 497, 543, Extending the Lesson 568, 609, 621, 659, 670, 701, 713 399, 409, 416, 444, 459, 511, 519, Prerequisite Skills, 5, 19, 27, 36, 528, 535, 543, 551, 558, 589, Changing dimensions (parameters) 43, 61, 74, 80, 87, 93, 100, 106, 622–623, 633, 685, 703, 704, 725, area, 495, 496, 599, 607, 608, 615, 125, 131, 138, 144, 150, 157, 177, 798–805 653

Index R81 Index R82 ounmatrix, Column Collinear, Closed sentence, triangles, Classifying angles, Classifying Circumference, Circumcenter, Circular cone, Circles, Circle graphs, Chords Chihuly, Dale, volume, 693,695,698,709,710, surface area, 647,653,658,695, perimeter, 495,496,507–509,599, midsegment ofatrapezoid,440 median ofatrapezoid,440 magnitude ofvectors(scalar),502 tangents, 552–553 segments, 624 segment relationships, 569–571 sectors, 623 secants, 561–564 radius, 522 pi, 524 locus, 522 intersecting, 523 inscribed polygons,537 inscribed angles,544–546 great, 165 graphing, 576 equations of,575–577 diameter, 522 concentric, 528 common internaltangents,558 common externaltangents,558 circumscribed, 537 circumference, 523,524 chords, 522,536–539 central angle,529 center, 522 area, 612 arcs, 530,536,537 sphere, 671 diameters and,537 circle, 522 arcs and,536 Index 712 708–710, 712 607, 608,615,653 minor, 530 major, 530 length of,532 degree measure, 530 520–589 6 666 531 238, 239 316 523, 524 506 67 30 178–180 Concept Summary, Concept maps, Concentric circles, Concave, Compound statement, Compound locus, Composition, Composite figures, Communication. Component form, Complementary angles, Common Misconceptions, write, 33,41,48,63,147,360,508, summarize, 154, state, 97,136,251, show, 571, name, 9,219,373, make achart,442, list, 91,420,442, identify, 467, find acounterexample, 154,210, explain, 25,41,71,78,84,91,147, draw andlabel,345, draw adiagram,434, draw acounterexample, 293, draw, 48,360,639 discuss, 502,508, determine, 33,63,136,407,427, describe, 16,71,97,103,128,142, compare andcontrast,78,162, compare, 380, choose, 103 Math Error; OpenEnded;Writing in 585, 586,628,629, 630,678,679, 514, 515,516,581, 582,583,584, 452, 453,454,455,456,512,513, 335, 336,392,393,394,395,396, 230, 274,275,276,332,333,334, 167, 168,169,170,227,228,229, 94, 115, 116, 117, 118, 119, 120, Find theError 498, 555,623,638,698. 140, 178,238,284,290,326,419, 525, 555 242, 387,467,493,717 651, 657,663,668,698,704,710 485, 532,539,577,613,625,639, 373, 387,407,434,449,472,478, 263, 284,311, 319,328,353,367, 162, 180,195,210,219,224,257, 555, 605 328, 414,525,532,564,619,698 485, 502,548,598,645,717 242, 257,270,301,311, 367,478, 45 471 199 See also 577 498 528 53, 54,55,56, 617 67 39, 107–108 Find the See also 22, 76, Consecutive InteriorAngles Consecutive interiorangles, Conjunction, Conjectures, Congruent Congruence Construction, Constructed Response. Cones, Conditional statements, Concurrent lines, Conclusion, Theorem, triangles, 192–194 solids, 707–708 segments, 15 angles, 31,108 transformations, 194 symmetric property of,108 right triangle(HA,HL,LL,LA), SSS, 186 ASA, 207 AAS, 188 volume, 697–698 surface area, 667 right, 666–667 oblique, 666 lateral area, 666 circular, 666 base, 666 axis, 666 altitude, 666 find centerof a circle, 541,577 equilateral triangle,542 copy asegment,15 copy anangle,31 congruent triangles circle circumscribed abouta circle inscribedinatriangle, bisect asegment,24 bisect anangle,32,33,237 altitudes ofatriangle,237 Preparing forStandardized Tests Assessment 680, 681,682,720,721,722. 214–215 560 circumscribed about atriangle, by SSS,200 by SAS,202 207 by ASA, triangle, 559 559 638 75 134 22, 62,63,64,115, 324 68 15 238 See 116 128 See kite, 438 Cotangent, 370 Deductive reasoning, 82, 117 median of a trapezoid, 441 Counterexample, 63–65, 77–81, 93, Degree, 29 median of a triangle, 236 121, 196, 242, 387, 422, 429, 457, parallel lines, 151 Degree distance, 351 467, 493, 607, 675 perpendicular bisectors of sides Degree measure of triangle, 236 CPCTC, 192 of angles, 30 perpendicular line through a Critical Thinking, 11, 18, 27, 35, 42, of arcs, 530 point not on the line, 44 43, 50, 65, 73, 79, 86, 93, 99, 104, perpendicular line through a Density, 693 113, 130, 138, 144, 149, 163, 182, point on the line, 44 190, 197, 205, 212, 220, 226, 245, rectangle, 425 Diagonals, 404 253, 260, 265, 272, 286, 296, 305, regular hexagon, 542 Diameters 314, 321, 330, 347, 355, 362, 369, rhombus, 433 chords, 537 375, 382, 389, 408, 416, 422, 429, separating a segment into of circles, 522 436, 444, 450, 468, 473, 480, 481, proportional parts, 314 radius, 522 487, 496, 505, 511, 527, 534, 541, square, 435 of spheres, 671

542, 551, 557, 566, 567, 573, 579, Index tangent to a circle from a point 599, 608, 614, 616, 620, 627, 641, outside the circle, 554 Dilations, 490–493 647, 653, 658, 664, 669, 676, 693, tangent to a circle through a center for, 490 700, 706, 712, 719 point on the circle, 556 in coordinate plane, 492 scale factor, 490 trapezoid, 444 Cross-Curriculum Connections. See in space, 716 trisect a segment, 311 also Applications; More About biology, 93, 201, 272, 330, 347, Contradiction, proof by, 255–257 Dimension, 7 404, 467 Contrapositive, 77 chemistry, 65, 83, 405 Direction of a vector, 498 earth science, 69, 79, 113, 374, Converse, 77 Direct isometry, 481 504, 579, 675, 689 Converse of the Pythagorean geography, 27, 104, 355, 620 Disjunction, 68 Theorem, 351 geology, 647, 700 Distance history, 107, 223, 265, 303, 526, Convex, 45 between a point and a line, 159, 572, 636, 664, 700 160 Coordinate geometry, 47, 48, 49, 74, life science, 34, 42, 225, 270, 375 between parallel lines, 160, 161 162, 163, 180, 194, 201, 241, 242, physical science, 181, 260, 369, between two points, 21 243, 244, 252, 287, 294, 295, 302, 377, 382, 501, 557, 579, 675, in coordinate plane, 21 305, 306, 311, 313, 352, 354, 359, 705 formulas, 21, 715 368, 369, 390, 415, 420, 421, 422, physics, 35, 98, 154, 204, 375, 579, on number line, 21 426, 428, 429, 432, 434, 437, 440, 614, 693 in space, 715 442, 443, 444, 445, 447, 448, 467, science, 351, 375, 422, 463, 510 468, 472, 473, 474, 479, 480, 481, Dodecagon, 46 Cross products, 283 488, 495, 497, 528, 597, 599, 600, Dodecahedron, 638 603, 605, 606, 614, 616, 618, 619, Cross section, 639, 640, 641, 648 620, 621, 642 Cubes, 638, 664 E Coordinate plane, 597 Cylinders, 638 Edges, 637 Coordinate proofs. See Proof altitude of, 655 lateral, 649, 660 axis, 655 Eiffel, Gustave, 298 Coordinates in space, 714–715 base, 655 Corner view, 636 lateral area, 655 Enrichment. See Critical Thinking; locus, 658 Extending the Lesson Corollaries, 188, 218, 263, 309, 477 oblique, 655, 691 Equality, properties of, 94 right, 655 Corresponding angles, 128 surface area, 656 Equal vectors, 499 volume, 690–691 Corresponding Angles Postulate, Equations 133 of circles, 575–577 Corresponding Parts of Congruent D linear, 145–147 solving linear, 737–738 Triangles (CPCTC), 192 Decagon, 46 solving quadratic by factoring, Cosecant, 370 Decision making. See Critical 750–751 Thinking systems of, 742–743 Cosine, 364 Law of Cosines, 385–387 Deductive argument, 94 Equiangular triangles, 178

Index R83 Index R84 Formulas. Formal proof, Foldables™ StudyOrganizer, Flow proof. Exterior AngleInequality Extending theLesson, Find theError, Extremes, Extra Practice, Exterior ofanangle, Exterior AngleTheorem, Exterior AngleSumTheorem, Exterior angles, Extended Response. Euler’s formula, Escher, M.C., Faces, Error Analysis. Equilateral triangles, Equidistant, lateral area formula,640 Euler’s distance Cavalieri’s Principle,691 area arc length,532 cover 521, 593,635,687 125, 177,235,281,341,403,461, 691, 704 493, 539,571,605,625,657,674, 301, 345,353,380,420,427,472, 142, 188,203,251,263,284,292, Theorem, 608, 642,654,658,729 297, 315,370,474,481,528,558, for Standardized Tests lateral, 649,660 Misconceptions; FindtheError Index of acone,667 in space,715 in acoordinate plane, 21 of atriangle,602 of atrapezoid,602 of asector, 623 of arhombus, 603 of aregular polygon,610 of aparallelogram,596 of acircle, 612 637, 649,660 283 e loinsideback See also See 160 248 289, 483 95 754–781 9, 48,84,111, 128, See Proof 128, 186,187 640 F Common See 29 179, 218 1 1, 19,43, Preparing 186 5, 61, 405 Frustum, Free Response. Fractals, Geometry Activity Geometry, typesof Geometric proof,typesof. Geometric probability, Geometric mean, Geometer’s Sketchpad. volume surface area sum ofcentralangles,529 r probability, 20,622 perimeter, 46 midpoint Standardized Tests self-similarity, 325 fractal tree, 328 Congruence inRightTriangles, Comparing Magnitudeand Circumference Ratio,524 Bisectors, Medians,and Bisect an Angle, 32 Area ofaTriangle, 601 Area ofaParallelogram,595 Area ofaCircle, 611 Angles ofTriangles, 184 Angle Relationships,38 Angle-Angle-Side Congruence, spherical, 165–166,430 plane Euclidean,165,166 non-Euclidean, 165,166 Proof Geometry Software Investigations ecursive, 327 of asphere, 702 of apyramid,696 of aprism,689 of acylinder, 690 of acone,697 of asphere, 673 of aregular pyramid,661 of aprism,650 of acylinder, 656 of acone,667 in space,715 in coordinate plane,22 of aregular pyramid,661 of aprism,650 of acylinder, 655 214, 215 Components of Vectors, 501 Altitudes, 236 208 325, 326 664 See G 342–344 Preparing for 20, 622–624 See See Golden rectangle, Golden ratio, Glide reflections, Geometry SoftwareInvestigation V T T T T T T Surface Area ofaSphere, 672 Sum oftheExterior Angles ofa Similar Triangles, 298 The SierpinskiTriangle, 324 110Right Angles, The Pythagorean Theorem, 349 Properties ofParallelograms,411 Probability andSegment Non-Euclidean Geometry, 165, Modeling thePythagorean Modeling IntersectingPlanes,8 Midpoint ofaSegment,22 Median ofaTrapezoid, 441 Measure ofInscribed Angles, 544 Matrix Logic,88 Locus andSpheres, 677 Kites, 438 Investigating theVolume ofa Intersecting Chords, 569 Inscribed andCircumscribed Inequalities forSidesand Angles Equilateral Triangles, 179 Draw aRectangularPrism,126 Constructing Perpendiculars,44 Congruent Chords andDistance, T Right Triangles Formed bythe Reflections inIntersectingLines, Quadrilaterals, 448 Measuring Polygons,51–52 Angles andParallelLines,132 The Ambiguous CaseoftheLaw Adding SegmentMeasures, esting foraParallelogram,417 essellations ofRegular essellations and angents andRadii,552 rigonometric Ratios,365 rigonometric Identities,391 ransformations, 462 olume ofaRectangularPrism, 688 Polygons, 483 T Polygon, 406 Measure, 20 166 Theorem, 28 Pyramid, 696 T of Triangles, 559,560 538 Altitude, 343 477 of Sines,384 101 ransformations, 489 riangles, 559,560 429 474 429 Graphing Indirect isometry, 481 497, 505, 511, 527, 535, 543, 551, circles, 576 557, 567, 571, 579, 599, 609, Indirect measurement, 300, 372, 379 lines, 145–147 615, 621, 627, 641, 647, 653, 659, ordered pairs, 728–729 Indirect proof. See proof 665, 669, 675, 693, 701, 705, 713, ordered triples, 719 Indirect reasoning, 255 719 using intercepts and slope, 741 www.geometryonline.com/ Inductive reasoning, 62, 64, 115 standardized_test, 59, 123, 173, Graphing calculator, 366, 524, 576, Inequalities 233, 279, 339, 399, 459, 519, 589, 667, 703 633, 685, 725 investigation, 158 definition of, 247 properties for real numbers, 247 www.geometryonline.com/ Great circles, 165, 671 solving, 739–740 vocabulary_review, 53, 115, 167, for triangles, 247–250 227, 274, 332, 392, 452, 512, 581, Gridded Response. See Preparing 628, 678, 720 for Standardized Tests Informal proof, 90 www.geometryonline.com/ webquest, 23, 65, 155, 164, 216, Grid In. See Assessment Inscribed angles, 544–546 241, 325, 390, 444, 527, 580, 618, Inscribed Angle Theorem, 544 703, 719 Index H Inscribed polygons, 537, 547, 548, Invariant points, 481 612 Inverse, 77 HA. See Hypotenuse-Angle Integers Investigations. See Geometry Hemisphere, 672 evaluating expressions, 736 Activity; Geometry Software volume of, 703 operations with, 732–735 Investigation; Graphing Heptagon, 46 Calculator Investigation; Interior angles, 128 Spreadsheet Investigation; Hexagon, 46 Interior Angle Sum Theorem, 404 WebQuest Hexahedron (cube), 638 Irregular figures Interior of an angle, 29 area, 617, 618 Hinge Theorem, 267, 268 Internet Connections in the coordinate plane, 618 HL. See Hypotenuse-Leg www.geometryonline.com/ Irregular polygon, 46, 618 careers, 10, 92, 163, 209, 272, 305, Isometry, 463, 470, 476 Homework Help, 9, 17, 25, 34, 42, 355, 422, 487, 573, 631, 683, 723 64, 72, 78, 85, 92, 97, 104, 112, 129, direct, 481 www.geometryonline.com/ indirect, 481 136, 148, 155, 162, 181, 204, 211, chapter_test, 57, 121, 171, 231, 219, 243, 252, 258, 271, 285, 293, 277, 337, 397, 457, 517, 587, 631, Isosceles trapezoid, 439 302, 312, 320, 328, 354, 368, 374, 683, 723 Isosceles triangles, 179, 216–17 381, 388, 407, 415, 421, 428, 434, www.geometryonline.com/data_ 442, 450, 467, 473, 479, 486, 494, update, 18, 86, 143, 190, 286, Isosceles Triangle Theorem, 216 503, 509, 526, 533, 540, 549, 556, 375, 422, 474, 527, 607, 646, 712 Iteration, 325 572, 578, 598, 606, 613, 619, 640, www.geometryonline.com/extra_ nongeometric, 327 646, 651, 657, 663, 668, 674, 692, examples, 7, 15, 23, 31, 39, 47, 69, 699, 704, 711, 717 77, 91, 127, 135, 141, 153, 161, Hypotenuse, 344, 345, 350 179, 187, 193, 201, 209, 217, 223, 239, 257, 263, 269, 283, 291, 299, K Hypotenuse-Angle (HA) 309, 317, 343, 359, 365, 373, 379, Congruence Theorem, 215 387, 405, 413, 419, 425, 433, 441, Key Concepts, 6, 15, 21, 22, 29, 30, 449, 465, 471, 477, 491, 507, 523, 31, 37, 39, 40, 45, 46, 67, 68, 75, Hypotenuse-Leg (HL) Congruence 77, 82, 90, 134, 139, 152, 159, 161, Postulate, 215 531, 537, 545, 553, 563, 577, 597, 603, 611, 617, 623, 637, 645, 651, 178, 179, 192, 215, 222, 247, 255, Hypothesis, 75 655, 661, 667, 673, 689, 697, 703, 283, 342, 377, 385, 411, 412, 424, 709, 715 431, 477, 490, 491, 498, 499, 500, www.geometryonline.com/ 501, 524, 529, 530, 532, 575, 596, I other_calculator_keystrokes, 158 602, 603, 610, 612, 622, 623, 650, 655, 656, 661, 667, 673, 690, 691, Icosahedron, 638 www.geometryonline.com/ self_check_quiz, 11, 19, 27, 35, 696, 697, 702, 707, 715 If-then statement, 75 43, 49, 65, 73, 93, 99, 105, 113, Keystrokes. See Graphing calculator; Incenter, 240 131, 137, 143, 149, 157, 163, 183, Study Tip, graphing calculator 191, 197, 205, 213, 221, 225, 245, Incenter Theorem, 240 253, 259, 265, 273, 287, 297, 305, Kites (quadrilateral), 438 315, 323, 331, 347, 355, 363, 369, Included angle, 201 Koch curve, 326 375, 383, 389, 409, 415, 423, 429, Included side, 207 437, 445, 451, 469, 475, 481, 487, Koch’s snowflake, 326

Index R85 Index R86 Leg-Angle (LA)Congruence Law ofSyllogism, Leg-Leg (LL)Congruence Law ofSines, Law ofDetachment, Law ofCosines, Lateral faces Lateral edges Lateral area LA. Locus, LL. Line segments, Lines, Line ofsymmetry, Line ofreflection, Linear pair, Linear equations, Theorem, cases forsolvingtriangles,380 ambiguous case,384 Theorem, of pyramids,660 of prisms,649 of pyramids,660 of prisms,649 of pyramids,660 of prisms,649 of cylinders,655 of cones,666 compound, 577 circle, 522 658, 671,677,719 transversal, 126 tangent, 552,553 of symmetry, 466 slope, 139,140 skew, 126,127 of reflection, 463 point ofconcurrency, 238 perpendicular, 40,140 parallel, 126,140 naming, 6 equations of,145–147 distance from pointtoline, concurrent, 238 auxiliary, 135 Index See to aplane,43 proving, 151–153 slope-intercept form,145,146 point-slope form,145,146 262 See 6–8 1 Leg-Leg Leg-Angle 1, 238,239,310,522,577, 37 215 215 377–379 13 385–387 L 145–147 463 466 83 82 Measurements Means, inproportion, Mean, geometric, Matrix logic, Matrices Look Back, Mathematics, historyof, Mandelbrot, Benoit, Major arc, Magnitude Logical Reasoning. Logically equivalent, Logic, volume, 689–703 r surface area, 645–673 segment, 13 precision, 14 perimeter, 47 indirect, 300,372,379 degree, 29 circumference, 523 changing unitsofmeasure, area, 604–618 arc, 530 angle, 30 vertex, 506 translation, 506 transformations with,506–508,716 ro r operations with,752–753 column, 506 265, 329,637,691 715, 716 602, 604,614,690,700,702,708, 447, 463,484,532,556,596,597, 180, 309,326,373,378,406,425, Thinking of avector, 498 of rotational symmetry, 478 valid conclusions,82 truth tables,70-71 proof, 90 matrix, 88 Law ofSyllogism,83 Law ofDetachment,82 spheres, 677 parallel lines,310 intersecting lines,239 cylinder, 658 elative error, 19 eflection, 507 730–731 tation, 507,508 67–71, 116 530 47, 133,141,151,161, 88 M 342 See 325 77 283 Critical 152, 156, More About. Mixed Review. arc, Minor Midsegment, Midpoint Theorem, Midpoint Formula Midpoint, Median, school rings,550 school, 73 sayings, 541 rivers, 504 r railroads, 374 Pulitzer Prize,86 Plato, 638 physics, 35,98 photography, 318 orienteering, 244 music, 65,480 mosaics, 196 miniatures, 711 maps, 149,310,351 landmarks, 566 kites, 212 John Playfair, 156 irrigation, 534 igloos, 675 history, 107,664 highways, 113 health, 270 golden rectangle, 429 geology, 647 food, 705 flags, 436 entertainment, 286 Eiffel Tower, 298 drawing, 415 design, 105,220,435,443 Dale Chihuly, 316 construction, 137 calculus, 620 buildings, 389 Blaise Pascal,329 billiards, 468 basketball, 712 baseball, 205,673 aviation, 382,501 astronomy, 369,557 art, 418,440,559 architecture, 181,408,451,615 aircraft, 495 Cross-Curriculum Connections in space,715 number line,22 on coordinate plane,22 of triangle,240,241 of trapezoid,440 ecreation, 18,718 440 22 530 See also 308 See Review 91 Applications; seasons, 79 Open Response. See Preparing for coordinate, 728–729 shopping, 256 Standardized Tests naming, 6 space travel, 579 parallel, 126 Open Sentence, 67 speed skating, 190 Platonic solids, 638 sports, 294 Opposite rays, 29 dodecahedron, 638 steeplechase, 225 Order of rotational symmetry, 478 hexahedron (cube), 638 tennis, 626 icosahedron, 638 tepees, 669 Ordered pairs, graphing, 728–729 octahedron, 638 tourism, 652 Ordered triple, 714 tetrahedron, 638 towers, 304 traffic signs, 49 Orthocenter, 241 Playfair, John, 152, 156 travel, 253, 375 treehouses, 250 Orthogonal drawing, 636 Point of concurrency, 238 triangle tiling, 361 Point of symmetry, 466 volcanoes, 700 windows, 426 P Point of tangency, 552 Paragraph proofs. See Proof Index Multiple Choice. See Assessment Points, 6 Parallel lines, 126 collinear, 6–8 N alternate exterior angles, 134 coplanar, 6–8 alternate interior angles, 134 graphing, 728–729 Negation, 67 consecutive interior angles, 133 naming, 6 Nets, 50, 643 corresponding angles, 134 of symmetry, 466 for a solid, 644 perpendicular lines, 134, 159-164 Point-slope form, 145 surface area and, 645 Parallel planes, 126 Newman, Barnett, 440 Polygons, 45–48 Parallel vectors, 499 area of, 604, 595–598, 601–603, n-gon, 46 Parallelograms, 140, 411–414 610–611, 617–618 Nonagon, 46 area of, 595–598 circumscribed, 555 concave, 45 Non-Euclidean geometry, 165, 166 base, 595 conditions for, 417, 418 convex, 45 on coordinate plane, 420, 597 on coordinate plane, 47, 48, 49, O diagonals, 413 74, 162, 163, 180, 194, 201, 241, 242, 243, 244, 252, 287, 294, 295, Oblique height of, 411 302, 305, 306, 311, 313, 352, 354, cone, 666 properties of, 418 359, 368, 369, 390, 415, 420, 421, cylinder, 655 tests for, 419 422, 426, 428, 429, 432, 434, 437, prism, 649 Parallel Postulate, 152 440, 442, 443, 444, 445, 447, 448, Obtuse angle, 30 Pascal, Blaise, 329 467, 468, 472, 473, 474, 479, 480, 481, 488, 495, 497, 528, 597, 599, Obtuse triangle, 178 Pascal’s Triangle, 327–329 600, 603, 605, 606, 614, 616, 618, Octagons, 46 Pentagons, 46 619, 620, 621, 642 diagonals of, 404 Octahedron, 638 Perimeters, 46 hierarchy of, 446 on coordinate plane, 47 Online Research inscribed, 537, 547, 548, 612 of rectangles, 46, 732–733 Career Choices, 10, 92, 163, 209, irregular, 46, 618 of similar triangles, 316 272, 305, 355, 422, 487, 573, 693 perimeter of, 46–47 of squares, 46, 732–733 Data Update, 18, 86, 143, 190, 286, regular, 46 375, 422, 474, 527, 607, 646, 712 Perpendicular, 40 apothem, 610 sum of exterior angles of, 406 Open Ended, 9, 16, 25, 33, 41, 48, 63, Perpendicular bisector, 238 sum of interior angles of, 404 71, 78, 84, 91, 97, 103, 111, 128, Perpendicular Transversal 136, 142, 147, 154, 162, 180, 188, Polyhedron (polyhedra), 637, 638. Theorem, 134 195, 203, 210, 219, 224, 242, 251, See also Platonic Solids; Prisms; 257, 263, 270, 284, 293, 301, 311, Perspective Pyramids 319, 328, 345, 353, 360, 367, 373, one-point, 10 edges, 637 387, 407, 414, 420, 427, 434, 442, two-point, 11 faces, 637 449, 467, 472, 478, 485, 493, 502, view, 636 prism, 637 bases, 637 508, 525, 532, 539, 548, 555, 564, Pi (), 524 572, 577, 598, 605, 613, 619, 625, pyramid, 637 639, 645, 651, 657, 663, 668, 674, Planes regular, 637 (See also Platonic 691, 698, 710, 717 as a locus, 719 solids)

Index R87 Index R88 Preparing forStandardizedTests, Prefixes, meaningof, Precision, Practice Quiz. Postulates, Polynomials, Practice ChapterTest. Prerequisite Skills. r T Student-Produced Response,798 Student-Produced Questions, Short Response,795,802–805 Selected Response,796 Open Response,802 Multiple Choice,795,796–797 Gridded Response,795,798–801 Grid In,798 Free Response,802 Extended Response,795,806–810 Constructed Response,802,806 795–810 Assessment Side-Side-Side congruence (SSS), Side-Angle-Side congruence 102 Segment Addition, Ruler, 101 Protractor, 107 Parallel, 152 133 Corresponding Angles, Angle-Side-Angle congruence 107 Angle Addition, 298, 477,604,617 dividing, 748–749 multiplying, 746–747 Getting ReadyfortheNext factoring tosolveequations, changing unitsofmeasure, area ofrectangles andsquares, algebraic expressions, 736 Assessment Index egular prism,637 est-Taking Tips, 795, 797,800, triangular, 637 r pentagonal, 637 802 201 (SAS), 202 (ASA), 207 297, 306,315,323, 348,356,363, 191, 206,213,221, 245,266,287, 100, 106,131,138,144,157,183, Lesson, 11, 19,27,43,74,80,93, 750–751 730–731 732–733 804, 810 ectangular, 637 14 89, 90,118, 141,151,215, See Assessment See also 594 See Proof, Projects. Problem Solving(additional), Probability, Prisms, indirect, 255–260, 266,272,277, flow, 187,190,197,202,203,204, coordinate, 222 by contradiction,255–260,266, algebraic, 94,95,97,98,100,256, additional, 782–794 782–794 with linesegments,20 with circle segments,624 with sectors,623 geometric, 622–624 arcs and,622 See also systems oflinearequations, square roots andradicals, solving linearequations, solving inequalities,739–740 polynomials, 746–749 perimeter ofrectangles and matrices, 752–753 integers, 734–735 graphing usingintercepts and graphing ordered pairs,728–729 Getting Started,5,61,125,177, volume of,688,689 surface area of,650 right, 649 r oblique, 649,654 lateral faces,649 lateral edges,649 lateral area, 649 bases, 637,649 egular, 637 475, 556 322, 376,439,442, 444,541 209, 210,212,213,229,231,269, 231, 233,355 with triangles,222–226,230, 456, 457,459,469,475 with quadrilaterals,447–451, 272, 277,475,556 350 742–743 744–745 737–738 squares, 732–733 slope, 741 634, 686 235, 281,341,403,461,521,593, 713 600, 609,616,670,694,701,706, 528, 535,543,551,558,568,574, 437, 445,469,475,482,497,505, 370, 376,383,409,416,423,430, 90 649 See Applications 20, 265,527,700,705. W ebQuest Pythagorean triples, Pythagorean Theorem, Pythagorean identity, Squares; Trapezoids See also Quadrilaterals, Protractor, Pyramids, Protractor Postulate, Proportions, Proportional PerimetersTheorem, Properties ofinequality, forreal Properties ofequality, forreal primitive, 354 Distance Formula,21 converse of,351 two-column (formal),95–98, paragraph (informal),90–93,96, volume, 696,697 surface area, 661 slant height,660 r lateral faces,660 lateral edges,660 lateral area, 660 base, 660 altitude, 660 r coordinate proofs with, solving, 284 means, 283 geometric mean,342 extremes, 283 definition of,283 cross products, 283 316 numbers, numbers, egular elationships among, 435 546, 549,550,566,573 439, 442,444,497,528,534,536, 270–273, 304,322,326,348,389, 242–244, 249,252,263,264,268, 210, 212,213,229,231,239, 182, 190,197,202,203,204,209, 134, 137,150,153–156,164,171, 102–105, 109,111–113, 121,131, 548, 550,567,568,573 418, 431,435,444,495,522,545, 317, 319,348,355,390,414,415, 243, 253,267,268,299,304,307, 190, 208,210,211, 215,221,239, 102, 108,121,131,137,163,182, 447–449 Parallelograms; Rhombus; 637, 660,661 30 247 94 282–284 See also Q 352 107 391 Rectangles. 350 R polyhedron, 637 S Radical expressions, 744–745 prism, 637 Saccheri quadrilateral, 430 pyramid, 660 Radius tessellations, 483, 484 Same side interior angles. See circles, 522 Consecutive interior angles diameter, 522 Related conditionals, 77 SAS. See Side-Angle-Side spheres, 671 Relative error, 19 Rate of change, 140 Scalar, 501 Remote interior angles, 186 Ratios, 282 Scalar multiplication, 501 Research, 11, 73, 156, 181, 246, 259, extended, 282 Scale factors, 290 330, 347, 429, 473, 594, 620, 654, trigonometric, 364 for dilation, 490 706. See also Online Research unit, 282 on maps, 292 Rays, 29 Resultant (vector), 500 similar solids, 709 naming, 29 Review Scalene triangle, 179 opposite, 29

Lesson-by-Lesson, 53–56, Index Secant Reading Math, 6, 29, 45, 46, 75, 186, 155–120, 167–170, 227–230, of circles, 561–564 207, 283, 411, 432, 441, 464, 470, 274–276, 332–336, 392–397, function, 370 483, 522, 536, 637, 649, 666 452–456, 512–516, 581–586, 628–630, 678–682, 720–722 Sector, 623 Reading Mathematics Mixed, 19, 27, 36, 50, 66, 74, 80, Biconditional Statements, 81 Segment Addition Postulate, 102 87, 93, 100, 106, 114, 131, 138, Describing What You See, 12 144, 150, 157, 164, 183, 191, 206, Segment bisector, 24 Hierarchy of Polygons, 446 213, 221, 226, 245, 254, 260, 266, Making Concept maps, 199 Segments, 13 273, 287, 297, 306, 315, 323, 331, Math Words and Everyday adding, 101 348, 356, 363, 370, 376, 383, 390, Words, 246 bisector, 24 416, 423, 437, 451, 469, 482, 488, Prefixes, 594 chords, 569, 570 497, 505, 511, 528, 535, 543, 551, of circles, 624 Real-world applications. See 558, 568, 574, 580, 600, 609, 616, congruent, 13, 15, 102 Applications; More About 627, 642, 648, 659, 665, 676, 694, measuring, 13 701, 706, 713, 719 Reasoning, 62, 63. See also Critical midpoint, 22 Thinking Rhombus, 431, 432. See also notation for, 13 deductive, 82 Parallelograms perpendicular bisector, 238 indirect, 255 area, 602, 603 secant, 570–571 inductive, 62, 64, 115 properties, 431 tangent, 570–571 Reciprocal identity, 391 Right Selected Response. See Preparing Rectangles, 424–427 angle, 108 for Standardized Tests cone, 666 area, 732 Self-similarity, 325, 326 on coordinate plane, 426–427 cylinder, 555 diagonals, 425 prism, 649 Semicircles, 530 perimeter, 46, 732 Right triangles, 178, 350 Semi-regular tessellations, 484 properties, 424 30°-60°-90°, 357–359 Short Response. See Assessment; Recursive formulas, 327 45°-45°-90°, 357–359 congruence of, 214, 215 See also Preparing for Reflection matrix, 507 geometric mean in, 343, 344 Standardized Tests Reflection symmetry, 642 special, 357–359 Side-Angle-Side (SAS) Congruence Postulate, 201 Reflections, 463–465 Rotational symmetry, 478 composition of, 471 magnitude, 478 Side-Angle-Side (SAS) in coordinate axes, 464 order, 478 Inequality/Hinge Theorem, 267 in coordinate plane, 465 Rotation matrix, 507 Side-Angle-Side (SAS) Similarity glide, 474 Theorem, 299 line of, 463 Rotations, 476–478 matrix of, 507 angle, 476 Sides in a point, 463, 465 center, 476 of angle, 29 rotations, as composition of, 477, as composition of reflections, of triangle, 178 478 477, 478 of polygon, 45 matrices of, 507 Regular Side-Side-Side (SSS) Congruence polygon, 46 Ruler Postulate, 101 Postulate, 186, 202

Index R89 Index R90 Spheres, Side-Side-Side (SSS)Inequality Special righttriangles. Space Solids. Slope, Slant height, Skew lines, Side-Side-Side (SSS)Similarity Sine, Simple closedcurves, Similar triangles, Similar solids, Similarity transformations, Similar figures, Sierpinski Triangle, chord, 671 T Theorem, vertex matrices,716 translations in,715–716 midpoint formula,715 graphing points,714–719 distance formula,715 dilation, 716 coordinates in,714–719 volume of,688 symmetry of,642 surface area of,645,650,656,667, spheres, 638 similar, 707,708 pyramids, 660 prisms, 637 nets for, 645 lateral area, 649,655,660,666 graphing inspace,714 frustum, 664 cylinders, 638 cross sectionof,639 congruent, 707,708 cones, 638 dimensional figures slope-intercept form,145 rate ofchange,140 point-slope form,145 perpendicular lines,141 parallel lines,141 Law ofSines,377–384 Theorem, scale factors,290 enlargement, 291 Index riangles 672 364 139, 140 See also 638 127 268 299 660 Polyhedron; Three- 707, 708 615 298–300 324, 325 46, 618 See 491 Study Tips Study organizer. Student-Produced Response. Student-Produced Questions. Statements, Standard position, Standardized Test Practice. SSS. Square roots, Squares Spreadsheet Investigation Spherical geometry, area ofarhombus, 603 altitudes ofaright triangle,343 adding angle measures, 32 absolute value,563 triangle,359 30°-60°-90° Study Organizer Preparing forStandardized Tests Preparing forStandardized Tests truth value,67–69 truth tables,70,71 negation, 67 logically equivalent,77 conditional, 75–77 compound, 67 biconditional, 81 Assessment properties of,432 perimeter, 46,722–723 area, 722–723 Prisms, 695 Fibonacci SequenceandRatios, Explore SimilarSolids,708 Angles ofPolygons,410 volume, 702,703 tangent, 671 surface area, 672,673 radius, 671 properties of,671 locus and,677 hemisphere, 672 great circles, 671 diameter, 671 truth value,76 r inverse, 77 if-then, 75–77 hypothesis, 75 converse, 77 contrapositive, 77 conclusion, 75 disjunction, 68,69 conjunction, 68,69 288 elated, 77 See , 432,433 Side-Side-Side 67 744–745 See 498 165, 166,430 Foldables™ See See See Look Back,47, 108,133,141,151, locus, 239,310,577 Law ofCosines,386 isosceles trapezoid,439 isometry dilation,491 isometric dotpaper, 643 interpreting figures, 40 inscribed polygons,547 inscribed andcircumscribed, 537 information from figures, 15 inequality, 261 inequalities intriangles,249,76 if-then statements,76 identifying tangents,553 identifying segments,127 identifying corresponding parts, helping youremember, 570 great circles, 672 graphing calculator, 242,367,508, formulas, 655 flow proofs, 202 finding thecenterofacircle, 541 estimation, 618 equivalent ratios,365 an equivalentproportion, 379 equation ofacircle, 575 eliminating fractions,240 eliminate thepossibilities,546 drawing tessellations,484 drawing nets,645 drawing inthree dimensions,714 drawing diagrams,89 Distance Formula,22,352 distance, 160 dimension, 7 corner viewdrawings,636 coordinate geometry, 420 conjectures, 62 congruent circles, 523 conditional statements,77,83 complementary and comparing numbers,333 comparing measures, 14 Commutative and Associative common misconceptions,22,76, classifying angles,30 circles andspheres, 671 choosing formsoflinear checking solutions,32,291 Cavalieri’s Principle,691 betweenness, 102 447, 463,484,532, 556,596,597, 161, 185,309,326, 378,406,425, 290 degree mode,366 576 supplementary angles,39 Properties, 94 498, 555,623,638,698 140, 178,238,284,290,326,419, equations, 146 602, 604, 700, 702, 707, 708, 715, using a ruler, 13 Tessellations, 483, 485 716 using fractions, 308 regular, 483, 484 making connections, 656, 667 using variables, 545 semi-regular, 484, 485 means and extremes, 242 validity, 82 uniform, 484 measuring the shortest distance, value of pi, 524 Test preparation. See Assessment 159 Venn diagrams, 69 medians as bisectors, 240 vertex angles, 223 Test-Taking Tips. See Assessment mental math, 95 volume and area, 689 modeling, 639 writing equations, 146 Tetrahedron, 638 naming angles, 30 Supplementary angles, 39, 107 Theorems. See pages R1–R8 for a naming arcs, 530 inscribed quadrilaterals, 548 complete list. naming congruent triangles, 194 linear pairs, 107–108 corollary, 188 naming figures, 40 parallel lines, 134 definition of, 90 negative slopes, 141 perpendicular lines, 134 obtuse angles, 377 Third Angle Theorem, 186 overlapping triangles, 209, 299, Supplement Theorem, 108 Three-dimensional figures, Index 307 363–369. See also Space patty paper, 38 Surface area, 644 cones, 667 corner view, 636 placement of figures, 222 models and, 643 problem solving, 448, 611 cylinders, 656 nets and, 645 nets for, 644 proof, 431 orthogonal drawings, 636 proving lines parallel, 153 prisms, 650 pyramids, 661 perspective view, 636 Pythagorean Theorem, 21 slicing, 639 radii and diameters, 523 spheres, 672, 673 rationalizing denominators, 358 Symbols. See also inside back cover Tiffany, Louis Comfort, 282 reading math, 617 angle, 29 Tolerance, 14 rectangles and parallelograms, 424 arc, 530 recursion on the graphing congruence, 15 Transformations, 462 calculator, 327 conjunction (and), 68 congruence, 194 right prisms, 650 degree, 29 dilation, 490–493 rounding, 378 disjunction (or), 68 isometry, 436, 463, 470, 476, 481 same side interior angles, 128 implies, 75 matrices, 506–508, 716 SAS inequality, 267 negation (not), 67 reflection, 463–465 shadow problems, 300 parallel, 126 rotation, 476–478 shortest distance to a line, 263 perpendicular, 40 similarity, 491 side and angle in the Law of pi (), 524 in space, 714–718 Cosines, 385 triangle, 178 translation, 470–471 simplifying radicals, 345 Symmetry slope, 139 line of, 466 Translation matrix, 506, 715–716 SOH-CAH-TOA (mnemonic point of, 466 Translations, 470, 471 device), 364 rotational, 478 in coordinate plane, 470 spreadsheets, 26 in solids, 642 with matrices, 506 square and rhombus, 433 Symmetry by repeated reflections, 471 square roots, 344, 690 line of, 466, 467 in space, 715 standardized tests, 709 point of, 466, 467 with vectors, 500 storing values in calculator reflection, 642 memory, 667 Transversals, 126 rotational, 478 symbols for angles and proportional parts, 309, 310 inequalities, 248 Systems of Equations, 742–743 Trapezoids, 439–441 tangent, 366 area, 602, 603 tangent lines, 552 base angles, 439 three-dimensional drawings, 8 T bases, 439 three parallel lines, 309 isosceles, 439 Tangent transformations, 194 legs, 439 to a circle, 552, 553 translation matrix, 506 median, 440 function, 364 transversals, 127 properties, 439 truth tables, 72 and slope of a line, 366 truth value of a statement, 255 to a sphere, 671 Triangle inequality, 261–263 units, 596 Tangrams, 421 Triangle Inequality Theorem, 261 units of measure, 14 units of time, 292 Tautology, 70 Triangle Midsegment Theorem, 308

Index R91 Index R92 T T riangles, riangle ProportionalityTheorem, SAS inequality(HingeTheorem), right, 178 proportional parts,307–310 proving trianglescongruent perpendicular bisectorsofsides, parts ofsimilar, 316–318 orthocenter, 241 obtuse, 178 midsegment, 308 medians, 240,241 LL, 215 LA, 215 isosceles, 179,216,217 inequalities for, 247–250 included side,207 included angle,201 incenter, 240 HL, 215 HA, 215 exterior angles,248 equilateral, 179,218 equiangular, 178 CPCTC, 192 corresponding parts,192 coordinate proofs, 222,223 congruent, 192–194 classifying, 178–180 circumcenter, 238,239 ASA, 207 AAS, 188 area, 601,602 angles, 185–188 angle bisectors,239,240 altitude, 241,242 acute, 178 converse of,308 307 Index 267, 268 special, 357–359 geometric meanin,343,344 45°-45°-90°, 357 30°-60°-90°, 358,359 SSS, 201 SAS, 202 LA, LL,215 215 HA, HL ASA, 207 AAS, 188 238, 239 vertex angle,216 base angles,216 174–399 T T T T V TODAY Snapshots, USA Uniform tessellations, Undefined terms, T T T T T riangular numbers, rigonometric identity, rigonometric ratio, rigonometry, riptych, runcation, ruth tables, ruth value, wo-column proof. arying dimensions, 411, 474,531,591,614,653,705 143, 175,206,259,296,347,401, tessellations, 361 SSS inequality, 268 solving atriangle,378 similar, 298–300 side-angle relationships, 249,250 scalene, 179 trigonometric ratios,364 Law ofSines,377–379 Law ofCosines,385–387 identities, 391 area oftriangle,608 709, 710 special segments,317 299 side-side-side similarity(SSS), 299 side-angle-side similarity(SAS), parts of,316–318 298–300 angle-angle similarity(AA), tangent, 364 sine, 364 secant, 370 evaluating withcalculator, 366 cosine, 370 cotangent, 370 cosecant, 370 599 664 67 70–71 364–366 U V 7 See 364–370 62 599, 647,695, 484 Proof 391 3, 16,63, W W W V V V V V V V V V olume, ertical AnglesTheorem, ertical angles, ertex matrix, ertex angle, ertex enn diagrams, ectors, isualization, riting inMath, ork backward, ebQuest, spheres, 702,703 pyramids, 696,697 prisms, 689 oblique figures, 698 hemisphere, 703 cylinders, 690,691 cones, 697,698 719 469, 527,580,591,618,662,703, 218, 241,325,347,390,401,444, 706, 713,719 648, 653,658,664,669,693,701, 579, 600,608,616,620,627,641, 527, 534,542,551,558,567,574, 469, 474,481,487,496,505,511, 409, 416,422,430,436,444,451, 348, 356,362,369,375,382,389, 260, 273,286,305,314,322,330, 198, 205,213,221,226,245,253, 138, 144,149,157,164,183,191, 50, 66,74,79,86,93,105,114, 130, of atriangle,178 of apolygon,45 of anangle,29 standard positionfor, 498 standard position,498 scalar multiplication,501 r parallel, 499 magnitude, 498 equal, 499 direction, 498 on coordinate plane,498–502 component form,498 adding, 500 esultant, 500 498–502 688 3, 23,65,155,164,175, 216 506, 716 7, 9,10 37, 110 69–70 W 90 1 1, 19,27,35,43, 110