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Geometry Notes G.2 Parallel Lines, Transversals, Angles Mrs. Grieser Name: Date
Geometry Notes G.2 Parallel Lines, Transversals, Angles Mrs. Grieser Name: ________________________________________ Date: ______________ Block: ________ Identifying Pairs of Lines and Angles Lines that intersect are coplanar Lines that do not intersect o are parallel (coplanar) OR o are skew (not coplanar) In the figure, name: o intersecting lines:___________ o parallel lines:_______________ o skew lines:________________ o parallel planes:_____________ Example: Think of each segment in the figure as part of a line. Find: Line(s) parallel to CD and containing point A _________ Line(s) skew to and containing point A _________ Line(s) perpendicular to and containing point A _________ Plane(s) parallel to plane EFG and containing point A _________ Parallel and Perpendicular Lines If two lines are in the same plane, then they are either parallel or intersect a point. There exist how many lines through a point not on a line? __________ Only __________ of them is parallel to the line. Parallel Postulate If there is a line and a point not on a line, then there is exactly one line through the point parallel to the given line. Perpendicular Postulate If there is a line and a point not on a line, then there is exactly one line through the point parallel to the given line. Angles and Transversals Interior angles are on the INSIDE of the two lines Exterior angles are on the OUTSIDE of the two lines Alternate angles are on EITHER SIDE of the transversal Consecutive angles are on the SAME SIDE of the transversal Corresponding angles are in the same position on each of the two lines Alternate interior angles lie on either side of the transversal inside the two lines Alternate exterior angles lie on either side of the transversal outside the two lines Consecutive interior angles lie on the same side of the transversal inside the two lines (same side interior) Geometry Notes G.2 Parallel Lines, Transversals, Angles Mrs. -
Angle Chasing
Angle Chasing Ray Li June 12, 2017 1 Facts you should know 1. Let ABC be a triangle and extend BC past C to D: Show that \ACD = \BAC + \ABC: 2. Let ABC be a triangle with \C = 90: Show that the circumcenter is the midpoint of AB: 3. Let ABC be a triangle with orthocenter H and feet of the altitudes D; E; F . Prove that H is the incenter of 4DEF . 4. Let ABC be a triangle with orthocenter H and feet of the altitudes D; E; F . Prove (i) that A; E; F; H lie on a circle diameter AH and (ii) that B; E; F; C lie on a circle with diameter BC. 5. Let ABC be a triangle with circumcenter O and orthocenter H: Show that \BAH = \CAO: 6. Let ABC be a triangle with circumcenter O and orthocenter H and let AH and AO meet the circumcircle at D and E, respectively. Show (i) that H and D are symmetric with respect to BC; and (ii) that H and E are symmetric with respect to the midpoint BC: 7. Let ABC be a triangle with altitudes AD; BE; and CF: Let M be the midpoint of side BC. Show that ME and MF are tangent to the circumcircle of AEF: 8. Let ABC be a triangle with incenter I, A-excenter Ia, and D the midpoint of arc BC not containing A on the circumcircle. Show that DI = DIa = DB = DC: 9. Let ABC be a triangle with incenter I and D the midpoint of arc BC not containing A on the circumcircle. -
Proofs with Perpendicular Lines
3.4 Proofs with Perpendicular Lines EEssentialssential QQuestionuestion What conjectures can you make about perpendicular lines? Writing Conjectures Work with a partner. Fold a piece of paper D in half twice. Label points on the two creases, as shown. a. Write a conjecture about AB— and CD — . Justify your conjecture. b. Write a conjecture about AO— and OB — . AOB Justify your conjecture. C Exploring a Segment Bisector Work with a partner. Fold and crease a piece A of paper, as shown. Label the ends of the crease as A and B. a. Fold the paper again so that point A coincides with point B. Crease the paper on that fold. b. Unfold the paper and examine the four angles formed by the two creases. What can you conclude about the four angles? B Writing a Conjecture CONSTRUCTING Work with a partner. VIABLE a. Draw AB — , as shown. A ARGUMENTS b. Draw an arc with center A on each To be prof cient in math, side of AB — . Using the same compass you need to make setting, draw an arc with center B conjectures and build a on each side of AB— . Label the C O D logical progression of intersections of the arcs C and D. statements to explore the c. Draw CD — . Label its intersection truth of your conjectures. — with AB as O. Write a conjecture B about the resulting diagram. Justify your conjecture. CCommunicateommunicate YourYour AnswerAnswer 4. What conjectures can you make about perpendicular lines? 5. In Exploration 3, f nd AO and OB when AB = 4 units. -
Point of Concurrency the Three Perpendicular Bisectors of a Triangle Intersect at a Single Point
3.1 Special Segments and Centers of Classifications of Triangles: Triangles By Side: 1. Equilateral: A triangle with three I CAN... congruent sides. Define and recognize perpendicular bisectors, angle bisectors, medians, 2. Isosceles: A triangle with at least and altitudes. two congruent sides. 3. Scalene: A triangle with three sides Define and recognize points of having different lengths. (no sides are concurrency. congruent) Jul 249:36 AM Jul 249:36 AM Classifications of Triangles: Special Segments and Centers in Triangles By angle A Perpendicular Bisector is a segment or line 1. Acute: A triangle with three acute that passes through the midpoint of a side and angles. is perpendicular to that side. 2. Obtuse: A triangle with one obtuse angle. 3. Right: A triangle with one right angle 4. Equiangular: A triangle with three congruent angles Jul 249:36 AM Jul 249:36 AM Point of Concurrency The three perpendicular bisectors of a triangle intersect at a single point. Two lines intersect at a point. The point of When three or more lines intersect at the concurrency of the same point, it is called a "Point of perpendicular Concurrency." bisectors is called the circumcenter. Jul 249:36 AM Jul 249:36 AM 1 Circumcenter Properties An angle bisector is a segment that divides 1. The circumcenter is an angle into two congruent angles. the center of the circumscribed circle. BD is an angle bisector. 2. The circumcenter is equidistant to each of the triangles vertices. m∠ABD= m∠DBC Jul 249:36 AM Jul 249:36 AM The three angle bisectors of a triangle Incenter properties intersect at a single point. -
THE GEOMETRY of PYRAMIDS One of the More Interesting Solid
THE GEOMETRY OF PYRAMIDS One of the more interesting solid structures which has fascinated individuals for thousands of years going all the way back to the ancient Egyptians is the pyramid. It is a structure in which one takes a closed curve in the x-y plane and connects straight lines between every point on this curve and a fixed point P above the centroid of the curve. Classical pyramids such as the structures at Giza have square bases and lateral sides close in form to equilateral triangles. When the closed curve becomes a circle one obtains a cone and this cone becomes a cylindrical rod when point P is moved to infinity. It is our purpose here to discuss the properties of all N sided pyramids including their volume and surface area using only elementary calculus and geometry. Our starting point will be the following sketch- The base represents a regular N sided polygon with side length ‘a’ . The angle between neighboring radial lines r (shown in red) connecting the polygon vertices with its centroid is θ=2π/N. From this it follows, by the law of cosines, that the length r=a/sqrt[2(1- cos(θ))] . The area of the iscosolis triangle of sides r-a-r is- a a 2 a 2 1 cos( ) A r 2 T 2 4 4 (1 cos( ) From this we have that the area of the N sided polygon and hence the pyramid base will be- 2 2 1 cos( ) Na A N base 2 4 1 cos( ) N 2 It readily follows from this result that a square base N=4 has area Abase=a and a hexagon 2 base N=6 yields Abase= 3sqrt(3)a /2. -
Extremal Problems for Convex Polygons ∗
Extremal Problems for Convex Polygons ∗ Charles Audet Ecole´ Polytechnique de Montreal´ Pierre Hansen HEC Montreal´ Fred´ eric´ Messine ENSEEIHT-IRIT Abstract. Consider a convex polygon Vn with n sides, perimeter Pn, diameter Dn, area An, sum of distances between vertices Sn and width Wn. Minimizing or maximizing any of these quantities while fixing another defines ten pairs of extremal polygon problems (one of which usually has a trivial solution or no solution at all). We survey research on these problems, which uses geometrical reasoning increasingly complemented by global optimization meth- ods. Numerous open problems are mentioned, as well as series of test problems for global optimization and nonlinear programming codes. Keywords: polygon, perimeter, diameter, area, sum of distances, width, isoperimeter problem, isodiametric problem. 1. Introduction Plane geometry is replete with extremal problems, many of which are de- scribed in the book of Croft, Falconer and Guy [12] on Unsolved problems in geometry. Traditionally, such problems have been solved, some since the Greeks, by geometrical reasoning. In the last four decades, this approach has been increasingly complemented by global optimization methods. This allowed solution of larger instances than could be solved by any one of these two approaches alone. Probably the best known type of such problems are circle packing ones: given a geometrical form such as a unit square, a unit-side triangle or a unit- diameter circle, find the maximum radius and configuration of n circles which can be packed in its interior (see [46] for a recent survey and the site [44] for a census of exact and approximate results with up to 300 circles). -
Lesson 3: Rectangles Inscribed in Circles
NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 3 M5 GEOMETRY Lesson 3: Rectangles Inscribed in Circles Student Outcomes . Inscribe a rectangle in a circle. Understand the symmetries of inscribed rectangles across a diameter. Lesson Notes Have students use a compass and straightedge to locate the center of the circle provided. If necessary, remind students of their work in Module 1 on constructing a perpendicular to a segment and of their work in Lesson 1 in this module on Thales’ theorem. Standards addressed with this lesson are G-C.A.2 and G-C.A.3. Students should be made aware that figures are not drawn to scale. Classwork Scaffolding: Opening Exercise (9 minutes) Display steps to construct a perpendicular line at a point. Students follow the steps provided and use a compass and straightedge to find the center of a circle. This exercise reminds students about constructions previously . Draw a segment through the studied that are needed in this lesson and later in this module. point, and, using a compass, mark a point equidistant on Opening Exercise each side of the point. Using only a compass and straightedge, find the location of the center of the circle below. Label the endpoints of the Follow the steps provided. segment 퐴 and 퐵. Draw chord 푨푩̅̅̅̅. Draw circle 퐴 with center 퐴 . Construct a chord perpendicular to 푨푩̅̅̅̅ at and radius ̅퐴퐵̅̅̅. endpoint 푩. Draw circle 퐵 with center 퐵 . Mark the point of intersection of the perpendicular chord and the circle as point and radius ̅퐵퐴̅̅̅. 푪. Label the points of intersection . -
Regents Exam Questions G.GPE.A.2: Locus 1 Name: ______
Regents Exam Questions G.GPE.A.2: Locus 1 Name: ________________________ www.jmap.org G.GPE.A.2: Locus 1 1 If point P lies on line , which diagram represents 3 The locus of points equidistant from two sides of the locus of points 3 centimeters from point P? an acute scalene triangle is 1) an angle bisector 2) an altitude 3) a median 4) the third side 1) 4 Chantrice is pulling a wagon along a smooth, horizontal street. The path of the center of one of the wagon wheels is best described as 1) a circle 2) 2) a line perpendicular to the road 3) a line parallel to the road 4) two parallel lines 3) 5 Which equation represents the locus of points 4 units from the origin? 1) x = 4 2 2 4) 2) x + y = 4 3) x + y = 16 4) x 2 + y 2 = 16 2 In the accompanying diagram, line 1 is parallel to line 2 . 6 Which equation represents the locus of all points 5 units below the x-axis? 1) x = −5 2) x = 5 3) y = −5 4) y = 5 Which term describes the locus of all points that are equidistant from line 1 and line 2 ? 1) line 2) circle 3) point 4) rectangle 1 Regents Exam Questions G.GPE.A.2: Locus 1 Name: ________________________ www.jmap.org 7 The locus of points equidistant from the points 9 Dan is sketching a map of the location of his house (4,−5) and (4,7) is the line whose equation is and his friend Matthew's house on a set of 1) y = 1 coordinate axes. -
Median and Altitude of a Triangle Goal: • to Use Properties of the Medians
Median and Altitude of a Triangle Goal: • To use properties of the medians of a triangle. • To use properties of the altitudes of a triangle. Median of a Triangle Median of a Triangle – a segment whose endpoints are the vertex of a triangle and the midpoint of the opposite side. Vertex Median Median of an Obtuse Triangle A D Point of concurrency “P” or centroid E P C B F Medians of a Triangle The medians of a triangle intersect at a point that is two-thirds of the distance from each vertex to the midpoint of the opposite side. A D If P is the centroid of ABC, then AP=2 AF E 3 P C CP=22CE and BP= BD B F 33 Example - Medians of a Triangle P is the centroid of ABC. PF 5 Find AF and AP A D E P C B F 5 Median of an Acute Triangle A Point of concurrency “P” or centroid E D P C B F Median of a Right Triangle A F Point of concurrency “P” or centroid E P C B D The three medians of an obtuse, acute, and a right triangle always meet inside the triangle. Altitude of a Triangle Altitude of a triangle – the perpendicular segment from the vertex to the opposite side or to the line that contains the opposite side A altitude C B Altitude of an Acute Triangle A Point of concurrency “P” or orthocenter P C B The point of concurrency called the orthocenter lies inside the triangle. Altitude of a Right Triangle The two legs are the altitudes A The point of concurrency called the orthocenter lies on the triangle. -
The Geometry of Diagonal Groups
The geometry of diagonal groups R. A. Baileya, Peter J. Camerona, Cheryl E. Praegerb, Csaba Schneiderc aUniversity of St Andrews, St Andrews, UK bUniversity of Western Australia, Perth, Australia cUniversidade Federal de Minas Gerais, Belo Horizonte, Brazil Abstract Diagonal groups are one of the classes of finite primitive permutation groups occurring in the conclusion of the O'Nan{Scott theorem. Several of the other classes have been described as the automorphism groups of geometric or combi- natorial structures such as affine spaces or Cartesian decompositions, but such structures for diagonal groups have not been studied in general. The main purpose of this paper is to describe and characterise such struct- ures, which we call diagonal semilattices. Unlike the diagonal groups in the O'Nan{Scott theorem, which are defined over finite characteristically simple groups, our construction works over arbitrary groups, finite or infinite. A diagonal semilattice depends on a dimension m and a group T . For m = 2, it is a Latin square, the Cayley table of T , though in fact any Latin square satisfies our combinatorial axioms. However, for m > 3, the group T emerges naturally and uniquely from the axioms. (The situation somewhat resembles projective geometry, where projective planes exist in great profusion but higher- dimensional structures are coordinatised by an algebraic object, a division ring.) A diagonal semilattice is contained in the partition lattice on a set Ω, and we provide an introduction to the calculus of partitions. Many of the concepts and constructions come from experimental design in statistics. We also determine when a diagonal group can be primitive, or quasiprimitive (these conditions turn out to be equivalent for diagonal groups). -
Lateral and Surface Area of Right Prisms 1 Jorge Is Trying to Wrap a Present That Is in a Box Shaped As a Right Prism
CHAPTER 11 You will need Lateral and Surface • a ruler A • a calculator Area of Right Prisms c GOAL Calculate lateral area and surface area of right prisms. Learn about the Math A prism is a polyhedron (solid whose faces are polygons) whose bases are congruent and parallel. When trying to identify a right prism, ask yourself if this solid could have right prism been created by placing many congruent sheets of paper on prism that has bases aligned one above the top of each other. If so, this is a right prism. Some examples other and has lateral of right prisms are shown below. faces that are rectangles Triangular prism Rectangular prism The surface area of a right prism can be calculated using the following formula: SA 5 2B 1 hP, where B is the area of the base, h is the height of the prism, and P is the perimeter of the base. The lateral area of a figure is the area of the non-base faces lateral area only. When a prism has its bases facing up and down, the area of the non-base faces of a figure lateral area is the area of the vertical faces. (For a rectangular prism, any pair of opposite faces can be bases.) The lateral area of a right prism can be calculated by multiplying the perimeter of the base by the height of the prism. This is summarized by the formula: LA 5 hP. Copyright © 2009 by Nelson Education Ltd. Reproduction permitted for classrooms 11A Lateral and Surface Area of Right Prisms 1 Jorge is trying to wrap a present that is in a box shaped as a right prism. -
Simple Polygons Scribe: Michael Goldwasser
CS268: Geometric Algorithms Handout #5 Design and Analysis Original Handout #15 Stanford University Tuesday, 25 February 1992 Original Lecture #6: 28 January 1991 Topics: Triangulating Simple Polygons Scribe: Michael Goldwasser Algorithms for triangulating polygons are important tools throughout computa- tional geometry. Many problems involving polygons are simplified by partitioning the complex polygon into triangles, and then working with the individual triangles. The applications of such algorithms are well documented in papers involving visibility, motion planning, and computer graphics. The following notes give an introduction to triangulations and many related definitions and basic lemmas. Most of the definitions are based on a simple polygon, P, containing n edges, and hence n vertices. However, many of the definitions and results can be extended to a general arrangement of n line segments. 1 Diagonals Definition 1. Given a simple polygon, P, a diagonal is a line segment between two non-adjacent vertices that lies entirely within the interior of the polygon. Lemma 2. Every simple polygon with jPj > 3 contains a diagonal. Proof: Consider some vertex v. If v has a diagonal, it’s party time. If not then the only vertices visible from v are its neighbors. Therefore v must see some single edge beyond its neighbors that entirely spans the sector of visibility, and therefore v must be a convex vertex. Now consider the two neighbors of v. Since jPj > 3, these cannot be neighbors of each other, however they must be visible from each other because of the above situation, and thus the segment connecting them is indeed a diagonal.