Isosceles Triangles on a Geoboard
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Section 5.5 Right Triangle Trigonometry 385
Section 5.5 Right Triangle Trigonometry 385 Section 5.5 Right Triangle Trigonometry In section 5.3 we were introduced to the sine and cosine function as ratios of the sides of a triangle drawn inside a circle, and spent the rest of that section discussing the role of those functions in finding points on the circle. In this section, we return to the triangle, and explore the applications of the trigonometric functions to right triangles where circles may not be involved. Recall that we defined sine and cosine as (x, y) y sin( θ ) = r r y x cos( θ ) = θ r x Separating the triangle from the circle, we can make equivalent but more general definitions of the sine, cosine, and tangent on a right triangle. On the right triangle, we will label the hypotenuse as well as the side opposite the angle and the side adjacent (next to) the angle. Right Triangle Relationships Given a right triangle with an angle of θ opposite sin( θ) = hypotenuse hypotenuse opposite adjacent cos( θ) = hypotenuse θ opposite adjacent tan( θ) = adjacent A common mnemonic for remembering these relationships is SohCahToa, formed from the first letters of “Sine is opposite over hypotenuse, Cosine is adjacent over hypotenuse, Tangent is opposite over adjacent.” 386 Chapter 5 Example 1 Given the triangle shown, find the value for cos( α) . The side adjacent to the angle is 15, and the 17 hypotenuse of the triangle is 17, so 8 adjacent 15 cos( α) = = α hypotenuse 17 15 When working with general right triangles, the same rules apply regardless of the orientation of the triangle. -
Geometry Honors Mid-Year Exam Terms and Definitions Blue Class 1
Geometry Honors Mid-Year Exam Terms and Definitions Blue Class 1. Acute angle: Angle whose measure is greater than 0° and less than 90°. 2. Adjacent angles: Two angles that have a common side and a common vertex. 3. Alternate interior angles: A pair of angles in the interior of a figure formed by two lines and a transversal, lying on alternate sides of the transversal and having different vertices. 4. Altitude: Perpendicular segment from a vertex of a triangle to the opposite side or the line containing the opposite side. 5. Angle: A figure formed by two rays with a common endpoint. 6. Angle bisector: Ray that divides an angle into two congruent angles and bisects the angle. 7. Base Angles: Two angles not included in the legs of an isosceles triangle. 8. Bisect: To divide a segment or an angle into two congruent parts. 9. Coincide: To lie on top of the other. A line can coincide another line. 10. Collinear: Lying on the same line. 11. Complimentary: Two angle’s whose sum is 90°. 12. Concave Polygon: Polygon in which at least one interior angle measures more than 180° (at least one segment connecting two vertices is outside the polygon). 13. Conclusion: A result of summary of all the work that has been completed. The part of a conditional statement that occurs after the word “then”. 14. Congruent parts: Two or more parts that only have the same measure. In CPCTC, the parts of the congruent triangles are congruent. 15. Congruent triangles: Two triangles are congruent if and only if all of their corresponding parts are congruent. -
The Construction, by Euclid, of the Regular Pentagon
THE CONSTRUCTION, BY EUCLID, OF THE REGULAR PENTAGON Jo˜ao Bosco Pitombeira de CARVALHO Instituto de Matem´atica, Universidade Federal do Rio de Janeiro, Cidade Universit´aria, Ilha do Fund˜ao, Rio de Janeiro, Brazil. e-mail: [email protected] ABSTRACT We present a modern account of Ptolemy’s construction of the regular pentagon, as found in a well-known book on the history of ancient mathematics (Aaboe [1]), and discuss how anachronistic it is from a historical point of view. We then carefully present Euclid’s original construction of the regular pentagon, which shows the power of the method of equivalence of areas. We also propose how to use the ideas of this paper in several contexts. Key-words: Regular pentagon, regular constructible polygons, history of Greek mathe- matics, equivalence of areas in Greek mathematics. 1 Introduction This paper presents Euclid’s construction of the regular pentagon, a highlight of the Elements, comparing it with the widely known construction of Ptolemy, as presented by Aaboe [1]. This gives rise to a discussion on how to view Greek mathematics and shows the care on must have when adopting adapting ancient mathematics to modern styles of presentation, in order to preserve not only content but the very way ancient mathematicians thought and viewed mathematics. 1 The material here presented can be used for several purposes. First of all, in courses for prospective teachers interested in using historical sources in their classrooms. In several places, for example Brazil, the history of mathematics is becoming commonplace in the curricula of courses for prospective teachers, and so one needs materials that will awaken awareness of the need to approach ancient mathematics as much as possible in its own terms, and not in some pasteurized downgraded versions. -
Refer to the Figure. 1. If Name Two Congruent Angles. SOLUTION: Isosceles Triangle Theorem States That If Two Sides of T
4-6 Isosceles and Equilateral Triangles Refer to the figure. 1. If name two congruent angles. SOLUTION: Isosceles Triangle Theorem states that if two sides of the triangle are congruent, then the angles opposite those sides are congruent. Therefore, in triangle ABC, ANSWER: BAC and BCA 2. If EAC ECA, name two congruent segments. SOLUTION: Converse of Isosceles Triangle Theorem states that if two angles of a triangle are congruent, then the sides opposite those angles are congruent. Therefore, in triangle EAC, ANSWER: Find each measure. 3. FH SOLUTION: By the Triangle Angle-Sum Theorem, Since the measures of all the three angles are 60°; the triangle must be equiangular. All the equiangular triangles are equilateral. Therefore, FH = GH = 12. ANSWER: 12 eSolutions4. m ManualMRP - Powered by Cognero Page 1 SOLUTION: Since all the sides are congruent, is an equilateral triangle. Each angle of an equilateral triangle measures 60°. Therefore, m MRP = 60°. ANSWER: 60 SENSE-MAKING Find the value of each variable. 5. SOLUTION: In the figure, . Therefore, triangle RST is an isosceles triangle. By the Converse of Isosceles Triangle Theorem, That is, . ANSWER: 12 6. SOLUTION: In the figure, Therefore, triangle WXY is an isosceles triangle. By the Isosceles Triangle Theorem, . ANSWER: 16 7. PROOF Write a two-column proof. Given: is isosceles; bisects ABC. Prove: SOLUTION: ANSWER: 8. ROLLER COASTERS The roller coaster track appears to be composed of congruent triangles. A portion of the track is shown. a. If and are perpendicular to is isosceles with base , and prove that b. If VR = 2.5 meters and QR = 2 meters, find the distance between and Explain your reasoning. -
Geometrygeometry
Park Forest Math Team Meet #3 GeometryGeometry Self-study Packet Problem Categories for this Meet: 1. Mystery: Problem solving 2. Geometry: Angle measures in plane figures including supplements and complements 3. Number Theory: Divisibility rules, factors, primes, composites 4. Arithmetic: Order of operations; mean, median, mode; rounding; statistics 5. Algebra: Simplifying and evaluating expressions; solving equations with 1 unknown including identities Important Information you need to know about GEOMETRY… Properties of Polygons, Pythagorean Theorem Formulas for Polygons where n means the number of sides: • Exterior Angle Measurement of a Regular Polygon: 360÷n • Sum of Interior Angles: 180(n – 2) • Interior Angle Measurement of a regular polygon: • An interior angle and an exterior angle of a regular polygon always add up to 180° Interior angle Exterior angle Diagonals of a Polygon where n stands for the number of vertices (which is equal to the number of sides): • • A diagonal is a segment that connects one vertex of a polygon to another vertex that is not directly next to it. The dashed lines represent some of the diagonals of this pentagon. Pythagorean Theorem • a2 + b2 = c2 • a and b are the legs of the triangle and c is the hypotenuse (the side opposite the right angle) c a b • Common Right triangles are ones with sides 3, 4, 5, with sides 5, 12, 13, with sides 7, 24, 25, and multiples thereof—Memorize these! Category 2 50th anniversary edition Geometry 26 Y Meet #3 - January, 2014 W 1) How many cm long is segment 6 XY ? All measurements are in centimeters (cm). -
Similar Quadrilaterals Cui, Kadaveru, Lee, Maheshwari Page 1
Similar Quadrilaterals Cui, Kadaveru, Lee, Maheshwari Page 1 Similar Quadrilaterals Authors Guangqi Cui, Akshaj Kadaveru, Joshua Lee, Sagar Maheshwari Special thanks to Cosmin Pohoata and the AMSP Cornell 2014 Geometric Proofs Class B0 C0 B A A0 D0 C D Additional thanks to Justin Stevens and David Altizio for the LATEX Template Similar Quadrilaterals Cui, Kadaveru, Lee, Maheshwari Page 2 Contents 1 Introduction 3 2 Interesting Property 4 3 Example Problems 5 4 Practice Problems 11 Similar Quadrilaterals Cui, Kadaveru, Lee, Maheshwari Page 3 1 Introduction Similar quadrilaterals are a very useful but relatively unknown tool used to solve olympiad geometry problems. It usually goes unnoticed due to the confinement of geometric education to the geometry of the triangle and other conventional methods of problem solving. Also, it is only in very special cases where pairs of similar quadrilaterals exist, and proofs using these qualities usually shorten what would have otherwise been an unnecessarily long proof. The most common method of finding such quadrilaterals involves finding one pair of adjacent sides with identical ratios, and three pairs of congruent angles. We will call this SSAAA Similarity. 0 0 0 0 Example 1.1. (SSAAA Similarity) Two quadrilaterals ABCD and A B C D satisfy \A = AB BC A0, B = B0, C = C0, and = . Show that ABCD and A0B0C0D0 are similar. \ \ \ \ \ A0B0 B0C0 B0 C0 B A A0 D0 C D 0 0 0 0 0 0 Solution. Notice 4ABC and 4A B C are similar from SAS similarity. Therefore \C A D = 0 0 0 0 0 0 0 0 0 0 \A − \B A C = \A − \BAC = \CAD. -
Right Triangles and the Pythagorean Theorem Related?
Activity Assess 9-6 EXPLORE & REASON Right Triangles and Consider △ ABC with altitude CD‾ as shown. the Pythagorean B Theorem D PearsonRealize.com A 45 C 5√2 I CAN… prove the Pythagorean Theorem using A. What is the area of △ ABC? Of △ACD? Explain your answers. similarity and establish the relationships in special right B. Find the lengths of AD‾ and AB‾ . triangles. C. Look for Relationships Divide the length of the hypotenuse of △ ABC VOCABULARY by the length of one of its sides. Divide the length of the hypotenuse of △ACD by the length of one of its sides. Make a conjecture that explains • Pythagorean triple the results. ESSENTIAL QUESTION How are similarity in right triangles and the Pythagorean Theorem related? Remember that the Pythagorean Theorem and its converse describe how the side lengths of right triangles are related. THEOREM 9-8 Pythagorean Theorem If a triangle is a right triangle, If... △ABC is a right triangle. then the sum of the squares of the B lengths of the legs is equal to the square of the length of the hypotenuse. c a A C b 2 2 2 PROOF: SEE EXAMPLE 1. Then... a + b = c THEOREM 9-9 Converse of the Pythagorean Theorem 2 2 2 If the sum of the squares of the If... a + b = c lengths of two sides of a triangle is B equal to the square of the length of the third side, then the triangle is a right triangle. c a A C b PROOF: SEE EXERCISE 17. Then... △ABC is a right triangle. -
"Perfect Squares" on the Grid Below, You Can See That Each Square Has Sides with Integer Length
April 09, 2012 "Perfect Squares" On the grid below, you can see that each square has sides with integer length. The area of a square is the square of the length of a side. A = s2 The square of an integer is a perfect square! April 09, 2012 Square Roots and Irrational Numbers The inverse of squaring a number is finding a square root. The square-root radical, , indicates the nonnegative square root of a number. The number underneath the square root (radical) sign is called the radicand. Ex: 16 = 121 = 49 = 144 = The square of an integer results in a perfect square. Since squaring a number is multiplying it by itself, there are two integer values that will result in the same perfect square: the positive integer and its opposite. 2 Ex: If x = (perfect square), solutions would be x and (-x)!! Ex: a2 = 25 5 and -5 would make this equation true! April 09, 2012 If an integer is NOT a perfect square, its square root is irrational!!! Remember that an irrational number has a decimal form that is non- terminating, non-repeating, and thus cannot be written as a fraction. For an integer that is not a perfect square, you can estimate its square root on a number line. Ex: 8 Since 22 = 4, and 32 = 9, 8 would fall between integers 2 and 3 on a number line. -2 -1 0 1 2 3 4 5 April 09, 2012 Approximate where √40 falls on the number line: Approximate where √192 falls on the number line: April 09, 2012 Topic Extension: Simplest Radical Form Simplify the radicand so that it has no more perfect-square factors. -
Introduction to Geometry" ©2014 Aops Inc
Excerpt from "Introduction to Geometry" ©2014 AoPS Inc. www.artofproblemsolving.com CHAPTER 12. CIRCLES AND ANGLES (a) Consider the circle with diameter OP. Call this circle . Why must hit O in at least two di↵erent C C points? (b) Why is it impossible for to hit O in three di↵erent points? Hints: 530 C (c) Let the points where hits O be A and B. Prove that \PAO = \PBO = 90 . C ◦ (d) Prove that PA and PB are tangent to O. (e)? Now for the tricky part – proving that these are the only two tangents. Suppose D is on O such that PD is tangent to O. Why must D be on ? Hints: 478 C (f) Why does the previous part tell us that PA! and PB! are the only lines through P tangent to O? 12.3.6? O is tangent to all four sides of rhombus ABCD, AC = 24, and AB = 15. (a) Prove that AC and BD meet at O (i.e., prove that the intersection of the diagonals of ABCD is the center of the circle.) Hints: 439 (b) What is the area of O? Hints: 508 12.4 Problems Problems Problem 12.19: A quadrilateral is said to be a cyclic quadrilateral if a circle can be drawn that passes through all four of its vertices. Prove that if ABCD is a cyclic quadrilateral, then \A + \C = 180◦. C Problem 12.20: In the figure, PC is tangent to the circle and PD 70◦ bisects \CPE. Furthermore, CD = 70◦, DE = 50◦, and \DQE = P B Q D 40◦. -
Day 7 Classwork for Constructing a Regular Hexagon
Classwork for Constructing a Regular Hexagon, Inscribed Shapes, and Isosceles Triangle I. Review 1) Construct a perpendicular bisector to 퐴퐵̅̅̅̅. 2) Construct a perpendicular line through point P. P • A B 3) Construct an equilateral triangle with length CD. (LT 1c) C D II. Construct a Regular Hexagon (LT 1c) Hexagon A six-sided polygon Regular Hexagon A six-sided polygon with all sides the same length and all interior angles the same measure Begin by constructing an equilateral with the segment below. We can construct a regular hexagon by observing that it is made up of Then construct 5 more equilateral triangles with anchor point on point R 6 adjacent (sharing a side) equilateral for one arc and anchor point on the most recent point of intersection for the second ( intersecting) arc. triangles. R III. Construct a Regular Hexagon Inscribed in a Circle Inscribed (in a Circle) All vertices of the inscribed shape are points on the circle Inscribed Hexagon Hexagon with all 6 vertex points on a circle Construct a regular hexagon with side length 퐴퐵̅̅̅̅: A B 1) Copy the length of 퐴퐵̅̅̅̅ onto the compass. 2) Place metal tip of compass on point C and construct a circle. 3) Keep the same length on the compass. 4) Mark any point (randomly) on the circle. C 5) Place the metal tip on the randomly marked • point and mark an arc on the circle. 6) Lift the compass and place the metal tip on the last arc mark, and mark new arc on the circle. 7) Repeat until the arc mark lands on the original point. -
Math 1330 - Section 4.1 Special Triangles
Math 1330 - Section 4.1 Special Triangles In this section, we’ll work with some special triangles before moving on to defining the six trigonometric functions. Two special triangles are 30 − 60 − 90 triangles and 45 − 45 − 90 triangles. With little additional information, you should be able to find the lengths of all sides of one of these special triangles. First we’ll review some conventions when working with triangles. We label angles with capital letters and sides with lower case letters. We’ll use the same letter to refer to an angle and the side opposite it, although the angle will be a capital letter and the side will be lower case. In this section, we will work with right triangles. Right triangles have one angle which measures 90 degrees, and two other angles whose sum is 90 degrees. The side opposite the right angle is called the hypotenuse, and the other two sides are called the legs of the triangle. A labeled right triangle is shown below. B c a A C b Recall: When you are working with a right triangle, you can use the Pythagorean Theorem to help you find the length of an unknown side. In a right triangle ABC with right angle C, 2 2 2 a + b = c . Example: In right triangle ABC with right angle C, if AB = 12 and AC = 8, find BC. 1 Special Triangles In a 30 − 60 − 90 triangle, the lengths of the sides are proportional. If the shorter leg (the side opposite the 30 degree angle) has length a, then the longer leg has length 3a and the hypotenuse has length 2a. -
Brahmagupta's Derivation of the Area of a Cyclic Quadrilateral
Brahmagupta’s derivation of the area of a cyclic quadrilateral Satyanad Kichenassamy To cite this version: Satyanad Kichenassamy. Brahmagupta’s derivation of the area of a cyclic quadrilateral. Historia Mathematica, Elsevier, 2010, 37 (1), pp.28-61. 10.1016/j.hm.2009.08.004. halshs-00443604 HAL Id: halshs-00443604 https://halshs.archives-ouvertes.fr/halshs-00443604 Submitted on 12 Mar 2018 HAL is a multi-disciplinary open access L’archive ouverte pluridisciplinaire HAL, est archive for the deposit and dissemination of sci- destinée au dépôt et à la diffusion de documents entific research documents, whether they are pub- scientifiques de niveau recherche, publiés ou non, lished or not. The documents may come from émanant des établissements d’enseignement et de teaching and research institutions in France or recherche français ou étrangers, des laboratoires abroad, or from public or private research centers. publics ou privés. Historia Mathematica 37 (2010) 28{61 http://dx.doi.org/10.1016/j.hm.2009.08.004 Brahmagupta's derivation of the area of a cyclic quadrilateral Satyanad Kichenassamy1 Abstract. This paper shows that Propositions XII.21{27 of Brahmagupta's Br¯ahma- sphut.asiddh¯anta (628 a.d.) constitute a coherent mathematical discourse leading to the expression of the area of a cyclic quadrilateral in terms of its sides. The radius of the circumcircle is determined by considering two auxiliary quadrilaterals. Observing that a cyclic quadrilateral is split by a diagonal into two triangles with the same circumcenter and the same circumradius, the result follows, using the tools available to Brahmagupta. The expression for the diagonals (XII.28) is a consequence.