Geometry unit 6.6
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This document provides information about properties of kites and trapezoids. It includes examples of using properties of kites to solve problems involving finding missing angle measures. Examples also demonstrate using properties of isosceles trapezoids, such as the fact that base angles are congruent and legs are congruent. The document concludes with the trapezoid midsegment theorem and examples applying it to find missing segment lengths.
Geometry unit 6.6
- 1. UUNNIITT 66..66 TTRRAAPPEEZZOOIIDDSS AANNDD KKIITTEESS
- 2. Poperties of Kites and Trapezoids Warm Up Solve for x. 1. x2 + 38 = 3x2 – 12 2. 137 + x = 180 3. 4. Find FE. 5 or –5 43 156
- 3. Objectives Use properties of kites to solve problems. Use properties of trapezoids to solve problems.
- 4. Vocabulary kite trapezoid base of a trapezoid leg of a trapezoid base angle of a trapezoid isosceles trapezoid midsegment of a trapezoid
- 5. A kite is a quadrilateral with exactly two pairs of congruent consecutive sides.
- 7. Example 1: Problem-Solving Application Lucy is framing a kite with wooden dowels. She uses two dowels that measure 18 cm, one dowel that measures 30 cm, and two dowels that measure 27 cm. To complete the kite, she needs a dowel to place along . She has a dowel that is 36 cm long. About how much wood will she have left after cutting the last dowel?
- 8. Example 1 Continued 11 Understand the Problem The answer will be the amount of wood Lucy has left after cutting the dowel. 22 Make a Plan The diagonals of a kite are perpendicular, so the four triangles are right triangles. Let N represent the intersection of the diagonals. Use the Pythagorean Theorem and the properties of kites to find , and . Add these lengths to find the length of .
- 9. 33 Solve Example 1 Continued N bisects JM. Pythagorean Thm. Pythagorean Thm.
- 10. Example 1 Continued Lucy needs to cut the dowel to be 32.4 cm long. The amount of wood that will remain after the cut is, 36 – 32.4 » 3.6 cm Lucy will have 3.6 cm of wood left over after the cut.
- 11. Example 1 Continued 44 Look Back To estimate the length of the diagonal, change the side length into decimals and round. , and . The length of the diagonal is approximately 10 + 22 = 32. So the wood remaining is approximately 36 – 32 = 4. So 3.6 is a reasonable answer.
- 12. Check It Out! Example 1 What if...? Daryl is going to make a kite by doubling all the measures in the kite. What is the total amount of binding needed to cover the edges of his kite? How many packages of binding must Daryl buy?
- 13. Check It Out! Example 1 Continued 11 Understand the Problem The answer has two parts. • the total length of binding Daryl needs • the number of packages of binding Daryl must buy
- 14. Check It Out! Example 1 Continued 22 Make a Plan The diagonals of a kite are perpendicular, so the four triangles are right triangles. Use the Pythagorean Theorem and the properties of kites to find the unknown side lengths. Add these lengths to find the perimeter of the kite.
- 15. Check It Out! Example 1 Continued 33 Solve Pyth. Thm. Pyth. Thm. perimeter of PQRS =
- 16. Check It Out! Example 1 Continued Daryl needs approximately 191.3 inches of binding. One package of binding contains 2 yards, or 72 inches. packages of binding In order to have enough, Daryl must buy 3 packages of binding.
- 17. Check It Out! Example 1 Continued 44 Look Back To estimate the perimeter, change the side lengths into decimals and round. , and . The perimeter of the kite is approximately 2(54) + 2 (41) = 190. So 191.3 is a reasonable answer.
- 18. Example 2A: Using Properties of Kites In kite ABCD, mÐDAB = 54°, and mÐCDF = 52°. Find mÐBCD. Kite cons. sides @ ΔBCD is isos. 2 @ sides isos. Δ isos. Δ base Ðs @ Def. of @ Ð s Polygon Ð Sum Thm. ÐCBF @ ÐCDF mÐCBF = mÐCDF mÐBCD + mÐCBF + mÐCDF = 180°
- 19. Example 2A Continued Substitute mÐCDF for mÐCBF. Substitute 52 for mÐCBF. Subtract 104 from both sides. mÐBCD + mÐCBF + mÐCDF = 180° mÐBCD + mÐCBF + mÐCDF = 180° mÐBCD + 52° + 52° = 180° mÐBCD = 76°
- 20. Example 2B: Using Properties of Kites In kite ABCD, mÐDAB = 54°, and mÐCDF = 52°. Find mÐABC. Kite one pair opp. Ðs @ Def. of @ Ðs Polygon Ð Sum Thm. ÐADC @ ÐABC mÐADC = mÐABC mÐABC + mÐBCD + mÐADC + mÐDAB = 360° Substitute mÐABC for mÐADC. mÐABC + mÐBCD + mÐABC + mÐDAB = 360°
- 21. Example 2B Continued mÐABC + mÐBCD + mÐABC + mÐDAB = 360° mÐABC + 76° + mÐABC + 54° = 360° Substitute. Simplify. 2mÐABC = 230° mÐABC = 115° Solve.
- 22. Example 2C: Using Properties of Kites In kite ABCD, mÐDAB = 54°, and mÐCDF = 52°. Find mÐFDA. Kite one pair opp. Ðs @ Def. of @ Ðs Ð Add. Post. Substitute. Solve. ÐCDA @ ÐABC mÐCDA = mÐABC mÐCDF + mÐFDA = mÐABC 52° + mÐFDA = 115° mÐFDA = 63°
- 23. Check It Out! Example 2a In kite PQRS, mÐPQR = 78°, and mÐTRS = 59°. Find mÐQRT. Kite ® cons. sides @ ΔPQR is isos. 2 @ sides ® isos. Δ isos. Δ ® base Ðs @ Def. of @ Ðs ÐRPQ @ ÐPRQ mÐQPT = mÐQRT
- 24. Check It Out! Example 2a Continued Polygon Ð Sum Thm. Substitute 78 for mÐPQR. mÐPQR + mÐQRP + mÐQPR = 180° 78° + mÐQRT + mÐQPT = 180° 78° + mÐQRT + mÐQRT = 180° Substitute. 78° + 2mÐQRT = 180° 2mÐQRT = 102° mÐQRT = 51° Substitute. Subtract 78 from both sides. Divide by 2.
- 25. Check It Out! Example 2b In kite PQRS, mÐPQR = 78°, and mÐTRS = 59°. Find mÐQPS. Kite one pair opp. Ðs @ Ð Add. Post. Substitute. Substitute. ÐQPS @ ÐQRS mÐQPS = mÐQRT + mÐTRS mÐQPS = mÐQRT + 59° mÐQPS = 51° + 59° mÐQPS = 110°
- 26. Check It Out! Example 2c Polygon Ð Sum Thm. Def. of @ Ðs Substitute. Substitute. Simplify. In kite PQRS, mÐPQR = 78°, and mÐTRS = 59°. Find each mÐPSR. mÐSPT + mÐTRS + mÐRSP = 180° mÐSPT = mÐTRS mÐTRS + mÐTRS + mÐRSP = 180° 59° + 59° + mÐRSP = 180° mÐRSP = 62°
- 27. A trapezoid is a quadrilateral with exactly one pair of parallel sides. Each of the parallel sides is called a base. The nonparallel sides are called legs. Base angles of a trapezoid are two consecutive angles whose common side is a base.
- 28. If the legs of a trapezoid are congruent, the trapezoid is an isosceles trapezoid. The following theorems state the properties of an isosceles trapezoid.
- 30. Reading Math Theorem 6-6-5 is a biconditional statement. So it is true both “forward” and “backward.”
- 31. Example 3A: Using Properties of Isosceles Trapezoids Isos. trap. Ðs base @ Find mÐA. Same-Side Int. Ðs Thm. Substitute 100 for mÐC. Subtract 100 from both sides. Def. of @ Ðs Substitute 80 for mÐB mÐC + mÐB = 180° 100 + mÐB = 180 mÐB = 80° ÐA @ ÐB mÐA = mÐB mÐA = 80°
- 32. Example 3B: Using Properties of Isosceles Trapezoids KB = 21.9m and MF = 32.7. Find FB. Isos. trap. Ðs base @ Def. of @ segs. Substitute 32.7 for FM. Seg. Add. Post. Substitute 21.9 for KB and 32.7 for KJ. Subtract 21.9 from both sides. KJ = FM KJ = 32.7 KB + BJ = KJ 21.9 + BJ = 32.7 BJ = 10.8
- 33. Example 3B Continued Same line. Isos. trap. Ðs base @ Isos. trap. legs @ SAS CPCTC Vert. Ðs @ ÐKFJ @ ÐMJF ΔFKJ @ ΔJMF ÐBKF @ ÐBMJ ÐFBK @ ÐJBM
- 34. Example 3B Continued Isos. trap. legs @ AAS CPCTC Def. of @ segs. Substitute 10.8 for JB. ΔFBK @ ΔJBM FB = JB FB = 10.8
- 35. Check It Out! Example 3a Same-Side Int. Ðs Thm. Isos. trap. Ðs base @ Def. of @ Ðs Substitute 49 for mÐE. mÐF + mÐE = 180° ÐE @ ÐH mÐE = mÐH mÐF + 49° = 180° mÐF = 131° Simplify. Find mÐF.
- 36. Check It Out! Example 3b JN = 10.6, and NL = 14.8. Find KM. Isos. trap. Ðs base @ Def. of @ segs. Segment Add Postulate Substitute. Substitute and simplify. KM = JL JL = JN + NL KM = JN + NL KM = 10.6 + 14.8 = 25.4
- 37. Example 4A: Applying Conditions for Isosceles Trapezoids Find the value of a so that PQRS is isosceles. a = 9 or a = –9 Trap. with pair base Ðs @ isosc. trap. Def. of @ Ðs Substitute 2a2 – 54 for mÐS and a2 + 27 for mÐP. Subtract a2 from both sides and add 54 to both sides. Find the square root of both sides. ÐS @ ÐP mÐS = mÐP 2a2 – 54 = a2 + 27 a2 = 81
- 38. Example 4B: Applying Conditions for Isosceles Trapezoids AD = 12x – 11, and BC = 9x – 2. Find the value of x so that ABCD is isosceles. Diags. @ isosc. trap. Def. of @ segs. Substitute 12x – 11 for AD and 9x – 2 for BC. Subtract 9x from both sides and add 11 to both sides. Divide both sides by 3. AD = BC 12x – 11 = 9x – 2 3x = 9 x = 3
- 39. Check It Out! Example 4 Find the value of x so that PQST is isosceles. Trap. with pair base ÐQ @ ÐS Ðs @ isosc. trap. Def. of @ Ðs Substitute 2x2 + 19 for mÐQ and 4x2 – 13 for mÐS. Subtract 2x2 and add 13 to both sides. mÐQ = mÐS 2x2 + 19 = 4x2 – 13 32 = 2x2 x = 4 or x = –4 Divide by 2 and simplify.
- 40. The midsegment of a trapezoid is the segment whose endpoints are the midpoints of the legs. In Lesson 5-1, you studied the Triangle Midsegment Theorem. The Trapezoid Midsegment Theorem is similar to it.
- 42. Example 5: Finding Lengths Using Midsegments Find EF. Trap. Midsegment Thm. Substitute the given values. EF = 10.75 Solve.
- 43. Check It Out! Example 5 Find EH. Trap. Midsegment Thm. Substitute the given values. Simplify. 16.5 = 1 (25 + EH) 2 33 = 25 + EH Multiply both sides by 2. 13 = EH Subtract 25 from both sides.
- 44. Lesson Quiz: Part I 1. Erin is making a kite based on the pattern below. About how much binding does Erin need to cover the edges of the kite? about 191.2 in. In kite HJKL, mÐKLP = 72°, and mÐHJP = 49.5°. Find each measure. 2. mÐLHJ 3. mÐPKL 81° 18°
- 45. Lesson Quiz: Part II Use the diagram for Items 4 and 5. 4. mÐWZY = 61°. Find mÐWXY. 119° 5. XV = 4.6, and WY = 14.2. Find VZ. 9.6 6. Find LP. 18
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