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6-5 Trapezoids and Kites Holt Geometry Warm Up Warm Up Lesson Presentation Lesson Presentation Lesson Quiz Lesson Quiz Slide 2 Warm Up Solve for x. 1. x 2 + 38 = 3x 2 12 2. 137 + x = 180 3. 4. Find FE. 5 or 5 43 156 6.5 Trapezoids and Kites Slide 3 Use properties of kites to solve problems. Use properties of trapezoids to solve problems. Objectives 6.5 Trapezoids and Kites Slide 4 kite trapezoid base of a trapezoid leg of a trapezoid base angle of a trapezoid isosceles trapezoid midsegment of a trapezoid Vocabulary 6.5 Trapezoids and Kites Slide 5 A kite is a quadrilateral with exactly two pairs of congruent consecutive sides. 6.5 Trapezoids and Kites Slide 6 Slide 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? 6.5 Trapezoids and Kites Slide 8 Example 1 Continued 1 Understand the Problem The answer will be the amount of wood Lucy has left after cutting the dowel. 2 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. 6.5 Trapezoids and Kites Slide 9 Solve 3 N bisects JM. Pythagorean Thm. Example 1 Continued 6.5 Trapezoids and Kites Slide 10 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. Example 1 Continued 6.5 Trapezoids and Kites Slide 11 Look Back 4 Example 1 Continued 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. 6.5 Trapezoids and Kites Slide 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? 6.5 Trapezoids and Kites Slide 13 Check It Out! Example 1 Continued 1 Understand the Problem The answer has two parts. the total length of binding Daryl needs the number of packages of binding Daryl must buy 6.5 Trapezoids and Kites Slide 14 2 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. Check It Out! Example 1 Continued 6.5 Trapezoids and Kites Slide 15 Solve 3 Pyth. Thm. Check It Out! Example 1 Continued perimeter of PQRS = 6.5 Trapezoids and Kites Slide 16 Daryl needs approximately 191.3 inches of binding. One package of binding contains 2 yards, or 72 inches. In order to have enough, Daryl must buy 3 packages of binding. Check It Out! Example 1 Continued packages of binding 6.5 Trapezoids and Kites Slide 17 Look Back 4 Check It Out! Example 1 Continued 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. 6.5 Trapezoids and Kites Slide 18 Kite cons. sides Example 2A: Using Properties of Kites In kite ABCD, mDAB = 54, and mCDF = 52. Find mBCD. BCD is isos. 2 sides isos. isos. base s Def. of s Polygon Sum Thm. CBF CDF mCBF = mCDF mBCD + mCBF + mCDF = 180 6.5 Trapezoids and Kites Slide 19 Example 2A Continued Substitute mCDF for mCBF. Substitute 52 for mCBF. Subtract 104 from both sides. mBCD + mCBF + mCDF = 180 mBCD + 52 + 52 = 180 mBCD = 76 mBCD + mCBF + mCDF = 180 6.5 Trapezoids and Kites Slide 20 Kite one pair opp. s Example 2B: Using Properties of Kites Def. of s Polygon Sum Thm. In kite ABCD, mDAB = 54, and mCDF = 52. Find mABC. ADC ABC mADC = mABC mABC + mBCD + mADC + mDAB = 360 mABC + mBCD + mABC + mDAB = 360 Substitute mABC for mADC. 6.5 Trapezoids and Kites Slide 21 Example 2B Continued Substitute. Simplify. mABC + mBCD + mABC + mDAB = 360 mABC + 76 + mABC + 54 = 360 2mABC = 230 mABC = 115 Solve. 6.5 Trapezoids and Kites Slide 22 Kite one pair opp. s Example 2C: Using Properties of Kites Def. of s Add. Post. Substitute. Solve. In kite ABCD, mDAB = 54, and mCDF = 52. Find mFDA. CDA ABC mCDA = mABC mCDF + mFDA = mABC 52 + mFDA = 115 mFDA = 63 6.5 Trapezoids and Kites Slide 23 Check It Out! Example 2a In kite PQRS, mPQR = 78, and mTRS = 59. Find mQRT. Kite cons. sides PQR is isos. 2 sides isos. isos. base s Def. of s RPQ PRQ mQPT = mQRT 6.5 Trapezoids and Kites Slide 24 Check It Out! Example 2a Continued Polygon Sum Thm. Substitute 78 for mPQR. mPQR + mQRP + mQPR = 180 78 + mQRT + mQPT = 180 Substitute. 78 + mQRT + mQRT = 180 78 + 2mQRT = 180 2mQRT = 102 mQRT = 51 Substitute. Subtract 78 from both sides. Divide by 2. 6.5 Trapezoids and Kites Slide 25 Check It Out! Example 2b In kite PQRS, mPQR = 78, and mTRS = 59. Find mQPS. Kite one pair opp. s Add. Post. Substitute. QPS QRS mQPS = mQRT + mTRS mQPS = mQRT + 59 mQPS = 51 + 59 mQPS = 110 6.5 Trapezoids and Kites Slide 26 Check It Out! Example 2c Polygon Sum Thm. Def. of s Substitute. Simplify. In kite PQRS, mPQR = 78, and mTRS = 59. Find each mPSR. mSPT + mTRS + mRSP = 180 mSPT = mTRS mTRS + mTRS + mRSP = 180 59 + 59 + mRSP = 180 mRSP = 62 6.5 Trapezoids and Kites Slide 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. 6.5 Trapezoids and Kites Slide 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. 6.5 Trapezoids and Kites Slide 29 Slide 30 Theorem 6-6-5 is a biconditional statement. So it is true both forward and backward. Reading Math 6.5 Trapezoids and Kites Slide 31 Isos. trap. s base Example 3A: Using Properties of Isosceles Trapezoids Find mA. Same-Side Int. s Thm. Substitute 100 for mC. Subtract 100 from both sides. Def. of s Substitute 80 for mB mC + mB = 180 100 + mB = 180 mB = 80 A B mA = mB mA = 80 6.5 Trapezoids and Kites Slide 32 Example 3B: Using Properties of Isosceles Trapezoids KB = 21.9m and MF = 32.7. Find FB. Isos. trap. s 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 6.5 Trapezoids and Kites Slide 33 Example 3B Continued Same line. Isos. trap. s base Isos. trap. legs SAS CPCTC Vert. s KFJ MJF BKF BMJ FBK JBM FKJ JMF 6.5 Trapezoids and Kites Slide 34 Isos. trap. legs AAS CPCTC Def. of segs. Substitute 10.8 for JB. Example 3B Continued FBK JBM FB = JB FB = 10.8 6.5 Trapezoids and Kites Slide 35 Isos. trap. s base Same-Side Int. s Thm. Def. of s Substitute 49 for mE. mF + mE = 180 E H mE = mH mF = 131 mF + 49 = 180 Simplify. Check It Out! Example 3a Find mF. 6.5 Trapezoids and Kites Slide 36 Check It Out! Example 3b JN = 10.6, and NL = 14.8. Find KM. Def. of segs. Segment Add Postulate Substitute. Substitute and simplify. Isos. trap. s base KM = JL JL = JN + NL KM = JN + NL KM = 10.6 + 14.8 = 25.4 6.5 Trapezoids and Kites Slide 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 s isosc. trap. Def. of s Substitute 2a 2 54 for mS and a 2 + 27 for mP. Subtract a 2 from both sides and add 54 to both sides. Find the square root of both sides. S PS P mS = mP 2a 2 54 = a 2 + 27 a 2 = 81 6.5 Trapezoids and Kites Slide 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 6.5 Trapezoids and Kites Slide 39 Check It Out! Example 4 Find the value of x so that PQST is isosceles. Subtract 2x 2 and add 13 to both sides. x = 4 or x = 4 Divide by 2 and simplify. Trap. with pair base s isosc. trap. Q SQ S Def. of s Substitute 2x 2 + 19 for mQ and 4x 2 13 for mS. mQ = mS 2x 2 + 19 = 4x 2 13 32 = 2x 2 6.5 Trapezoids and Kites Slide 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. 6.5 Trapezoids and Kites Slide 41 Slide 42 Example 5: Finding Lengths Using Midsegments Find EF. Trap. Midsegment Thm. Substitute the given values. Solve. EF = 10.75 6.5 Trapezoids and Kites Slide 43 Check It Out! Example 5 Find EH. Trap. Midsegment Thm. Substitute the given values. Simplify. Multiply both sides by 2. 33 = 25 + EH Subtract 25 from both sides. 13 = EH 1 16.5 = ( 25 + EH ) 2 6.5 Trapezoids and Kites Slide 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? In kite HJKL, mKLP = 72, and mHJP = 49.5. Find each measure. 2. mLHJ3. mPKL about 191.2 in. 8118 6.5 Trapezoids and Kites Slide 45 Lesson Quiz: Part II Use the diagram for Items 4 and 5. 4. mWZY = 61. Find mWXY. 5. XV = 4.6, and WY = 14.2. Find VZ. 6. Find LP. 119 9.6 18 6.5 Trapezoids and Kites