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6-5. Trapezoids and Kites. Warm Up. Lesson Presentation. Lesson Quiz. Holt Geometry. 6.5 Trapezoids and Kites. Warm Up Solve for x. 1. x 2 + 38 = 3 x 2 – 12 2. 137 + x = 180 3. 4. Find FE. 5 or –5. 43. 156. 6.5 Trapezoids and Kites. Objectives.

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**6-5**Trapezoids and Kites Warm Up Lesson Presentation Lesson Quiz Holt Geometry**6.5 Trapezoids and Kites**Warm Up Solve for x. 1.x2 + 38 = 3x2 – 12 2. 137 + x = 180 3. 4. Find FE. 5 or –5 43 156**6.5 Trapezoids and Kites**Objectives Use properties of kites to solve problems. Use properties of trapezoids to solve problems.**6.5 Trapezoids and Kites**Vocabulary kite trapezoid base of a trapezoid leg of a trapezoid base angle of a trapezoid isosceles trapezoid midsegment of a trapezoid**6.5 Trapezoids and Kites**A kiteis a quadrilateral with exactly two pairs of congruent consecutive sides.**6.5 Trapezoids and Kites**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**1 Make a Plan Understand the Problem 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 . 2 Example 1 Continued The answer will be the amount of wood Lucy has left after cutting the dowel.**6.5 Trapezoids and Kites**3 Solve Example 1 Continued N bisects JM. Pythagorean Thm. Pythagorean Thm.**6.5 Trapezoids and Kites**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.**6.5 Trapezoids and Kites**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. 4 Example 1 Continued Look Back**6.5 Trapezoids and Kites**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**1 Understand the Problem Check It Out! Example 1 Continued 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**Make a Plan 2 Check It Out! Example 1 Continued 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.**6.5 Trapezoids and Kites**3 Solve perimeter of PQRS = Check It Out! Example 1 Continued Pyth. Thm. Pyth. Thm.**6.5 Trapezoids and Kites**packages of binding Check It Out! Example 1 Continued 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.**6.5 Trapezoids and Kites**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. 4 Check It Out! Example 1 Continued Look Back**6.5 Trapezoids and Kites**Example 2A: Using Properties of Kites In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mBCD. Kite cons. sides ∆BCD is isos. 2 sides isos. ∆ isos. ∆base s CBF CDF mCBF = mCDF Def. of s Polygon Sum Thm. mBCD + mCBF + mCDF = 180°**6.5 Trapezoids and Kites**Example 2A Continued mBCD + mCBF + mCDF = 180° Substitute mCDF for mCBF. mBCD + mCBF+ mCDF= 180° Substitute 52 for mCBF. mBCD + 52°+ 52° = 180° Subtract 104 from both sides. mBCD = 76°**6.5 Trapezoids and Kites**Example 2B: Using Properties of Kites In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mABC. ADC ABC Kite one pair opp. s Def. of s mADC = mABC Polygon Sum Thm. mABC + mBCD + mADC + mDAB = 360° Substitute mABC for mADC. mABC + mBCD + mABC+ mDAB = 360°**6.5 Trapezoids and Kites**Example 2B Continued mABC + mBCD + mABC + mDAB = 360° mABC + 76°+ mABC + 54° = 360° Substitute. 2mABC = 230° Simplify. mABC = 115° Solve.**6.5 Trapezoids and Kites**Example 2C: Using Properties of Kites In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mFDA. CDA ABC Kite one pair opp. s mCDA = mABC Def. of s mCDF + mFDA = mABC Add. Post. 52° + mFDA = 115° Substitute. mFDA = 63° Solve.**6.5 Trapezoids and Kites**Check It Out! Example 2a In kite PQRS, mPQR = 78°, and mTRS = 59°. Find mQRT. Kite cons. sides ∆PQR is isos. 2 sides isos. ∆ RPQ PRQ isos. ∆base s mQPT = mQRT Def. of s**6.5 Trapezoids and Kites**Check It Out! Example 2a Continued mPQR + mQRP + mQPR = 180° Polygon Sum Thm. Substitute 78 for mPQR. 78° + mQRT+ mQPT = 180° 78° + mQRT + mQRT = 180° Substitute. 78° + 2mQRT = 180° Substitute. Subtract 78 from both sides. 2mQRT = 102° mQRT = 51° Divide by 2.**6.5 Trapezoids and Kites**Check It Out! Example 2b In kite PQRS, mPQR = 78°, and mTRS = 59°. Find mQPS. QPS QRS Kite one pair opp. s Add. Post. mQPS = mQRT + mTRS mQPS = mQRT + 59° Substitute. mQPS = 51° + 59° Substitute. mQPS = 110°**6.5 Trapezoids and Kites**Check It Out! Example 2c In kite PQRS, mPQR = 78°, and mTRS = 59°. Find each mPSR. Polygon Sum Thm. mSPT + mTRS + mRSP = 180° mSPT = mTRS Def. of s mTRS + mTRS + mRSP = 180° Substitute. 59° + 59°+ mRSP = 180° Substitute. Simplify. mRSP = 62°**6.5 Trapezoids and Kites**A trapezoidis 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 anglesof a trapezoid are two consecutive angles whose common side is a base.**6.5 Trapezoids and Kites**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**Reading Math Theorem 6-6-5 is a biconditional statement. So it is true both “forward” and “backward.”**6.5 Trapezoids and Kites**Example 3A: Using Properties of Isosceles Trapezoids Find mA. mC + mB = 180° Same-Side Int. s Thm. 100 + mB = 180 Substitute 100 for mC. mB = 80° Subtract 100 from both sides. A B Isos. trap. s base mA = mB Def. of s mA = 80° Substitute 80 for mB**6.5 Trapezoids and Kites**Example 3B: Using Properties of Isosceles Trapezoids KB = 21.9m and MF = 32.7. Find FB. Isos. trap. s base KJ = FM Def. of segs. KJ = 32.7 Substitute 32.7 for FM. Seg. Add. Post. KB + BJ = KJ 21.9 + BJ = 32.7 Substitute 21.9 for KB and 32.7 for KJ. BJ = 10.8 Subtract 21.9 from both sides.**6.5 Trapezoids and Kites**Example 3B Continued Same line. KFJ MJF Isos. trap. s base Isos. trap. legs SAS ∆FKJ ∆JMF CPCTC BKF BMJ Vert. s FBK JBM**6.5 Trapezoids and Kites**Example 3B Continued Isos. trap. legs AAS ∆FBK ∆JBM CPCTC FB = JB Def. of segs. FB = 10.8 Substitute 10.8 for JB.**6.5 Trapezoids and Kites**Check It Out! Example 3a Find mF. mF + mE = 180° Same-Side Int. s Thm. E H Isos. trap. s base mE = mH Def. of s mF + 49°= 180° Substitute 49 for mE. mF = 131° Simplify.**6.5 Trapezoids and Kites**Check It Out! Example 3b JN = 10.6, and NL = 14.8. Find KM. Isos. trap. s base Def. of segs. KM = JL JL = JN + NL Segment Add Postulate KM = JN + NL Substitute. KM = 10.6 + 14.8 = 25.4 Substitute and simplify.**6.5 Trapezoids and Kites**Example 4A: Applying Conditions for Isosceles Trapezoids Find the value of a so that PQRS is isosceles. Trap. with pair base s isosc. trap. S P mS = mP Def. of s Substitute 2a2 – 54 for mS and a2 + 27 for mP. 2a2 – 54 = a2 + 27 Subtract a2 from both sides and add 54 to both sides. a2 = 81 a = 9 or a = –9 Find the square root of both sides.**6.5 Trapezoids and Kites**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. AD = BC Substitute 12x – 11 for AD and 9x – 2 for BC. 12x – 11 = 9x – 2 Subtract 9x from both sides and add 11 to both sides. 3x = 9 x = 3 Divide both sides by 3.**6.5 Trapezoids and Kites**Check It Out! Example 4 Find the value of x so that PQST is isosceles. Trap. with pair base s isosc. trap. Q S mQ = mS Def. of s Substitute 2x2 + 19 for mQ and 4x2 – 13 for mS. 2x2 + 19 = 4x2 – 13 Subtract 2x2 and add 13 to both sides. 32 = 2x2 Divide by 2 and simplify. x = 4 or x = –4**6.5 Trapezoids and Kites**The midsegment of a trapezoidis 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**Example 5: Finding Lengths Using Midsegments Find EF. Trap. Midsegment Thm. Substitute the given values. Solve. EF = 10.75**6.5 Trapezoids and Kites**1 16.5 = (25 + EH) 2 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**6.5 Trapezoids and Kites**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. mLHJ 3. mPKL about 191.2 in. 81° 18°**6.5 Trapezoids and Kites**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

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