Properties of Kites 6 6 Properties of Kites

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Properties of Kites 6 -6 Properties of Kites and Trapezoids 6 -6 and Trapezoids

Properties of Kites 6 -6 Properties of Kites and Trapezoids 6 -6 and Trapezoids Warm Up Lesson Presentation Lesson Quiz Holt Mc. Dougal Geometry

6 -6 Properties of Kites and Trapezoids Warm Up Solve for x. 1. x

6 -6 Properties of Kites and Trapezoids Warm Up Solve for x. 1. x 2 + 38 = 3 x 2 – 12 5 or – 5 2. 137 + x = 180 43 3. 156 4. Find FE. Holt Mc. Dougal Geometry

6 -6 Properties of Kites and Trapezoids Objectives Use properties of kites to solve

6 -6 Properties of Kites and Trapezoids Objectives Use properties of kites to solve problems. Use properties of trapezoids to solve problems. Holt Mc. Dougal Geometry

6 -6 Properties of Kites and Trapezoids Vocabulary kite trapezoid base of a trapezoid

6 -6 Properties of Kites and Trapezoids Vocabulary kite trapezoid base of a trapezoid leg of a trapezoid base angle of a trapezoid isosceles trapezoid midsegment of a trapezoid Holt Mc. Dougal Geometry

6 -6 Properties of Kites and Trapezoids A kite is a quadrilateral with exactly

6 -6 Properties of Kites and Trapezoids A kite is a quadrilateral with exactly two pairs of congruent consecutive sides. Holt Mc. Dougal Geometry

6 -6 Properties of Kites and Trapezoids Holt Mc. Dougal Geometry

6 -6 Properties of Kites and Trapezoids Holt Mc. Dougal Geometry

6 -6 Properties of Kites and Trapezoids Example 1: Problem-Solving Application Lucy is framing

6 -6 Properties of Kites and Trapezoids 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. About how much wood will she need for the last dowel? Holt Mc. Dougal Geometry

6 -6 Properties of Kites and Trapezoids Example 1 Continued Solve: N bisects JM.

6 -6 Properties of Kites and Trapezoids Example 1 Continued Solve: N bisects JM. Pythagorean Thm. Holt Mc. Dougal Geometry

6 -6 Properties of Kites and Trapezoids Example 2 A: Using Properties of Kites

6 -6 Properties of Kites and Trapezoids Example 2 A: Using Properties of Kites In kite ABCD, m DAB = 54°, and m CDF = 52°. Find m BCD. Kite cons. sides ∆BCD is isos. 2 sides isos. ∆ CBF CDF isos. ∆ base s m CBF = m CDF Def. of s m BCD + m CBF + m CDF = 180° Polygon Sum Thm. Holt Mc. Dougal Geometry

6 -6 Properties of Kites and Trapezoids Example 2 A Continued m BCD +

6 -6 Properties of Kites and Trapezoids Example 2 A Continued m BCD + m CBF + m CDF = 180° Substitute m CDF m BCD + m CDF = 180° for m CBF. m BCD + 52° = 180° m BCD = 76° Holt Mc. Dougal Geometry Substitute 52 for m CDF. Subtract 104 from both sides.

6 -6 Properties of Kites and Trapezoids Example 2 B: Using Properties of Kites

6 -6 Properties of Kites and Trapezoids Example 2 B: Using Properties of Kites In kite ABCD, m DAB = 54°, and m CDF = 52°. Find m ABC. ADC ABC Kite one pair opp. s Def. of s Polygon Sum Thm. m ABC + m BCD + m ADC + m DAB = 360° m ADC = m ABC Substitute m ABC for m ADC. m ABC + m BCD + m ABC + m DAB = 360° Holt Mc. Dougal Geometry

6 -6 Properties of Kites and Trapezoids Example 2 B Continued m ABC +

6 -6 Properties of Kites and Trapezoids Example 2 B Continued m ABC + m BCD + m ABC + m DAB = 360° m ABC + 76° + m ABC + 54° = 360° 2 m ABC = 230° m ABC = 115° Holt Mc. Dougal Geometry Substitute. Simplify. Solve.

6 -6 Properties of Kites and Trapezoids Example 2 C: Using Properties of Kites

6 -6 Properties of Kites and Trapezoids Example 2 C: Using Properties of Kites In kite ABCD, m DAB = 54°, and m CDF = 52°. Find m FDA. CDA ABC Kite one pair opp. s m CDA = m ABC Def. of s m CDF + m FDA = m ABC Add. Post. 52° + m FDA = 115° m FDA = 63° Holt Mc. Dougal Geometry Substitute. Solve.

6 -6 Properties of Kites and Trapezoids Check It Out! Example 2 a In

6 -6 Properties of Kites and Trapezoids Check It Out! Example 2 a In kite PQRS, m PQR = 78°, and m TRS = 59°. Find m QRT. Kite cons. sides ∆PQR is isos. RPQ PRQ m QPT = m QRT Holt Mc. Dougal Geometry 2 sides isos. ∆ base s Def. of s

6 -6 Properties of Kites and Trapezoids Check It Out! Example 2 a Continued

6 -6 Properties of Kites and Trapezoids Check It Out! Example 2 a Continued m PQR + m QRP + m QPR = 180° Polygon Sum Thm. 78° + m QRT + m QPT = 180° Substitute 78 for m PQR. 78° + m QRT = 180° Substitute. 78° + 2 m QRT = 180° Substitute. 2 m QRT = 102° Subtract 78 from both sides. m QRT = 51° Holt Mc. Dougal Geometry Divide by 2.

6 -6 Properties of Kites and Trapezoids Check It Out! Example 2 b In

6 -6 Properties of Kites and Trapezoids Check It Out! Example 2 b In kite PQRS, m PQR = 78°, and m TRS = 59°. Find m QPS QRS Kite one pair opp. s m QPS = m QRT + m TRS Add. Post. m QPS = m QRT + 59° Substitute. m QPS = 51° + 59° m QPS = 110° Substitute. Holt Mc. Dougal Geometry

6 -6 Properties of Kites and Trapezoids Check It Out! Example 2 c In

6 -6 Properties of Kites and Trapezoids Check It Out! Example 2 c In kite PQRS, m PQR = 78°, and m TRS = 59°. Find each m PSR. m SPT + m TRS + m RSP = 180° Polygon Sum Thm. m SPT = m TRS Def. of s m TRS + m RSP = 180° Substitute. 59° + m RSP = 180° Substitute. Simplify. m RSP = 62° Holt Mc. Dougal Geometry

6 -6 Properties of Kites and Trapezoids A trapezoid is a quadrilateral with exactly

6 -6 Properties of Kites and Trapezoids 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. Holt Mc. Dougal Geometry

6 -6 Properties of Kites and Trapezoids If the legs of a trapezoid are

6 -6 Properties of Kites and Trapezoids If the legs of a trapezoid are congruent, the trapezoid is an isosceles trapezoid. The following theorems state the properties of an isosceles trapezoid. Holt Mc. Dougal Geometry

6 -6 Properties of Kites and Trapezoids Holt Mc. Dougal Geometry

6 -6 Properties of Kites and Trapezoids Holt Mc. Dougal Geometry

6 -6 Properties of Kites and Trapezoids Reading Math Theorem 6 -6 -5 is

6 -6 Properties of Kites and Trapezoids Reading Math Theorem 6 -6 -5 is a biconditional statement. So it is true both “forward” and “backward. ” Holt Mc. Dougal Geometry

6 -6 Properties of Kites and Trapezoids Example 3 A: Using Properties of Isosceles

6 -6 Properties of Kites and Trapezoids Example 3 A: Using Properties of Isosceles Trapezoids Find m A. m C + m B = 180° 100 + m B = 180 Same-Side Int. s Thm. Substitute 100 for m C. m B = 80° A B Subtract 100 from both sides. Isos. trap. s base m A = m B Def. of s m A = 80° Substitute 80 for m B Holt Mc. Dougal Geometry

6 -6 Properties of Kites and Trapezoids Example 3 B: Using Properties of Isosceles

6 -6 Properties of Kites and Trapezoids Example 3 B: Using Properties of Isosceles Trapezoids KB = 21. 9 and MF = 32. 7. Find FB. Isos. trap. s base KJ = FM Def. of segs. KJ = 32. 7 Substitute 32. 7 for FM. KB + BJ = KJ Seg. Add. Post. 21. 9 + BJ = 32. 7 Substitute 21. 9 for KB and 32. 7 for KJ. BJ = 10. 8 Subtract 21. 9 from both sides. Holt Mc. Dougal Geometry

6 -6 Properties of Kites and Trapezoids Example 3 B Continued Same line. KFJ

6 -6 Properties of Kites and Trapezoids Example 3 B Continued Same line. KFJ MJF Isos. trap. s base Isos. trap. legs ∆FKJ ∆JMF SAS BKF BMJ CPCTC FBK JBM Vert. s Holt Mc. Dougal Geometry

6 -6 Properties of Kites and Trapezoids Example 3 B Continued Isos. trap. legs

6 -6 Properties of Kites and Trapezoids Example 3 B Continued Isos. trap. legs ∆FBK ∆JBM AAS CPCTC FB = JB Def. of segs. FB = 10. 8 Substitute 10. 8 for JB. Holt Mc. Dougal Geometry

6 -6 Properties of Kites and Trapezoids Check It Out! Example 3 a Find

6 -6 Properties of Kites and Trapezoids Check It Out! Example 3 a Find m F + m E = 180° E H m E = m H m F + 49° = 180° m F = 131° Holt Mc. Dougal Geometry Same-Side Int. s Thm. Isos. trap. s base Def. of s Substitute 49 for m E. Simplify.

6 -6 Properties of Kites and Trapezoids Check It Out! Example 3 b JN

6 -6 Properties of Kites and Trapezoids Check It Out! Example 3 b JN = 10. 6, and NL = 14. 8. Find KM. Isos. trap. s base KM = JL JL = JN + NL Def. of segs. KM = JN + NL Substitute. Segment Add Postulate KM = 10. 6 + 14. 8 = 25. 4 Substitute and simplify. Holt Mc. Dougal Geometry

6 -6 Properties of Kites and Trapezoids Example 4 A: Applying Conditions for Isosceles

6 -6 Properties of Kites and Trapezoids Example 4 A: Applying Conditions for Isosceles Trapezoids Find the value of a so that PQRS is isosceles. Trap. with pair base s isosc. trap. S P m S = m P 2 a 2 – 54 = a 2 Def. of s Substitute 2 a 2 – 54 for m S and + 27 2 a + 27 for m P. = 81 a = 9 or a = – 9 Holt Mc. Dougal Geometry Subtract a 2 from both sides and add 54 to both sides. Find the square root of both sides.

6 -6 Properties of Kites and Trapezoids Example 4 B: Applying Conditions for Isosceles

6 -6 Properties of Kites and Trapezoids Example 4 B: Applying Conditions for Isosceles Trapezoids AD = 12 x – 11, and BC = 9 x – 2. Find the value of x so that ABCD is isosceles. Diags. isosc. trap. AD = BC Def. of segs. Substitute 12 x – 11 for AD and 12 x – 11 = 9 x – 2 for BC. 3 x = 9 x=3 Holt Mc. Dougal Geometry Subtract 9 x from both sides and add 11 to both sides. Divide both sides by 3.

6 -6 Properties of Kites and Trapezoids Check It Out! Example 4 Find the

6 -6 Properties of Kites and Trapezoids Check It Out! Example 4 Find the value of x so that PQST is isosceles. Q S m Q = m S Trap. with pair base s isosc. trap. Def. of s 2 + 19 for m Q Substitute 2 x 2 x 2 + 19 = 4 x 2 – 13 and 4 x 2 – 13 for m S. 32 = 2 x 2 x = 4 or x = – 4 Holt Mc. Dougal Geometry Subtract 2 x 2 and add 13 to both sides. Divide by 2 and simplify.

6 -6 Properties of Kites and Trapezoids The midsegment of a trapezoid is the

6 -6 Properties of Kites and Trapezoids 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. Holt Mc. Dougal Geometry

6 -6 Properties of Kites and Trapezoids Holt Mc. Dougal Geometry

6 -6 Properties of Kites and Trapezoids Holt Mc. Dougal Geometry

6 -6 Properties of Kites and Trapezoids Example 5: Finding Lengths Using Midsegments Find

6 -6 Properties of Kites and Trapezoids Example 5: Finding Lengths Using Midsegments Find EF. Trap. Midsegment Thm. Substitute the given values. EF = 10. 75 Holt Mc. Dougal Geometry Solve.

6 -6 Properties of Kites and Trapezoids Check It Out! Example 5 Find EH.

6 -6 Properties of Kites and Trapezoids Check It Out! Example 5 Find EH. Trap. Midsegment Thm. 1 16. 5 = 2 (25 + EH) Substitute the given values. Simplify. 33 = 25 + EH Multiply both sides by 2. 13 = EH Subtract 25 from both sides. Holt Mc. Dougal Geometry

6 -6 Properties of Kites and Trapezoids Lesson Quiz: Part I 1. Erin is

6 -6 Properties of Kites and Trapezoids 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, m KLP = 72°, and m HJP = 49. 5°. Find each measure. 2. m LHJ 81° Holt Mc. Dougal Geometry 3. m PKL 18°

6 -6 Properties of Kites and Trapezoids Lesson Quiz: Part II Use the diagram

6 -6 Properties of Kites and Trapezoids Lesson Quiz: Part II Use the diagram for Items 4 and 5. 4. m WZY = 61°. Find m WXY. 119° 5. XV = 4. 6, and WY = 14. 2. Find VZ. 9. 6 6. Find LP. 18 Holt Mc. Dougal Geometry

6 -6 Properties of Kites and Trapezoids Classwork/Homework • 6. 6# 1, 2, 4

6 -6 Properties of Kites and Trapezoids Classwork/Homework • 6. 6# 1, 2, 4 -12, 14 -18, 27 -32 • Properties of Trapezoids WS Holt Mc. Dougal Geometry