Properties of Kites 6 6 Properties of Kites

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

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 Geometry

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

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 Geometry

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

6 -6 Properties of Kites and Trapezoids Holt Geometry

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. She has a dowel that is 36 cm long. About how much wood will she have left after cutting the last dowel? Holt Geometry

6 -6 Properties of Kites and Trapezoids 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. Holt Geometry

6 -6 Properties of Kites and Trapezoids Example 1 Continued 3 Solve N bisects JM. Pythagorean Thm. Holt Geometry

6 -6 Properties of Kites and Trapezoids 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. Holt Geometry

6 -6 Properties of Kites and Trapezoids Example 1 Continued 4 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. Holt Geometry

6 -6 Properties of Kites and Trapezoids 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? Holt Geometry

6 -6 Properties of Kites and Trapezoids 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 Holt Geometry

6 -6 Properties of Kites and Trapezoids Check It Out! Example 1 Continued 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. Holt Geometry

6 -6 Properties of Kites and Trapezoids Check It Out! Example 1 Continued 3 Solve Pyth. Thm. perimeter of PQRS = Holt Geometry

6 -6 Properties of Kites and Trapezoids 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. Holt Geometry

6 -6 Properties of Kites and Trapezoids Check It Out! Example 1 Continued 4 Look Back To estimate the perimeter, change the side lengths into decimals and round. , and kite is approximately . The perimeter of the 2(54) + 2 (41) = 190. So 191. 3 is a reasonable answer. Holt Geometry

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 Geometry

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

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 Geometry

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 Geometry Substitute. Simplify. Solve.

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 Geometry Substitute. Solve.

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 Geometry 2 sides isos. ∆ base s Def. of s

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 Geometry Divide by 2.

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 Geometry

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 Geometry

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 Geometry

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 Geometry

6 -6 Properties of Kites and Trapezoids Holt Geometry

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 Geometry

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 Holt Geometry 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

6 -6 Properties of Kites and Trapezoids Example 3 B: Using Properties of Isosceles Trapezoids KB = 21. 9 m 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 Geometry

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 Geometry

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

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 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 = 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 Geometry

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 Substitute 2 a 2 – 54 for m S and + 27 2 a + 27 for m P. = 81 a = 9 or a = – 9 Holt Geometry Def. of s 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 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 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 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 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 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 Geometry

6 -6 Properties of Kites and Trapezoids Holt Geometry

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 Geometry Solve.

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 Geometry

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 Holt Geometry 81° 3. m PKL 18°

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 Geometry