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

Warm Up Solve for x. 1. x2 + 38 = 3x2 – 12 2. 137 + x = 180 3. 4. Find FE. 5 or –5 43 156

Objectives Use properties of kites to solve problems. Use properties of trapezoids to solve problems.

A kite is a quadrilateral with exactly two pairs of congruent consecutive sides.

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?

Understand the Problem 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 .

Example 1 Continued Solve 3 N bisects JM. Pythagorean Thm. Pythagorean Thm.

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.

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.

Example 2A: 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.

Example 2A Continued Substitute mCDF for mCBF. mBCD + mCBF + mCDF = 180° Substitute 52 for mCBF. mBCD + 52° + 52° = 180° Subtract 104 from both sides. mBCD = 76°

Example 2: Using Properties of Kites In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mABC. ADC  ABC Kite  one pair opp. s  mADC = mABC Def. of  s Polygon  Sum Thm. mABC + mBCD + mADC + mDAB = 360° Substitute mABC for mADC. mABC + mBCD + mABC + mDAB = 360°

Example 2B Continued mABC + mBCD + mABC + mDAB = 360° mABC + 76° + mABC + 54° = 360° Substitute. 2mABC = 230° Simplify. mABC = 115° Solve.

Example 2C: 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° Substitute. mFDA = 63° Solve.

trapezoid A ______________is a quadrilateral with exactly one pair of parallel sides. Each of the parallel sides is called a ________. The nonparallel sides are called . of a trapezoid are two consecutie angles whose common side is a base. base legs Base angles

If the legs of a trapezoid are congruent, the trapezoid is an . The following theorems state the properties of an isosceles trapezoid. isosceles trapezoid

Example 3A: Using Properties of Isosceles Trapezoids Find mA. mC + mB = 180° Same-Side Int. s Thm. 100 + mB = 180 Substitute 100 for mC. mB = 80° Subtract 100 from both sides. A  B Isos. trap. s base  mA = mB Def. of  s mA = 80° Substitute 80 for mB

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. 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.

Example 3B Continued Same line. KFJ  MJF Isos. trap.  s base  Isos. trap.  legs  ∆FKJ  ∆JMF SAS CPCTC BKF  BMJ FBK  JBM Vert. s 

Example 3B Continued Isos. trap.  legs  ∆FBK  ∆JBM AAS CPCTC FB = JB Def. of  segs. FB = 10.8 Substitute 10.8 for JB.

Check It Out! Example 3C Find mF. mF + mE = 180° Same-Side Int. s Thm. E  H Isos. trap. s base  mE = mH Def. of  s mF + 49° = 180° Substitute 49 for mE. mF = 131° Simplify.

Check It Out! Example 3D JN = 10.6, and NL = 14.8. Find KM. Isos. trap. s base  KM = JL Def. of  segs. JL = JN + NL Segment Add Postulate KM = JN + NL Substitute. KM = 10.6 + 14.8 = 25.4 Substitute and simplify.

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 mS = mP Def. of  s Substitute 2a2 – 54 for mS and a2 + 27 for mP. 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.

The midsegment of a trapezoid is the segment whose endpoints are the midpoints of the legs. In

Example 5: Finding Lengths Using Midsegments Find EF. Trap. Midsegment Thm. Substitute the given values. EF = 10.75 Solve.

Substitute the given values. Check It Out! Example 5 Find EH. Trap. Midsegment Thm. 1 16.5 = (25 + EH) 2 Substitute the given values. Simplify. 33 = 25 + EH Multiply both sides by 2. 13 = EH Subtract 25 from both sides.