Pearson Unit 1 Topic 6: Polygons and Quadrilaterals 6-6: Trapezoids and Kites Pearson Texas Geometry ©2016 Holt Geometry Texas ©2007

TEKS Focus: (5)(A) Investigate patterns to make conjectures about geometric relationships, including angles formed by parallel lines cut by a transversal, criteria required for triangle congruence, special segments of triangles, diagonals of quadrilaterals, interior and exterior angles of polygons, and special segments and angles of circles choosing from a variety of tools. (1)(F) Analyze mathematical relationships to connect and communicate mathematical ideas. (1)(C) Select tools, including real objects, manipulatives paper and pencil, and technology as appropriate, and techniques, including mental math, estimations, and number sense as appropriate, to solve problems.

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.

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

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

Example: 1 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? Round to the nearest tenth.

Understand the Problem Example: 1cont. 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: 1cont. 3 Solve N bisects JM. Pythagorean Thm.

Example: 1cont. 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. 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: 2 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 Polygon  Sum Thm. mBCD + mCBF + mCDF = 180°

Example: 2 continued mBCD + mCBF + mCDF = 180° 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: 3 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 mABC + mBCD + mADC + mDAB = 360° Polygon  Sum Thm. mABC + mBCD + mABC + mDAB = 360° Substitute mABC for mADC.

Example: 3 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: 4 In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mFDA. Kite  one pair opp. s are bisected by diag. BAF  FAD Def. of  bisector mFAD = (1/2) mDAB mFAD = (1/2) (54) = 27° Substitution and simplify Diagonals of a kite are perpendicular mAFD = 90° mFAD + mAFD + mFDA = 180° Triangle SumThm. 27 + 90 + mFDA = 180° Substitute mFAD for mAFD mFDA = 63° Simplify

Example: 5 mC + mB = 180° Same-Side Int. s Thm. 100 + mB = 180 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: 6 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. Subtract 21.9 from both sides. BJ = 10.8

Example: 6 cont. Same line. KFJ  MJF Isos. trap.  s base  Isos. trap.  legs  ∆FKJ  ∆JMF SAS CPCTC BKF  BMJ FBK  JBM Vert. s  Isos. trap.  legs  ∆FBK  ∆JBM AAS CPCTC FB = JB Def. of  segs. FB = 10.8 Substitute 10.8 for JB.

Example: 7 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.

Example: 8 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.

Example: 9 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.

Example: 10 Find EF. Trap. Midsegment Thm. Substitute the given values. Solve. EF = 10.75

Example: 11 Find EH. Trap. Midsegment Thm. 1 16.5 = (25 + EH) Substitute the given values. Simplify. 33 = 25 + EH Multiply both sides by 2. 8 = EH Subtract 25 from both sides.

Example: 12 QR is the midsegment of Trapezoid LMNP. Find x. x + 2 = ½ (4x – 10 + 8) 2(x + 2) = 4x – 2 2x + 4 = 4x – 2 4 = 2x – 2 6 = 2x 3 = x