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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. Trapezoids and Kites. 9-4. A kite is a quadrilateral with exactly two pairs of congruent consecutive sides.

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Solve for x.

1.x2 + 38 = 3x2 – 12

2. 137 + x = 180

3.

4. Find FE.

5 or –5

43

156

### Trapezoids and Kites

9-4

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

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.

9-17

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.

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?

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.

Solve

Example 1 Continued

N bisects JM.

Pythagorean Thm.

Pythagorean Thm.

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.

In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mBCD.

Kite cons. sides 

∆BCD is isos.

2  sides isos. ∆

isos. ∆base s 

CBF  CDF

mCBF = mCDF

Def. of  s

Polygon  Sum Thm.

mBCD + mCBF + mCDF = 180°

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°

In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mABC.

Kite  one pair opp. s 

Def. of s

Polygon  Sum Thm.

mABC + mBCD + mADC + mDAB = 360°

mABC + mBCD + mABC+ mDAB = 360°

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.

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

52° + mFDA = 115°

Substitute.

mFDA = 63°

Solve.

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

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

21.9 + BJ = 32.7

Substitute 21.9 for KB and 32.7 for KJ.

BJ = 10.8

Subtract 21.9 from both sides.

Same line.

KFJ  MJF

Isos. trap.  s base 

Isos. trap.  legs

SAS

∆FKJ  ∆JMF

CPCTC

BKF  BMJ

Vert. s

FBK  JBM

Isos. trap.  legs 

AAS

∆FBK  ∆JBM

CPCTC

FB = JB

Def. of  segs.

FB = 10.8

Substitute 10.8 for JB.

Check It Out! Example 7

JN = 10.6, and NL = 14.8. Find KM.

Isos. trap. s base 

Def. of segs.

KM = JL

JL = JN + NL

KM = JN + NL

Substitute.

KM = 10.6 + 14.8 = 25.4

Substitute and simplify.

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.

AD = 12x – 11, and BC = 9x – 2. Find the value of x so that ABCD is isosceles.

Diags.  isosc. trap.

Def. of segs.

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.

Find EF.

Trap. Midsegment Thm.

Substitute the given values.

Solve.

EF = 10.75

16.5 = (25 + EH)

2

Check It Out! Example 11

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

Pg 473 #1-6, 13-16, 24, 28-30