Warm Up Solve for x. 1. x 2 + 38 = 3 x 2 – 12 2. 137 + x = 180 3. 4. Find FE .

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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 .

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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 .

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Warm Up

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.

9-18

9-16

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.

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?

1

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.

3

Solve

Example 1 Continued

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 2: Using Properties of Kites

In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mBCD.

Kite cons. sides

∆BCD is isos.

2 sides isos. ∆

isos. ∆base s

CBF CDF

mCBF = mCDF

Def. of s

Polygon Sum Thm.

mBCD + mCBF + mCDF = 180°

Example 2 Continued

mBCD + mCBF + mCDF = 180°

Substitute mCDF for mCBF.

mBCD + mCBF+ mCDF= 180°

Substitute 52 for mCBF.

mBCD + 52°+ 52° = 180°

Subtract 104 from both sides.

mBCD = 76°

Example 3: Using Properties of Kites

In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mABC.

ADC ABC

Kite one pair opp. s

Def. of s

mADC = mABC

Polygon Sum Thm.

mABC + mBCD + mADC + mDAB = 360°

Substitute mABC for mADC.

mABC + mBCD + mABC+ mDAB = 360°

Example 3 Continued

mABC + mBCD + mABC + mDAB = 360°

mABC + 76°+ mABC + 54° = 360°

Substitute.

2mABC = 230°

Simplify.

mABC = 115°

Solve.

Example 4: Using Properties of Kites

In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mFDA.

CDA ABC

Kite one pair opp. s

mCDA = mABC

Def. of s

mCDF + mFDA = mABC

Add. Post.

52° + mFDA = 115°

Substitute.

mFDA = 63°

Solve.

Example 5: Using Properties of Isosceles Trapezoids

Find mA.

mC + mB = 180°

Same-Side Int. s Thm.

100 + mB = 180

Substitute 100 for mC.

mB = 80°

Subtract 100 from both sides.

A B

Isos. trap. s base

mA = mB

Def. of s

mA = 80°

Substitute 80 for mB

Example 6: 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.

Seg. Add. Post.

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.

Example 6 Continued

Same line.

KFJ MJF

Isos. trap. s base

Isos. trap. legs

SAS

∆FKJ ∆JMF

CPCTC

BKF BMJ

Vert. s

FBK JBM

Example 6 Continued

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

Segment Add Postulate

KM = JN + NL

Substitute.

KM = 10.6 + 14.8 = 25.4

Substitute and simplify.

Example 8: Applying Conditions for Isosceles Trapezoids

Find the value of a so that PQRS is isosceles.

Trap. with pair base s isosc. trap.

S P

mS = mP

Def. of s

Substitute 2a2 – 54 for mS and a2 + 27 for mP.

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: Applying Conditions for Isosceles Trapezoids

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: Finding Lengths Using Midsegments

Find EF.

Trap. Midsegment Thm.

Substitute the given values.

Solve.

EF = 10.75

1

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