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Properties of Kites and Trapezoids. 6-6. Holt Geometry. 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. A kite is a quadrilateral with exactly two pairs of congruent consecutive sides. Example 1: Problem-Solving Application.

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Presentation Transcript
slide1

Properties of Kites

and Trapezoids

6-6

Holt Geometry

slide2

Warm Up

Solve for x.

1.x2 + 38 = 3x2 – 12

2. 137 + x = 180

3.

4. Find FE.

5 or –5

43

156

slide5

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?

slide6

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. ∆

isos. ∆base s 

CBF  CDF

mCBF = mCDF

Def. of  s

Polygon  Sum Thm.

mBCD + mCBF + mCDF = 180°

slide7

Example 2A 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°

slide8

Example 2B: 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

mADC = mABC

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°

slide9

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.

slide10

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.

slide11

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.

slide12

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

slide14

Reading Math

Theorem 6-6-5 is a biconditional statement. So it is true both “forward” and “backward.”

slide15

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

slide16

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.

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.

slide17

Example 3B Continued

Same line.

KFJ  MJF

Isos. trap.  s base 

Isos. trap.  legs

SAS

∆FKJ  ∆JMF

CPCTC

BKF  BMJ

Vert. s

FBK  JBM

slide18

Example 3B Continued

Isos. trap.  legs 

AAS

∆FBK  ∆JBM

CPCTC

FB = JB

Def. of  segs.

FB = 10.8

Substitute 10.8 for JB.

slide19

Check It Out! Example 3a

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.

slide20

Check It Out! Example 3b

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.

slide21

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.

slide22

Example 4B: 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.

slide23

Check It Out! Example 4

Find the value of x so that PQST is isosceles.

Trap. with pair base s  isosc. trap.

Q  S

mQ = mS

Def. of s

Substitute 2x2 + 19 for mQ and 4x2 – 13 for mS.

2x2 + 19 = 4x2 – 13

Subtract 2x2 and add 13 to both sides.

32 = 2x2

Divide by 2 and simplify.

x = 4 or x = –4

slide24

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.

slide26

Example 5: Finding Lengths Using Midsegments

Find EF.

Trap. Midsegment Thm.

Substitute the given values.

Solve.

EF = 10.75

slide27

1

16.5 = (25 + EH)

2

Check It Out! Example 5

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

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