Properties of Kites
Download
1 / 27

6-6 - PowerPoint PPT Presentation


  • 67 Views
  • Uploaded on

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.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about ' 6-6' - britanni-farmer


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
6 6

Properties of Kites

and Trapezoids

6-6

Holt Geometry


6 6

Warm Up

Solve for x.

1.x2 + 38 = 3x2 – 12

2. 137 + x = 180

3.

4. Find FE.

5 or –5

43

156


6 6

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


6 6

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?


6 6

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°


6 6

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°


6 6

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°


6 6

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.


6 6

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.


6 6

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.


6 6

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


6 6

Reading Math an

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


6 6

Example 3A: Using Properties of Isosceles Trapezoids an

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


6 6

Example 3B: Using Properties of Isosceles Trapezoids an

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.


6 6

Example 3B Continued an

Same line.

KFJ  MJF

Isos. trap.  s base 

Isos. trap.  legs

SAS

∆FKJ  ∆JMF

CPCTC

BKF  BMJ

Vert. s

FBK  JBM


6 6

Example 3B Continued an

Isos. trap.  legs 

AAS

∆FBK  ∆JBM

CPCTC

FB = JB

Def. of  segs.

FB = 10.8

Substitute 10.8 for JB.


6 6

Check It Out! an 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.


6 6

Check It Out! an 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.


6 6

Example 4A: Applying Conditions for Isosceles Trapezoids an

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.


6 6

Example 4B: Applying Conditions for Isosceles Trapezoids an

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.


6 6

Check It Out! an 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


6 6

The an 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.


6 6

Example 5: Finding Lengths Using Midsegments an

Find EF.

Trap. Midsegment Thm.

Substitute the given values.

Solve.

EF = 10.75


6 6

1 an

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


ad