6.6 Properties of Kites and Trapezoids - Ephrata High · PDF fileHolt McDougal Geometry 6-6 Properties of Kites and Trapezoids A kite is a quadrilateral with exactly two pairs of congruent

Holt McDougal Geometry

6-6 Properties of Kites and Trapezoids

Toolbox

pg. 432 (14; 16; 24; 28-29, 41; 42; 48 why4,

ch. 45)

pg. 432 (18-22 even, 30-34 even, 35, 43, 47

why4, ch. 51)



How do you use properties of kites to solve problems?

How do you use properties of trapezoids to solve problems?

Essential Questions



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





Kite � cons. sides ≅

Example 1A: Using Properties of Kites

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

∆BCD is isos. 2 ≅ sides �isos. ∆

isos. ∆ �base ∠∠∠∠s ≅

Def. of ≅ ∠∠∠∠ s

Polygon ∠∠∠∠ Sum Thm.

∠CBF ≅ ∠CDF

m∠CBF = m∠CDF

m∠BCD + m∠CBF + m∠CDF = 180°



Example 1A Continued

Substitute m∠CDF for m∠CBF.

Substitute 52 for m∠CDF.

Subtract 104 from both sides.

m∠BCD + m∠CDF + m∠CDF = 180°

m∠BCD + 52° + 52° = 180°

m∠BCD = 76°

m∠BCD + m∠CBF + m∠CDF = 180°



Kite � one pair opp. ∠∠∠∠s ≅

Example 1B: Using Properties of Kites

Def. of ≅ ∠∠∠∠s

Polygon ∠∠∠∠ Sum Thm.

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

∠ADC ≅ ∠ABC

m∠ADC = m∠ABC

m∠ABC + m∠BCD + m∠ADC + m∠DAB = 360°

m∠ABC + m∠BCD + m∠ABC + m∠DAB = 360°

Substitute m∠ABC for m∠ADC.



Example 1B Continued

Substitute.

Simplify.

m∠ABC + m∠BCD + m∠ABC + m∠DAB = 360°

m∠ABC + 76° + m∠ABC + 54° = 360°

2m∠ABC = 230°

m∠ABC = 115° Solve.



Kite � one pair opp. ∠∠∠∠s ≅

Example 1C: Using Properties of Kites


∠∠∠∠ Add. Post.

Substitute.

Solve.

In kite ABCD, m∠∠∠∠DAB = 54°, and m∠∠∠∠CDF = 52°. Find m∠∠∠∠FDA.

∠CDA ≅ ∠ABC

m∠CDA = m∠ABC

m∠CDF + m∠FDA = m∠ABC

52° + m∠FDA = 115°

m∠FDA = 63°



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.



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





Isos.� trap. ∠∠∠∠s base ≅

Example 2A: Using Properties of Isosceles

Trapezoids

Find m∠∠∠∠A.

Same-Side Int. ∠∠∠∠s Thm.

Substitute 100 for m∠∠∠∠C.

Subtract 100 from both sides.


Substitute 80 for m∠∠∠∠B

m∠C + m∠B = 180°

100 + m∠B = 180

m∠B = 80°

∠A ≅ ∠B

m∠A = m∠B

m∠A = 80°



Example 2B: Using Properties of Isosceles

Trapezoids

KB = 21.9 and MF = 32.7.

Find FB.

Isos. � trap. ∠∠∠∠s base ≅

Def. of ≅ segs.

Substitute 32.7 for FM.

Seg. Add. Post.

Substitute 21.9 for KB and 32.7 for KJ.

Subtract 21.9 from both sides.

KJ = FM

KJ = 32.7

KB + BJ = KJ

21.9 + FB = 32.7

FB = 10.8



Example 3A: Applying Conditions for Isosceles

Trapezoids

Find the value of a so that PQRS

is isosceles.

a = 9 or a = –9

Trap. with pair base ∠∠∠∠s � ≅ isosc. trap.


Substitute 2a2 – 54 for m∠∠∠∠S and a2 + 27 for m∠∠∠∠P.

Subtract a2 from both sides and add 54 to both sides.

Find the square root of both sides.

∠S ≅ ∠P

m∠S = m∠P

2a2 – 54 = a2 + 27

a2 = 81



Example 3B: 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.

Substitute 12x – 11 for AD and 9x – 2 for BC.

Subtract 9x from both sides and add 11 to both sides.

Divide both sides by 3.

AD = BC

12x – 11 = 9x – 2

3x = 9

x = 3



The midsegment of a trapezoid is 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 4: Finding Lengths Using Midsegments

Find EF.

Trap. Midsegment Thm.

Substitute the given values.

Solve.EF = 10.75

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6.6 Properties of Kites and Trapezoids - Ephrata High · PDF fileHolt McDougal Geometry 6-6 Properties of Kites and Trapezoids A kite is a quadrilateral with exactly two pairs of congruent