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Name ________________________________________ Date ___________________ Class __________________
LESSON
6-6
Problem Solving
Properties of Kites and Trapezoids
Use the figure of the kite for Exercises 1 and 2.
1. What is AD to the nearest tenth?
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2. What is the perimeter of the kite to the
nearest tenth?
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3. In kite STUV, m∠TUW = 35° and
m∠WSV = 21°. What is the measure
of ∠UVS?
4. A car window is in the shape of a trapezoid.
When the window is halfway down, the
top is KL , the midsegment of FGHJ.
If KL = 23 inches, what is GH?
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Choose the best answer.
6. In kite KLMN, find the measure of ∠M.
5. Trapezoid PQRS has base angles that
measure (9r + 21)° and (15r − 21)°.
Find the value of r so that PQRS is
isosceles.
A 3
B 5
C 7
D 14
F 100.5°
H 122°
G 101°
J 130°
7. In the design, eight isosceles trapezoids
surround a regular octagon. What is the
measure of ∠B in trapezoid ABCD?
A 35°
B 45°
C 55°
D 65°
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
6-49
Holt McDougal Geometry
found from BA and AE by using the
Pythagorean Theorem. BD is the sum of
1
BE and ED. The area is (AC)(BD).
2
3. No; possible answer: there is no way to
use the Pythagorean Theorem to find the
length of AE, and thus AC, with the
information provided.
4. Possible answer: It is given that QU is
perpendicular to PT and SR is
perpendicular to SR , so ∠PTU and
∠QUT are right angles. It is also given that
PQ is parallel to SR , so the same-side
interior angles ∠PTU and ∠TPQ are
supplementary, as are ∠QUT and ∠PQU.
An angle supplementary to a right angle
must be a right angle, so ∠TPQ and
∠PQU are right angles. All four interior
angles of PQUT are right angles, so
PQUT is a rectangle by the definition of a
rectangle.
5. Possible answer: It is given that JKLM is
an isosceles trapezoid, so JK ≅ ML and
KL & JM . Base angles in an isosceles
trapezoid are congruent, so ∠J ≅ ∠M.
Corresponding angles are equal, so
∠NKL ≅ ∠J and ∠NLK ≅ ∠M. By the
Transitive Property of Congruence, ∠NKL
≅ ∠NLK. ∠NKL and ∠NLK are the base
angles of UNKL, so it is an isosceles
triangle. Thus NK ≅ NL . By the Segment
Addition Postulate, JN = JK + NK and MN
= ML + NL. By the Addition Property of
Equality and the definition of congruent
segments, JN = MN. Because JN and
MN have the same length, they are
congruent. So UJNM is isosceles.
Challenge
1. Possible answer: A dart is a concave
quadrilateral with exactly two pairs of
congruent consecutive sides.
2. Darts and kites both have exactly two
pairs of congruent consecutive sides.
Darts are concave, and kites are convex.
3. x = 11; y = 2
4. Possible answer: The lines are
perpendicular. The line containing the
first diagonal bisects the diagonal that
joins the fin angles.
5. Possible answer: The triangles are
congruent and obtuse.
6. Possible answer:
It is given that AB ≅ AD and BC ≅ DC.
By the Reflex. Prop. of ≅, AC ≅ AC. This
means that UABC ≅ UADC by SSS. So
∠B ≅ ∠D by CPCTC.
Problem Solving
1.
3.
5.
7.
23.1
124°
C
B
2. 74.0
4. 18 in.
6. G
Reading Strategies
1.
2.
3.
4.
5.
6.
Reteach
1. 43°
2. 55°
3. 70°
4. 56°
5. 2.8
6. x = ±4
7. z = 6
8. 21
9. 9.5
7.
10. 9.15
11. 10.8
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
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Holt McDougal Geometry