Lesson 5.3
Trapezoids and Kites
Homework: 5.3/1-8,19
QUIZ Wednesday 5.1 – 5.4
PROCEDURES for today:
1. OPEN TEXTBOOKS
2. Tools – patty paper(2), protractor, ruler
3. INVESTIGATIONs 1 & 2 – ALL steps
4. Complete the 4 kite conjectures &
the 3 trapezoid conjectures
Definition
Kite – a quadrilateral that has two pairs
of consecutive congruent sides, but
opposite sides are not congruent.
Perpendicular Diagonals of a Kite
If a quadrilateral is a kite, then its
diagonals are perpendicular.
C
B
D
A
AC  BD
Non-Vertex Angles of a Kite
If a quadrilateral is a kite, then nonvertex angles are congruent
C
D
B
A
A  C, B  D
Vertex diagonals bisect vertex angles
If a quadrilateral is a kite then the vertex
diagonal bisects the vertex angles.
C
B
D
A
Vertex diagonal bisects the
non-vertex diagonal
If a quadrilateral is a kite then the vertex diagonal
bisects the non-vertex diagonal
C
B
D
A
Trapezoid
Definition-a quadrilateral with exactly one
pair of parallel sides.
A
Base
› B
Leg
Leg
C
›
Base
D
Property of a Trapezoid
Leg Angles are Supplementary
A
›
B
<A + <C = 180
<B + <D = 180
C
›
D
Isosceles Trapezoid
Definition - A trapezoid with congruent legs.
Isosceles Trapezoid - Properties
1) Base Angles Are Congruent
2) Diagonals Are Congruent
Example
S
R
50
P
Q
PQRS is an isosceles trapezoid. Find m
P, m Q and mR.
m R = 50 since base angles are congruent
mP = 130 and mQ = 130 (consecutive angles
of parallel lines cut by a transversal are )
Find the measures of the
angles in trapezoid
48
m< A = 132
m< B = 132
m< D = 48
Find BE
AC = 17.5, AE = 9.6
E
Example
Find the side lengths of the kite.
X
12
W
20
12
U
12
Z
Y
Example Continued
X
12
W
20
We can use the Pythagorean Theorem to
find the side lengths.
122 + 202 = (WX)2
122 + 122 = (XY)2
144 + 400 = (WX)2
144 + 144 = (XY)2
544 = (WX)2
288 = (XY)2
WX = 4 34
XY =12 2
likewise WZ = 4 34
likewise ZY =12 2
12
U
12
Z
Y
Find the lengths of the sides of
the kite
W
4
Z
5
5
8
Y
X
Find the lengths of the sides of
kite to the nearest tenth
2
4
7
2
Example 3
J
Find mG and mJ.
H 132
Since GHJK is a kite G  J
So 2(mG) + 132 + 60 = 360
2(mG) =168
mG = 84 and mJ = 84
G
60
K
Try This!
 RSTU is a kite. Find mR, mS and mT.
S
R x+30
x
T
125
U
x +30 + 125 + 125 + x = 360
2x + 280 = 360
2x = 80
x = 40
So mR = 70, mT = 40 and mS = 125
Try These
2. m<C = x +12 and
m<B = 3x – 2, find x and the
measures of the 2 angles
1. If <A = 134, find m<D
A
base
leg
D
base
m<D = 46
A
B
leg
leg
C
D
base
B
leg
base
x = 42.5
m<C = 54.5
m<B = 125.5
C
Using Properties of Trapezoids
When working with a trapezoid, the height may be measured
anywhere between the two bases. Also, beware of "extra"
information. The 35 and 28 are not needed to compute this area.
1
Area of trapezoid = hb1  b2 
2
Find the area of this trapezoid.
A = ½ * 26 * (20 + 42)
A = 806
Using Properties of Trapezoids
Example 2
Find the area of a trapezoid
with bases of 10 in and 14
in, and a height of 5 in.
Using Properties of Kites
Area Kite = one-half product of diagonals
1
A  d1d 2
2
D
1
Area  AC  BD
2
A
B
C
Using Properties of Kites
Example 6
ABCD is a Kite.
A
a) Find the lengths of all
the sides.
D
2
E 4
4
4
b) Find the area of the Kite.
C
B
Venn Diagram:
http://teachers2.wcs.edu/high/rhs/staceyh/Geometry/Chapter%206%20Notes.ppt#435,22,6.2 – Properties of Parallelograms
Flow Chart:
Homework