Geometry Section 11-1/11-2

EXTENDING AREA, SURFACE AREA, AND
VOLUME
CHAPTER 11/12

AREAS OF PARALLELOGRAMS,
TRIANGLES, RHOMBI, AND
TRAPEZOIDS
SECTION 11-1 AND 11-2

ESSENTIAL QUESTIONS
• How do you find perimeters and areas of
parallelograms?
• How do you find perimeters and areas of
triangles?
• How do you find areas of trapezoids?
• How do you find areas of rhombi and kites?

VOCABULARY
1. Base of a Parallelogram:
2. Height of a Parallelogram:
3. Base of a Triangle:
4. Height of a Triangle:
5. Height of a Trapezoid:

VOCABULARY
Can be any side of a parallelogram

VOCABULARY
The perpendicular distance between
any two parallel bases of a parallelogram

VOCABULARY
Can be any side of a triangle

VOCABULARY
The length of a segment perpendicular to a
base to the opposite vertex

VOCABULARY
The length of a segment perpendicular to a
base to the opposite vertex
The perpendicular distance between bases

EXAMPLE 1
Find the perimeter and area of .!RSTU

EXAMPLE 1
P = 2l + 2w

EXAMPLE 1
P = 2l + 2w
P = 2(32) + 2(20)

EXAMPLE 1
P = 2l + 2w
P = 2(32) + 2(20)
P = 64 + 40

EXAMPLE 1
P = 2l + 2w
P = 2(32) + 2(20)
P = 64 + 40
P = 104 in.

EXAMPLE 1
P = 2l + 2w
P = 2(32) + 2(20)
P = 64 + 40
P = 104 in.
A = bh

EXAMPLE 1
P = 2l + 2w
P = 2(32) + 2(20)
P = 64 + 40
P = 104 in.
A = bh a2
+ b2
= c2

EXAMPLE 1
P = 2l + 2w
P = 2(32) + 2(20)
P = 64 + 40
P = 104 in.
A = bh a2
+ b2
= c2
a2
+122
= 202

EXAMPLE 1
P = 2l + 2w
P = 2(32) + 2(20)
P = 64 + 40
P = 104 in.
A = bh a2
+ b2
= c2
a2
+122
= 202
a2
+144 = 400

EXAMPLE 1
P = 2l + 2w
P = 2(32) + 2(20)
P = 64 + 40
P = 104 in.
A = bh a2
+ b2
= c2
a2
+122
= 202
a2
+144 = 400
−144 −144

EXAMPLE 1
P = 2l + 2w
P = 2(32) + 2(20)
P = 64 + 40
P = 104 in.
A = bh a2
+ b2
= c2
a2
+122
= 202
a2
+144 = 400
−144 −144
a2
= 256

EXAMPLE 1
P = 2l + 2w
P = 2(32) + 2(20)
P = 64 + 40
P = 104 in.
A = bh a2
+ b2
= c2
a2
+122
= 202
a2
+144 = 400
−144 −144
a2
= 256
a2
= 256

EXAMPLE 1
P = 2l + 2w
P = 2(32) + 2(20)
P = 64 + 40
P = 104 in.
A = bh a2
+ b2
= c2
a2
+122
= 202
a2
+144 = 400
−144 −144
a2
= 256
a2
= 256
a = 16

EXAMPLE 1
P = 2l + 2w
P = 2(32) + 2(20)
P = 64 + 40
P = 104 in.
A = bh a2
+ b2
= c2
a2
+122
= 202
a2
+144 = 400
−144 −144
a2
= 256
a2
= 256
a = 16
h = 16

EXAMPLE 1
P = 2l + 2w
P = 2(32) + 2(20)
P = 64 + 40
P = 104 in.
A = bh a2
+ b2
= c2
a2
+122
= 202
a2
+144 = 400
−144 −144
a2
= 256
a2
= 256
a = 16
h = 16
A = 32(16)

EXAMPLE 1
P = 2l + 2w
P = 2(32) + 2(20)
P = 64 + 40
P = 104 in.
A = bh a2
+ b2
= c2
a2
+122
= 202
a2
+144 = 400
−144 −144
a2
= 256
a2
= 256
a = 16
h = 16
A = 32(16)
A = 512 in2

EXAMPLE 2
Matt Mitarnowski needs to buy enough boards to make the frame
of the triangular sandbox shown and enough sand to cover the
bottom. If one of the boards is 3 feet long and one bag of sand
covers 9 square feet of the sandbox, how many boards and bags
does he need to buy?

EXAMPLE 2
P = a + b + c

EXAMPLE 2
P = a + b + c
P = 12+16 + 7.5

EXAMPLE 2
P = a + b + c
P = 12+16 + 7.5
P = 35.5 ft

EXAMPLE 2
P = a + b + c
P = 12+16 + 7.5
P = 35.5 ft
35.5
3

EXAMPLE 2
P = a + b + c
P = 12+16 + 7.5
P = 35.5 ft
35.5
3
≈ 11.83

EXAMPLE 2
P = a + b + c
P = 12+16 + 7.5
P = 35.5 ft
35.5
3
≈ 11.83 Matt needs 12 boards.

EXAMPLE 2
P = a + b + c
P = 12+16 + 7.5
P = 35.5 ft
35.5
3
≈ 11.83
A = 1
2
bh
Matt needs 12 boards.

EXAMPLE 2
P = a + b + c
P = 12+16 + 7.5
P = 35.5 ft
35.5
3
≈ 11.83
A = 1
2
bh
A = 1
2
(12)(9)

EXAMPLE 2
P = a + b + c
P = 12+16 + 7.5
P = 35.5 ft
35.5
3
≈ 11.83
A = 1
2
bh
A = 1
2
(12)(9)
A = 54 ft2

EXAMPLE 2
P = a + b + c
P = 12+16 + 7.5
P = 35.5 ft
35.5
3
≈ 11.83
A = 1
2
bh
A = 1
2
(12)(9)
A = 54 ft2
54
9

EXAMPLE 2
P = a + b + c
P = 12+16 + 7.5
P = 35.5 ft
35.5
3
≈ 11.83
A = 1
2
bh
A = 1
2
(12)(9)
A = 54 ft2
54
9
= 6

EXAMPLE 2
P = a + b + c
P = 12+16 + 7.5
P = 35.5 ft
35.5
3
≈ 11.83
A = 1
2
bh
A = 1
2
(12)(9)
A = 54 ft2
54
9
= 6
Matt needs 6 bags of
sand.

POSTULATE 11.2
If two figures are congruent, then they have the same area.

EXAMPLE 3
Find the area of the trapezoid.

EXAMPLE 3
A = 1
2
h(b1
+ b2
)

EXAMPLE 3
A = 1
2
h(b1
+ b2
)
A = 1
2
(1)(3 + 2.5)

EXAMPLE 3
A = 1
2
h(b1
+ b2
)
A = 1
2
(1)(3 + 2.5)
A = 1
2
(5.5)

EXAMPLE 3
A = 1
2
h(b1
+ b2
)
A = 1
2
(1)(3 + 2.5)
A = 1
2
(5.5)
A = 2.75 cm2

EXAMPLE 4
Fuzzy Jeff designed a deck shaped like the trapezoid shown. Find
the area of the deck.

EXAMPLE 4
a2
+ b2
= c2

EXAMPLE 4
a2
+ b2
= c2
42
+ b2
= 52

EXAMPLE 4
a2
+ b2
= c2
42
+ b2
= 52
16 + b2
= 25

EXAMPLE 4
a2
+ b2
= c2
42
+ b2
= 52
16 + b2
= 25
−16 −16

EXAMPLE 4
a2
+ b2
= c2
42
+ b2
= 52
16 + b2
= 25
−16 −16
b2
= 9

EXAMPLE 4
a2
+ b2
= c2
42
+ b2
= 52
16 + b2
= 25
−16 −16
b2
= 9
b2
= 9

EXAMPLE 4
a2
+ b2
= c2
42
+ b2
= 52
16 + b2
= 25
−16 −16
b2
= 9
b2
= 9 b = 3

EXAMPLE 4
a2
+ b2
= c2
42
+ b2
= 52
16 + b2
= 25
−16 −16
b2
= 9
b2
= 9 b = 3
b1
= 9; b2
= 9 − 3 = 6

EXAMPLE 4
a2
+ b2
= c2
42
+ b2
= 52
16 + b2
= 25
−16 −16
b2
= 9
b2
= 9 b = 3
b1
= 9; b2
= 9 − 3 = 6
A = 1
2
h(b1
+ b2
)

EXAMPLE 4
a2
+ b2
= c2
42
+ b2
= 52
16 + b2
= 25
−16 −16
b2
= 9
b2
= 9 b = 3
b1
= 9; b2
= 9 − 3 = 6
A = 1
2
h(b1
+ b2
)
A = 1
2
(4)(6 + 9)

EXAMPLE 4
a2
+ b2
= c2
42
+ b2
= 52
16 + b2
= 25
−16 −16
b2
= 9
b2
= 9 b = 3
b1
= 9; b2
= 9 − 3 = 6
A = 1
2
h(b1
+ b2
)
A = 1
2
(4)(6 + 9)
A = (2)(15)

EXAMPLE 4
a2
+ b2
= c2
42
+ b2
= 52
16 + b2
= 25
−16 −16
b2
= 9
b2
= 9 b = 3
b1
= 9; b2
= 9 − 3 = 6
A = 1
2
h(b1
+ b2
)
A = 1
2
(4)(6 + 9)
A = (2)(15)
A = 30 ft2

EXAMPLE 5
Find the area of each rhombus or kite.

EXAMPLE 5
A = 1
2
d1
d2

EXAMPLE 5
A = 1
2
d1
d2
A = 1
2
(7)(12)

EXAMPLE 5
A = 1
2
d1
d2
A = 1
2
(7)(12)
A = 42 ft2

EXAMPLE 5
A = 1
2
d1
d2
A = 1
2
(14)(18)

EXAMPLE 5
A = 1
2
d1
d2
A = 1
2
(14)(18)
A = 126 in2

EXAMPLE 6
One diagonal of a rhombus is half as long as the other diagonal. If
the area of the rhombus is 64 square inches, what are the lengths
of the diagonals?

EXAMPLE 6
of the diagonals?
A = 1
2
d1
d2

EXAMPLE 6
of the diagonals?
A = 1
2
d1
d2
d1
= x

EXAMPLE 6
of the diagonals?
A = 1
2
d1
d2
d1
= x
d2
= 1
2
x

EXAMPLE 6
of the diagonals?
A = 1
2
d1
d2
d1
= x
d2
= 1
2
x
64 = 1
2
(x)(1
2
x)

EXAMPLE 6
of the diagonals?
A = 1
2
d1
d2
d1
= x
d2
= 1
2
x
64 = 1
2
(x)(1
2
x)
64 = 1
4
x2

EXAMPLE 6
of the diagonals?
A = 1
2
d1
d2
d1
= x
d2
= 1
2
x
64 = 1
2
(x)(1
2
x)
64 = 1
4
x2
4(64) = ( 1
4
x2
)4

EXAMPLE 6
of the diagonals?
A = 1
2
d1
d2
d1
= x
d2
= 1
2
x
64 = 1
2
(x)(1
2
x)
64 = 1
4
x2
4(64) = ( 1
4
x2
)4
256 = x2

EXAMPLE 6
of the diagonals?
A = 1
2
d1
d2
d1
= x
d2
= 1
2
x
64 = 1
2
(x)(1
2
x)
64 = 1
4
x2
4(64) = ( 1
4
x2
)4
256 = x2
256 = x2

EXAMPLE 6
of the diagonals?
A = 1
2
d1
d2
d1
= x
d2
= 1
2
x
64 = 1
2
(x)(1
2
x)
64 = 1
4
x2
4(64) = ( 1
4
x2
)4
256 = x2
256 = x2
x = 16

EXAMPLE 6
of the diagonals?
A = 1
2
d1
d2
d1
= x
d2
= 1
2
x
64 = 1
2
(x)(1
2
x)
64 = 1
4
x2
4(64) = ( 1
4
x2
)4
256 = x2
256 = x2
x = 16
d1
= 16 in.

EXAMPLE 6
of the diagonals?
A = 1
2
d1
d2
d1
= x
d2
= 1
2
x
64 = 1
2
(x)(1
2
x)
64 = 1
4
x2
4(64) = ( 1
4
x2
)4
256 = x2
256 = x2
x = 16
d1
= 16 in.
d2
= 8 in.

PROBLEM SET
p. 767 #1, 2, 5-9 all; p. 777 #1-13 odd
“The best preparation for tomorrow is doing your best today.”
- H. Jackson Brown, Jr.

Geometry Section 11-1/11-2

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