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3.3 Problem Solving in Geometry
Perimeter (Circumference), Area, and
Volume
Ex. Find the area:
26m
18m
18m
37m
21.1m
A = ½h(a + b)
A = ½(18)(26 + 37)
A = ½(18)(63)
A = 9(63)
A = 567
The area is 567 m2
Ex. A rectangle has a width of 46cm and a
perimeter of 208cm. What is its length?
x = length
width = 46 cm
length = x
P = 2L + 2W
208 = 2(x) + 2(46)
208 = 2x + 92
208 – 92 = 2x + 92 – 92
116 = 2x
116 = 2x
2
2
58 = x
The length is 58 cm
Circles
diameter – distance across
circle through the
center
d
r
radius – distance from
center to edge of circle
(half the diameter)
Note:
π is approximately 3.14
Ex. Find the area and circumference
(a) in terms of π
(b) rounded to the nearest whole number
9m
A = πr2
A = π(9)2
A = π81
A = 81π
(a) area is 81π m2
C = 2πr
C = 2π(9)
C = 18π
(a) circumference is 18π m
C = 18π
C ≈ 18(3.14)
A = 81π
C ≈ 56.52 m
A ≈ 81(3.14)
(b) circumference is
A ≈ 254.34
approximately 57 m
(b) area is approximately 254 m2
Ex. A cylinder with radius 3 inches and height 4 inches
has its radius tripled. How many times greater is the
volume of the larger cylinder than the smaller cylinder?
Ex. A cylinder with radius 3 inches and height 4 inches
has its radius tripled. How many times greater is the
volume of the larger cylinder than the smaller cylinder?
4in
4in
3in
9in
V1 = πr2h
V1 = π(3)2(4)
V1 = π(9) (4)
V1 = 36π in3
V2 = πr2h
V2 = π(9)2(4)
V2 = π(81) (4)
V2 = 324π in3
9
V2 324

9
V1
36
1
9 times
Ex. A water reservoir is shaped like a rectangular solid with a base that is 50 yards by
30 yards, and a vertical height of 20 yards. At the start of a three-month period of no
rain, the reservoir was completely full. At the end of this period, the height of the
water was down to 6 yards. How much water was used in the three-month period?
Ex. A water reservoir is shaped like a rectangular solid with a base that is 50 yards by
30 yards, and a vertical height of 20 yards. At the start of a three-month period of no
rain, the reservoir was completely full. At the end of this period, the height of the
water was down to 6 yards. How much water was used in the three-month period?
20yd
6yd
30yd
50yd
Vstart = lwh
Vstart = (50)(30)(20)
Vstart = 30,000 yd3
Vend = lwh
Vend = (50)(30)(6)
Vend = 9,000 yd3
Vstart – Vend = 30,000 – 9,000
= 21,000 yd3
Angles of a Triangle
The sum of the interior
angles of a triangle is
180°.
C
A
B
A° + B° + C° = 180°
Ex. One angle of a triangle is three times as large as another.
The measure of the third angle is 40°more than that of the
smallest angle. Find the measure of each angle.
Ex. One angle of a triangle is three times as large as another.
The measure of the third angle is 40°more than that of the
smallest angle. Find the measure of each angle.
A° + B° + C° = 180°
3x
x + 40
1  x
2  3x
3  x  40
x
x + 3x + x + 40 = 180
5x + 40 = 180
5x + 40 – 40 = 180 – 40
5x = 140
5x = 140
5
5
x = 28
x = 28°
3x = 3(28) = 84°
x + 40 = 28 + 40 = 68°
28°, 84°and 68°
straight angle
right angle
180°
90°
complementary angles – 2 angles whose sum is 90°
If one angle is x
complementary angle is 90 – x
B
A
A + B = 90°
supplementary angles – 2 angles whose sum is 180°
B
A
A + B = 180°
angle
50°
17°
x°
comp
40°
73°
(90 – x)°
If one angle is x
supplementary angle is 180 – x
supp
130°
163°
(180 – x)°
Ex. Find the measure of an angle whose
supplement measures 39° more than twice its
complement.
Ex. Find the measure of an angle whose
supplement measures 39° more than twice its
complement.
angle = x
comp = 90 – x
supp = 180 – x
180 – x = 2(90 – x) + 39
180 – x = 180 – 2x + 39
180 – x = 219 – 2x
180 – x + 2x = 219 – 2x + 2x
180 + x = 219
180 + x – 180 = 219 – 180
x = 39
The angle is 39°
Groups
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