Math 9
Review of Surface Area and Volume
Journeys Page 218
Worksheet 4-8b
Name SOLUTIONS
1. Find the area of the figure.
Area of Triangle
bh
A=
2
(10.5)(7.2)
A=
2
A = 37.8 cm2
P=5(10.5) = 52.5cm
Area of Figure = 5(Triangle)
= 5(37.8)
= 189 cm2
Area of Circle
A = r 2
A = (3.14)(6.6)2
A = (3.14)(43.56)
A = 136.7784
Half of Circle = 68.39 cm2
Area of Triangle
bh
A=
2
(11.2)(12.5)
A=
2
A = 70 cm2
P=7(11.2) = 78.4cm
Area of Figure = 7(Triangle)
= 7(70)
= 490cm2
Area of Rectangle
Total Area of Figure
A = lw
= Circle + Rectangle
A = (13.2)(9.1)
= 68.39 + 120.12
A = 120.12 cm2
= 188.51 cm2
Perimeter = 9.1+9.1+13.2+20.7
Circumference/2 = (3.14)(13.2)/2 = 20.7
= 52.1cm
Area of Circle
Area of Triangle
Total Area of Figure
bh
A = r 2
A=
= Circle + Triangle
2
( 2.1)(1.9)
A = (3.14)(1.05)2
A=
= 1.730925 + 1.995
2
A = 3.14(1.1025)
A = 1.995 m2
= 3.725925 m2
2
A = 3.46185 m
Half of a Circle
= 1.730925 m2
Perimeter = 2.2+2.2+3.3
= 7.7m
Circum/2=(3.14)(2.1)/2 = 3.3
c 2  a 2  b 2  1.9 2  (1.05) 2 c=2.2
2. Find the surface area.
Top
Front
A = lw
A = lw
A = (2.5)(1.5) A = (2.5)(3)
A = 3.75
A = 7.5
Bottom
A = lw
A = (4)(4)
A = 16
Bottom
bh
A=
2
(3)( 4)
A=
2
A=6
Side
A = s2
A = (7)2
A = 49
Side
A = lw
A = (1.5)(3)
A = 4.5
Side
bh
A=
2
( 4)(3)
A=
2
A=6
Total Surface Area
= 2(top) + 2(front) + 2(side)
= 2( 3.75) + 2(7.5) + 2(4.5)
= 7.5 + 15 + 9
= 31.5 cm2
Total Surface Area
= Bottom + 4(sides)
= 16 + 4(6)
= 40 cm2
Side
Side
A = lw
A = lw
A = (8)(4) A = (8)(5)
Side
A = lw
A = (8)(3)
A = 32
A = 40
A = 24
Total Surface Area
= 6(side)
= 6(49)
= 294 cm2
Total Surface Area
= 2(bottom) + side + side + side
= 2(6) + 32 + 40 + 24
= 108 cm2
3. The wind sock at an airfield shows a pilot
which way the wind is blowing. A wind
sock is a frustum of a cone, open at both
ends. It is made of light fabric. How
much fabric is used for this wind sock?
Surface Area of Big Cone
A = rs
A = 3.14(30)(150)
A = 14130
Surface Area of Small Cone Total Surface Area
A = rs
= Big – Small
A = 3.14(10)(50)
= 14130 - 1570
A = 1570
= 12560 cm2
4. How much sheet metal is needed to make this industrial hopper? The top is open
and the bottom closed.
Area of Trapezoid
h(a  b)
A=
2
(1)(1.6  0.75)
A=
2
A = 1.175
bottom
A = lw
Sides
Total Surface area
A = lw
= 2(traps) + Bottom + 2(sides)
A = (0.75)(3) A = (1.1)(3)
= 2(1.175) + 2.25 + 2(3.3)
A = 2.25
= 11.2 m2
A = 3.3
5. The Evergreen Landscaping Company is constructing soil rollers from closed
cylinders filled with sand.
a) How much metal is needed to make one roller?
b) What is the volume of one roller to the nearest litre? ( 1 cm3 = 1 ml)
c) Sand to fill the cylinders is sold by the half tonne. How much sand should
be ordered for 9 rollers if 1 L of sand has a mass of 2 kg?
Area of Circle
Area of Curved surface
Total Surface Area
2
A = r
A = dh
= 2( r 2 ) + dh
2
A = 3.14(20)
A = 3.14(40)(65)
= 2(1256) + 8164
A = 1256
A = 8164
= 10676 cm2
V = r 2 h
V = 3.14(20)2(65)
V = 81640 cm3
1000 cm3 = 1 L
81640 cm3  1000 = 81.64 L
1 L = 2 kg
81.64 L x 2 kg = 163.28 kg x 9 rollers = 1469.52 kg
6. A tissue box measures 24.0 cm by 11.5 cm by 5.0 cm. If 50 cm2 are added for
overlap, what is the area of cardboard needed to make the box?
Bottom
A = lw
A = (24)(11.5)
Front
Side
A = lw
A = lw
A =(24)(5) A = (11.5)(5)
Total Surface Area
= 2(bottom) + 2(front) + 2(side) + overlap
= 2(276) + 2(120) + 2(57.5) + 50
A = 276
A = 120
A = 57.5
= 957 cm2
7. a) How much water will it take to fill this swimming pool? ( 1 m3 = 1000 L)
b) How long will it take to fill the pool at a rate of 5 L every 30 s?
V = lwh
V = (2.5)(10)(6)
V = 150 m3
1 m3 = 1 L
30 x30  900s
 15 min
150
 30
Fill pool =
5
8. The curved surface is composed of steel panels 2.5 m by 1.5 m. About how many
panels are needed for the roof?
Area of Curved Surface
A = dh
A = 3.14(10)(20)
A = 628
Half of a Curved Surface = 314
Steel panels
A = lw
A = (2.5)(1.5)
A = 3.75
314
 83.73 Therefore you need 84 steel panels.
3.75