1. x x 3 ft 36 in 24 yd = 72 ft 5 yd = 180 in 1 yd 1 yd
Examples
1. Convert 24 yards to feet Convert 5 yd 6 in to inches.
Start with what you know:
Start with what you know:
3 ft
36 in x x
24 yd
= 72 ft
5 yd
= 180 in
1 yd
1 yd
5 yd 6 in= in
multiply by the conversion fraction (units you want on the top)
= 186 in
multiply by the conversion fraction (units you want on the top)

2. x x x 12 in 2.54 cm 5 ft = 60 in 61 in = 154.94 cm 1 ft 1 in
Examples
3. Quentin is 5 ft. 1 in. tall. What is his height to the nearest centimetre?
Start with what you know:
12 in
2.54 cm x x
5 ft
= 60 in
61 in
= cm
1 ft
1 in
in =
61 in.
Quentin is cm tall
multiply by the conversion fraction (units you want on the top)
4. An indoor lacrosse goal is 4 ft. high. What is this measurement to the nearest tenth of a metre?
Start with what you know:
The goal is 1.2 m high x m
= m
4 ft
1 ft
multiply by the conversion fraction (units you want on the top)

3. The distance between Calgary and Red Deer is approximately 77 miles.
4. A map of Alberta has a scale of 1: The distance on the map between Calgary and Red Deer is inches.
What is the distance to the nearest mile? 1
1 ft
1 mi 3
( in) x x 4
12 in
5280 ft
a b/c or
= mi
The distance between Calgary and Red Deer is approximately 77 miles.

4. The top scale gives the whole number: (where the zero on the bottom scale points)
To get the decimal portion of the measurement find the first set of lines that match up mm .5

5. 7. Determine the surface area of this right rectangular pyramid to the nearest square inch.
A Base =
= (3)(8)
(length)(width)
= 24 in2
Need to find slant height (s)
(3)(9.85) bh
a2 + b2 = c2
A Sides = =
= s2 2 2
= s2
Blue
= in2
97 = s2
√97 = s
9.85 in = s
SA = (14.775) + 2(36.48)
= in2
(8)(9.12)
= 127 in2
a2 + b2 = c2 bh
A Front/Back = =
= s2 2 2
= s2
Red
83.25 = s2
= in2
√83.25 = s
9.12 in= s

6. 8. Determine the surface area of this right cone to the nearest square metre.
Need to find slant height (s)
a2 + b2 = c2
= s2
SACone =
πr2 + πrs
= s2
34 = s2
= π(3)2 + π(3)(5.83)
√34= s
5.83 m = s
= 9π π
= 26.49π =
SA = 83 m2

7. πr2 + πrs 400.2 = π(6)2 + π(6)s 400.2 = 36 π + 6πs 400.2 - 36π = 6πs
9. The surface area of a right cone is m2. The radius of the cone is 6.0 m. Determine the height of the cone to the nearest metre.
SACone =
πr2 + πrs
400.2 = π(6)2 + π(6)s
a2 + b2 = c2
400.2 = 36 π + 6πs
r2 + h2 = s2
-36π π
π = 6πs
62 + h2 =
36 + h2 = 6π 6π
h2 =
= s
h = √195.98 6π
h = 14 m
h = m
= s

8. SASphere = 4 SAHemi = 2 + = 3 = 3 = 1268.2 cm2
10. A hemisphere has radius 11.6 cm. What is the surface area of the hemisphere to the nearest tenth of a square centimeter?
SASphere = 4
SAHemi = 2 +
= 3
= 3
= cm2

9. (area of base)(height) 3 1 = (3.3)(6.8)(1.5) 3 = 11.22
11. Determine the volume of this composite object, which is a right square prism and a right rectangular pyramid, to the nearest tenth of a cubic metre. 1
VTOP =
(area of base)(height) 3 1
= (3.3)(6.8)(1.5) 3
= 11.22
VBOTTOM =
L x W x H
= (3.3)(6.8)(3.3) =
VTOTAL = =
VTOTAL = 85.3 m3

10. (area of base)(height) = r2h 4 VSPHERE = r3 = (5)2(12) 3 = 300 4
12. A pail of ice cream is cylindrical, with diameter 10 in. and height 12 in. A scoop makes a sphere of ice cream with diameter 2 in. How many full scoops of ice cream can be made from this pail?
VCYLINDER =
(area of base)(height)
= r2h 4
VSPHERE =
r3
= (5)2(12) 3
= 300 4
= (1)3 3 4
How many scoops of ice cream can be made?
=  3
= 225
225 full scoops