Examples1. Convert 24 yards to feet. 2. Convert 5 yd 6 in to inches.Start with what you know:

24 yd

multiply by the conversion fraction (units you want on the top)

3 ft1 ydx = 72 ft

Start with what you know:

5 yd

multiply by the conversion fraction (units you want on the top)

36 in1 ydx = 180 in

5 yd 6 in= 180 + 6 in= 186 in

Examples3. Quentin is 5 ft. 1 in. tall. What is his height to the nearest centimetre?Start with what you know:

5 ft

multiply by the conversion fraction (units you want on the top)

12 in1 ftx = 60 in

60 + 1 in = 61 in.

61 in2.54 cm

1 inx = 154.94 cm

Quentin is 155 cm tall

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:

4 ft

multiply by the conversion fraction (units you want on the top)

0.3048 m

1 ftx = 1.2192 mThe goal is 1.2 m high

4. A map of Alberta has a scale of 1: 1 505 000. The distance on the

map between Calgary and Red Deer is inches.

What is the distance to the nearest mile?

134

3 1 ft12 in

(1505000 in) x

= 77.19775884 mi

The distance between Calgary and Red Deer is approximately 77 miles.

14

1 mi5280 ft

x

a b/c or

The top scale gives the whole number: (where the zero on

the bottom scale points)

19 mm

To get the decimal portion of the measurement find the

first set of lines that match up

.5

7. Determine the surface area of this right rectangular pyramid to the nearest square inch.

(length)(width)

= 24 in2

Need to find slant height (s)

A Base = = (3)(8)

a2 + b2 = c2

SA = 24 + 2(14.775) + 2(36.48)

A Sides = 42 + 92 = s2

16 + 81 = s2

97 = s2

√97 = s9.85 in = s

bh2

(3)(9.85)2=

= 14.775 in2 Blue

a2 + b2 = c2

A Front/Back = 1.52 + 92 = s2

2.25 + 81 = s2

83.25 = s2

√83.25 = s9.12 in= s

bh2

(8)(9.12)2=

= 36.48 in2 Red

= 126.51 in2 = 127 in2

3 in. 8 in.

9 in.

8. Determine the surface area of this right cone to the nearest square metre.

Need to find slant height (s)

SA = 83 m2

SACone = πr2 + πrs

= π(3)2 + π(3)(5.83)

a2 + b2 = c2

32 + 52 = s2

9 + 25 = s2

34 = s2

√34= s5.83 m = s = 9π + 17.49π

= 26.49π= 83.221

3 m

5 m

9. The surface area of a right cone is 400.2 m2. The radius of the cone is 6.0 m. Determine the height of the cone to the nearest metre.

h = 14 m

-36π -36π

a2 + b2 = c2

r2 + h2 = s2

36 + h2 = 231.98

h2 = 195.98h = √195.98h = 13.999 m

400.2 - 36π = 6πs

15.231 = s

6π 6π62 + h2 = 15.2312

SACone = πr2 + πrs400.2 = π(6)2 + π(6)s400.2 = 36 π + 6πs

287.1026645 = s6π

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 2r

SAHemi = 2 2r 2r+

2r= 32)6.11(= 3

= 1268.2 cm2

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.

VTOP = (area of base)(height)31

= (3.3)(6.8)(1.5)31

= 11.22

VBOTTOM = L x W x H= (3.3)(6.8)(3.3)= 74.052

VTOTAL = 11.22 + 74.052 = 85.272VTOTAL = 85.3 m3

|

1.5

6.83.3

3.3

m

mm

m

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= (5)2(12) VSPHERE = r3

= (1)3

= How many scoops of ice cream can be made?

= 225

225 full scoops

= 30034

34

34