New Century Maths 12 Mathematics General 2 HSC course
Worked Solutions
Chapter 3: Area and volume
SkillCheck
1 a 6.73 cm has three significant figures: 6, 7 and 3
b 12 mL has two significant figures: 1 and 2
c 68.0 g has three significant figures: 6, 8 and 0
d 12 000 km rounded to the nearest 100 km has three significant figures: 1, 2 and 0
2 a 24.67 m correct to two significant figures is 25 m.
b 3.69 kg correct to two significant figures is 3.7 kg.
c 248 670 000 km correct to two significant figures is 250 000 000 km.
d 0.052 49 cm correct to two significant figures is 0.052 cm.
3 a 1 m2  1  1002 cm2
 10 000 cm2
b 1 m3  1  1003 cm3
 1 000 000 cm3
c 1 ha 10 000 m2
d 1 cm2  1  102 mm2
 100 mm 2
e 1 cm3  1 mL
f 1 m3  1 000 000cm3
 1 000 000  1000 L
 1000 L
4 a A = πr2
 π  32
 28.3 cm 2
b A=

1
2
1
2
(a + b)h
 4  7  5
 27.5 cm2
c A = πr2
 68 
 π 
 2 
 π  34 2
2
 3632 m 2
d A = bh
 0.8  0.4
 0.32 m 2
e A=
1
2
xy
1
 10  6
2
 30 cm 2

5 a A  π  R2  r 2 
 π(62  42 )
 63
θ
 2πr
360
135

 2  π  14
360
 32.99
b l
θ
 πr 2
360
112

 π  7.4 2
360
 54
c l
6 Area of one tile is 1  1  1 cm2 .
Area of 250 000 tiles is 250 000  1  250 000 cm 2
250 000cm2  250 000 1002 m 2
 25 m2
Exercise 3-01: Percentage error
1 The smallest unit on a scale used to measure 35.41 g is 0.01 g.
The correct answer is B.
1
2 Absolute error is   0.1 m  0.05 m.
2
The correct answer is C.
3 a The actual length of the sides lies between 14.8 m  0.5  0.1 m  14.75 m and
14.8 m  0.5  0.1 m  14.85 m.
b Absolute error  0.5  0.1 m  0.05 m
0.05
 100%
14.8
 0.3%
c Percentage error 
4 a Length lies between 10.3  0.5  0.1  10.25 m and 10.3  0.5  0.1  10.35 m.
b Width lies between 6.7  0.5  0.1  6.65 m and 6.7  0.5  0.1  6.75 m.
5 Absolute error = 0.5  0.1  0.05
0.05
 100%
42.8
 0.12%
Percentage error 
The correct answer is C.
6 a i Absolute error = 0.5  1  0.5 km
0.05
 100%
6452
 0.007 7495...%
ii Percentage error 
 7.750  103%
  0.007 750%
b i Absolute error = 0.5  0.1  0.05 m
0.05
 100%
6.9
 0.724 63...%
ii Percentage error 
 0.72%
c i Absolute error = 0.5  100  50 m
50
 100%
8000
 0.625%
ii Percentage error 
 0.63%
d i Absolute error = 0.5  0.01  0.005 mg
0.005
 100%
39.85
 0.012 547 05...%
ii Percentage error 
 0.012 55%
e i Absolute error = 0.5  1  0.5 ML
0.5
 100%
39.85
 0.204 0816...%
ii Percentage error 
 0.204%
f
i Absolute error = 0.5  1  0.5 cm
0.5
 100%
165
 0.303 03...%
ii Percentage error 
 0.303%
7 The most accurate measurement is 6452 km (in question 6 part a) because it has the
smallest percentage error.
8 a 12 000 km measured to the nearest 1000 km has two significant figures.
b Absolute error = 0.5  1000  500 km
c Percentage error 
500
 100%
12 000
 4.166 666 66...%
 4.17%
7 a Absolute error = 0.5  1  0.5 cm
0.5
 100%
3
 16.666 6666...%
Percentage error 
 17%
b Absolute error = 0.5  0.1  0.05 cm
0.5
 100%
3.0
 1.666 666 66...%
Percentage error 
 1.7%
10 Absolute error = 0.5  10  5 km
5
 100%
1740
 0.287 356...%
Percentage error 
 0.287%
11 a 300cm and 200 cm are both measured correct to the nearest 10 cm so there are two
significant figures in each measurement.
b Absolute error = 0.5  10  5 cm
5
 100%
200
 2.5%
Percentage error 
Exercise 3-02: Areas of composite shapes
1 Area = area of rectangle + area of triangle
1
= 45  58 +  45  23
2
 3127.5 m2
The correct answer is A.
2 a Shaded area  area of square  area of rectangle
 2.4  2.4  1.1  0.6
 5.1 m2
b Engraving cost  cost per square metre  area
 26.50  5.1
 $135.15
3 a Shaded area  area of outer rectangle  area of inner rectangle
 6  20  9  3
 93 m 2
b Area  area of rectangle  (2  area of square)
 8 6 2 2 2
 56 m2
c Area = bh
 49
 36 cm 2
d Shaded area  area of square  area of circle
 190  190  π  952
 7747.1263...
 7747 mm 2
e Shaded area  area of outer circle  area of inner circle
 π  7 2  π  2.52
 134.3030...
 134 m2
f
Area  area of rectangle  area of rectangle  area of rectangle
 2  5  5  15  7  6
 127 m2
g Shaded area  area of rectangle  area of trapezium
 10  6 
1
 2.5  6   7
2
 30.25
 30 m 2
h Area  area of circle  area of rectangle  area of circle
2
 7.8 
2
 π
  7.8  10.2  π  3.5
2


 88.859 11...
 89 m 2
4 a Area  area of outer rectangle  area of rectangle  area of rectangle
 10  7  3  2  3  4
 52 m 2
b Area  area of square  (2  area of circle)
 8  8  2  π  42
 164.530 96...
 165 m 2
c Height of triangle  102  62
8
Area  area of triangle  area of semicircle
1
1
 8  12   π  62
2
2
 104.548 66...

 105 m 2
d Area  area of rectangle  area of triangle
 9.6  2.8 
1
 9.6  1.3
2
 33.12
 33 m 2
e Area  area of rectangle  area of trapezium
 12  8 
1
12  6   8
2
 168 m2
f
Area  area of trapezium  area of trapezium
1
1
 6  11  4   5  11  4
2
2
2
 66 m

5 a i Shaded area  area of square  (2  area of circle)
 10  10  2  π  2.52
 60.730 09...
 60.7 cm 2
ii Shaded area  area of square (see answer for part b below)
 16  16
 256 cm 2
b The semicircles that lie outside of the square in part ii fit exactly into the
semicircular cut outs from within the square. So the area required is just the area of
the square.
6 a A = 2400 × 1800
 4 320 000 mm 2
 4 320 000  10002 m 2
 4.32 m 2
b d = 1800 ÷ 3
 600 mm = 0.6 m
c Area remaining  area of tin  (12  area of circle)
 4.32  12  π  0.32
 0.927 07...
 0.93 m 2
7 AM = 582  322
 48.37
Let N be the midpoint of AM.
AN  48.37  2
 24.19
RA  24.192  162
 29
a Area of farm  area of triangle MFA  area of triangle ARM
1
1
=  48.37  32   48.37  16
2
2
2
 1161 m
b Perimeter = MF + FA + AR + RM
 32  58  29  29
 148 m
8 a Area 
1
1
1
1
 15  21   33  21   12  14   36  14
2
2
2
2
 840 m2
12
14
21
21
15
b Area 
1
1
1
1
1
 30  16   (16  37)  68   42  37   88  58   52  58
2
2
2
2
2
 6879 m2
42
37
46
58
16
22
30
Exercise 3-03: Circular shapes
θ
 πr 2
360
70

 π  82
360
 39.1 cm2
1 A
The correct answer is C.
2
θ
 2πr
360
θ
15 
2 π6
360
θ
15 
 12π
360
360  15  θ  12π
l
5400
θ
12π
θ  143
The correct answer is D.
θ
 2πr
360
80

 2  π  12
360
 16.8 m
3 a l
θ
 2πr
360
90

 2  π  9.6
360
 15.1 m
b l
θ
 2πr
360
135

 2  π  86
360
 202.6 mm
c l
4 a A
θ
 πr 2
360
90
 π  5.82
360
 26 m2

θ
 πr 2
360
270

 π  7502
360
 1 325 359 m 2
b A
θ
 πr 2
360
180

 π  7502
360
 883 573 m 2
c A
θ
 πr 2
360
30

 π  52
360
 7 m2
d A
θ
 πr 2
360
120

 π  902
360
 8482 cm 2
e A
f
θ
 πr 2
360
75

 π  4802
360
 150 796 km 2
A
5 a A = π(R2 – r2)
 π 122  92 
 198 cm 2
b A = π(R2 – r2)
 π 162  12.82 
 225 m 2
c A = π(R2 – r2)
 π 15.82  7.42 
 612 mm 2
d A = π(R2 – r2)
 π  427 2  115.52 
 531 000 km 2
θ
 πr 2
360
45

 π  152
360
 88 cm 2
6 a A
θ
 2πr
360
45

 2  π  15
360
 12 cm
b l
7 a The shape of the cross-section is an annulus.
b A = π(R2 – r2)
 π  2.252  1.62 
 7.9 m 2
1
× π(R2 – r2)
2
1
  π 8.62  3.42
2
 98 m2
8 a A=


98
30
= 3.266 6 . . .
b Number of tins 
Round up to the nearest whole number of tins. Four tins of paint are required.
Cost  4  152
 $608
9 a A
θ
 πr 2
360
150
 π  182
360
 424 cm 2

θ
 2πr  r  r
360
150

 2  π  18  18  18
360
 83 cm
b P
A = π(R2 – r2)
10
140  π  R 2  2.52 
140
 R 2  6.25
π
140
 6.25  R 2
π
R 2  50.813 38...
R  50.813 38...
R  7.1 mm
Exercise 3-04: Simpson’s rule
h
 d f  4d m  d l 
3
3
  4  4  5  6
3
 30
1 A
The correct answer is C.
h
 d f  4d m  d l 
3
18
 15  4  21  12 
3
 666 m 2
2 a A
h
 d f  4d m  d l 
3
15
 16  4  24  28
3
 700 m2
b A
3 a A
h
 d f  4d m  d l 
3
5
 4.5  4  6  8
3
 61 m 2

b Cost 
61
 45
12
= $228.75
4 Using the first three measurements:
h
 d f  4d m  d l 
3
4
  0  4  7  8
3
 48
A
Using the second three measurements:
h
 d f  4d m  d l 
3
4
  8  4  12  10 
3
 88
A
Total area = 48 + 88 = 136 m2
5 a Using the first three measurements:
h
 d f  4d m  d l 
3
6
  2.5  4  4  1.8 
3
 40.6
A
Using the second three measurements:
h
 d f  4d m  d l 
3
6
 1.8  4  7.2  7 
3
 75.2
A
Total area = 40.6 + 75.2 = 115.8 m2
b Volume = area × depth
 115.8  1.2
 139 m3
 140 m (correct to two significant figures)
h
 d f  4d m  d l 
3
h
 0  4  y  0
3
4hy

3
6 A
The correct answer is D.
7 a
23
20
15
17
5
12
12
12
12
b Using the first three measurements:
h
 d f  4d m  d l 
3
12
  5  4  17  23
3
 384
A
Using the second three measurements:
h
 d f  4d m  d l 
3
12
  23  4  15  20 
3
 412
A
Total area = 384 + 412 = 796 m2
c Volume = area × depth
 796  20
 15 920 m3
Capacity  15 920  1000 litres
 15 920 000 litres
 15 920 000  1 000 000 megalitres
 15.92 megalitres
Exercise 3-05: Volume and surface area of a prism
1 V = Ah
 (1  0.6)  2
 1.2 m3
The correct answer is A.
2 a V = Ah
 (8  6)  5
 240 m3
b V = Ah
 (7.6  7.6)  7.6
 439 m3
c V = Ah
1

   9  5   14   12
2

 1176 m3
d V = Ah
  5.4  5.4   6.8
 198 m3
e V = Ah
1

   3  8   15
2


3
 180 m
f Find the height of the triangular face first.
h  102  62
10
h
6
 64
8
V  Ah
1

   12  8   20
2

3
 960 m
3 Surfacearea  1  2  2  1  0.6   2   2  0.6 
 5.6 m2
The correct answer is C.
4 Surfacearea  2  16.4  7.4  2   7.4  23.1  2  16.4  23.1
 1342 cm2
The correct answer is A.
5 a Surfacearea  2  8  6  2   6  5  2  8  5
 236 m2
b Surfacearea  6   7.6  7.6 
 347 m2
c Find the length of the slant edge of the trapezium first.
x  22  142
 200
 14.14
x
9
2
1

Surface area  2    5  9   14   2  14.14  12   9  12  5  12
2

2
 703 m
d Surfacearea  2  5.4  5.4  4   6.8  5.4 
 205 m2
e Find the length of the hypotenuse of the triangular face first.
c  32  82
c
8
3
 73
 8.54
1

Surface area  8.54  15  3  15  8  15  2    3  8 
2

2
 317 m
f The height of the triangular faces is 8, see the answer for question 2f.
1

Surface area  2  10  20   12  20  2    12  8 
2

2
 736 m
6 Find the height of the triangular faces first.
h  502  302
30
50
h
 1600
 40 cm
1

Surface area  2    0.6  0.4   2   0.5  1.5 
2

2
 1.7 m
7 a A  0.4   0.32  0.32  0.3  0.32   0.32  0.3  0.3  0.3
 0.6644 m2
P  1.02  0.3  0.3  0.32  0.32  0.32  0.40  0.94
 3.92 m
h  1.2 m (given)
S  2 A  Ph
 2  0.6644  3.92  1.2
 6 m2
b V = Ah
 0.6644  1.2
 0.80 m3 (correct to two significant figures)
8 a V = Ah
1

   3  14   4
2

 84 cm3
b Find the length of the hypotenuse of the triangular faces first.
c
3 cm
14 cm
c  32  142
 205
 14.32 cm
1

Surface area  3  4  4  14.32  4  14  2    14  3 
2

 167.28 cm 2
 167.28  10 000 m 2
 0.017 m 2
1
0.017
 298
Number of wedges that can be varnished with five pots  5 
Kobi can varnish 298 wedges with five pots.
Exercise 3-06: Volume and surface area of a cylinder
1 a V = πr2h
 π  152  10
 7068.6 cm3
1 2
πr h
2
1
  π  0.62  1.5
2
 0.8 m3
b V=
1 2
πr h
2
1
  π  22  5
2
 31.4 m3
c V=
d V = πR2h – πr2h
 π  5.52  200  π  42  200
 8953.5 cm3
2 a V = πr2h
 π  42  6.5
 327 cm3
b Engine capacity  6  326.72...
 1960 cm3
 2L
3 a A  2πr 2  2πrh
 2  π  72  2  π  7  12
 840 m2
b A  2πr 2  2πrh
 2  π  0.42  2  π  0.4  2
 6.0 m2
c A  2πr 2  2πrh
 2  π  62  2  π  6  24
 1100 cm2
d A  2πr 2  2πrh
 2  π  0.652  2  π  0.65  0.2
 3.5 m2
e A  2πr 2  2πrh
 2  π  2.42  2  π  2.4  2.4
 72 m2
f
A  2πr 2  2πrh
 2  π  0.852  2  π  0.85  1.3
 11 m2
4 Top layer : A  πr 2  2πrh
 π  152  2  π  15  12
 1837.83 cm2
Bottom layer : A  π  R 2  r 2   2πRh
 π  252  152   2  π  25  15
 3612.83 cm3
Area of icing  1837.83  3612.83
 5451cm 2
5 a V=
1 2
πr h
2
1
 π  12  2
2
 3.1 m3

1
2πrh  2πr 2   2  2

2
1
 2  π  1  2  2  π  12  2  2
2
 13.4 m2
b Surface area 


6 a Vblock   30  30   5
 4500 cm3
Vhole  πr 2 h
 π  52  5
 392.70 cm 3
Vconcrete  4500  392.70
 4107 cm3
The volume of the concrete block is 4107 cm3, to the nearest cm3.
b % removed 
392.70
 100%
4500
 8.7%
7 a Vhole  πr 2 h
 π  2.52  2
 39 m3
b Vtank  πr 2 h
 π  22  2
 25.133 m3
 25.133  1000 L
= 25133 L
8 Voriginal  πr 2 h
 π  52  h
 25πh
Vnew  πr 2 h
 π  102  h
 100πh
Vnew
100πh

Voriginal 25πh
4
The volume is four times larger.
The correct answer is C.
Exercise 3-07: Volume and surface area of a sphere
4 3
πr
3
4
  π  423
3
 310 000 km3
1 a V=
1
4
× πr3
2
3
1 4
   π  47.53
2 3
 224 000 m3
b V=
4 3
πr
3
4
  π  6003
3
 9.05  108 m3
c V=
1
4
× πr3
2
3
1 4
   π  783
2 3
 993 899 mm3
d V=
 993 899  1000 cm3
 994 cm3
e V=
4 3
πr
3
4
 π  83.53
3
 2 438 641 cm3

 2 438 641  1 000 000 m 3
 2.44 m3
4 3
πr
3
4
  π  143
3
11500 m3
f V=
2 a i Radius of Earth = 12 683 ÷ 2 = 6341.5 km
Surface area of Earth  4πr 2
 4  π  6341.52
 505 351 847.3... km 2
1
Area not covered by water   505 351847.3...
5
 101 070 369.5...
 1.011  108 km 2
4 3
πr
3
4
  π  6341.53
3
1.068  1012 km3
ii V =
b Mass per cubic kilometre 
5.974  1024
1.068  1012
 5.594  1012 t/km3
1
4
× πr3
2
3
1 4
   π  0.53
2 3
 0.2618 m3
3 a V=
 0.2618  1000 L
 262 L
1
 4πr 2
2
1
  4  π  0.52
2
 1.57 m2
b Surface area 
4 3
πr
3
4
  π  33
3
113 cm3
4 a Vbauble 
b Vbox  6  6  6
 216 cm3
Vair  Vbox  Vbauble
 216  113
 103 cm 3
c Surface area of one bauble = 4r2
= 4 ×  × 32
= 113.0973 ... cm2
Surface area of 100 baubles = 113.0973 ... × 100
= 11 309.73 ... cm2
= 11 309.73 ... ÷ 10 000 m2
= 1. 130973 ... m2
Cost to cover 100 baubles = 1. 130973 ... × $15
= $16.96460 ...
≈ $16.96
5
V=
4 3
πr
3
4 3
πr
3
400  3  4πr 3
400 
1200  4πr 3
1200
 r3
4π
r 3  95.4929...
r  3 95.4929...
r  4.5707...
r  4.6
The correct answer is A.
4 3
πr
3
4
  π  283
3
 92 000 cm3
6 a V=
b Surface area  4πr 2
 4  π  282
 9852 cm2
Each tile covers 1 cm2, so it takes 9852 tiles to cover the surface of the ball.
7 a Find the surface area of a sphere rather than that of two half spheres.
Surface area  4πr 2
 4  π  0.482
 2.9 m 2
b A = 2 × πr2
 2  π  0.482
 1.4 m2
c Find the volume of one sphere rather than two half spheres.
4 3
πr
3
4
  π  0.483
3
 0.46 m3
V
Exercise 3-08: Volumes of pyramids and cones
1
Ah
3
1
  9.32  12.5
3
 360 cm3
1 a V=
1
Ah
3
1
   8.3  4.7   16
3
 208 m3
b V=
1
Ah
3
1 1

    18  11  15
3 2

c V=
 495 mm3
1
Ah
3
1
   26  18   10
3
 1560 cm3
d V=
1
Ah
3
1
  162  20
3
 1707 m3
e V=
1
Ah
3
1 1

    6  8  6
3 2

f V=
 48 m3
1 2
πr h
3
1
  π  122  15
3
 2261.95 cm3
2 a V=
1 2
πr h
3
1
  π  6.42  8.2
3
 351.72 cm3
b V=
1 2
πr h
3
1
  π  0.6752  2.45
3
 1.17 m3
c V=
1 2
πr h
3
1
  π  302  65
3
 61 261.06 mm3
d V=
1
3 V = 2 × Ah
3
1
 2   62  4
3
 96 mm3
1 2
πr h
3
1
  π  8.22  20.5
3
 1440 cm3
4 a V=
b Capacity  1440  1000
 1.44 L
1
5 Vlarge cone  πr 2 h
3
1
  π  102  25
3
 2618 cm3
1
Vtip  πr 2 h
3
1
  π  42  10
3
 168 cm 3
Vfrustum  Vlarge cone  Vtip
 2618  168
 2450 cm3
1
Ah
3
1
  2302  137
3
 2 420 000 m3
6 V=
1
7 Vcone  πr 2 h
3
1
  π  52  4
3
 104.7 m3
Vcylinder  πr 2 h
 π  52  6
 471.2 m3
Vwheat  Vcone  Vcylinder
 104.7  471.2
 576 m3
Exercise 3-09: Volume and surface area of composite solids
1 4 3
 πr
2 3
1 4
   π  33
2 3
 56.54... cm3
1 a Vhemisphere 
1
Vcone  πr 2 h
3
1
  π  32  13
3
 122.52... cm3
Vice-cream  Vhemisphere  Vcone
 56.54...  122.52...
 179.1 cm3
1
 4πr 2
2
1
  4  π  32
2
 56.5 cm 2
b Surface area of hemisphere 
1
2 Vcone  πr 2 h
3
1
  π  3.52  11.5
3
147.52 cm3
Vcylinder  πr 2 h
 π  3.52  11
 423.33 cm3
Vgunpowder  Vcone  Vcylinder
 147.52  423.33
 571 cm3
3 a Vcan  πr 2 h
 π  37.52  225
 994 020 mm3
4 3
πr
3
4
  π  37.53
3
 220 893.2... mm3
b Vball 
Vthree balls  3  220 893.2...
 662 680 mm3
Vair  994 020  662 080
 331 341mm3
% air 
331341
 100%
994 020
 33.3%
c The square-based prism would have a base with side length 75 mm (which is the
diameter of a ball) and a height of 225 mm (which is the sum of the diameters of three
balls).
V  Ah
 752  225
 1 265 625 mm3
d The volume of the cylinder containing the balls is less than that of a square prism
containing the balls. The cylinder is the most efficient method of packaging because it
contains the least amount of air.
4 a V = Ah
 110  120   30
 1 716 000 mm3
b Vhole  πr 2 h
 π  102  130
 40 840.7 mm3
Vafter drilling  1 716 000  3  40 840.7
 1 593 478 mm3
 1 593 478  1000 cm3
 1593cm3
c Percentage removed 
3  40 840.7
 100%
1 716 000
 7%
5 a Vwhole  πr 2 h
 π  192  8.5
 9640 cm3
7
 9640
8
 8400 cm3
Vremaining 
b The top and bottom of the slice are both sectors with angle of 45o and radius 19 cm.
θ
 πr 2
360
45

 π  192
360
 141.8 cm 2
Asector 
The sides of the slice are both rectangles with length 19 cm and width 8.5 cm.
Arectangle  19  8.5
 161.5 cm2
The curved surface is a rectangle with width 8.5 cm and a length that is
circumference of a circle of radius 19 cm. Call this length l.
θ
 2πr
360
45

 2  π  19
360
 14.92 cm
l
Acurved surface  14.92  8.5
 126.8 cm2
Surface area  2  Asector  2  Arectangular sides  Acurved surface
 2  141.8  2  161.5  126.8
 730 cm 2
6 a Pool A:
A  πr 2  lw
 π  1.52  3  7
 28.069 m 2
45
of the
360
V  Ah
 28.069  1.5
 42.103 m3
 42.103  1000 L
 42 103 L
Pool B:
1
A  lw  (a  b)h
2
1
 6  1.5  (1.5  2)  4
2
2
 16 m
V  Ah
 16  3
 48 m3
 48  1000 L
 48 000 L
b Pool B.
Difference  48000  42103
 5897 L
c From part a: Afront  Aback  16 m 2
Aleft side  1.5  3
 4.5 m 2
Aright side  2  3
 6 m2
Aflat bottom  6  3
 18 m2
Asloped bottom  l  3
Find l using Pythagoras’ theorem:
4
l
0.5
l  42  0.52
 4.03 m
Asloped bottom  4.03  3
 12.1 m2
Tiled area  16  4.5  16  6  18.12.1
 73 m 2
7 a Find the height of the triangular cut-out.
10
20
h  20 2  10 2
 17.32 cm
h
Afront face  Arectangle  Atriangle
1
 40  68   20  17.32
2
 2546.8 cm 2
V  Ah
 2546.8  120
 305 616 cm3
 305 616  1 000 000 m3
 0.31 m3
b Afront face  2546.8cm2
Abottom face  68  120
 8160 cm2
Aside face  40  120
 4800 cm2
Atop white rectangle  24  120
 2880 cm2
Atop green rectangle  20  120
 2400 cm2
Surface area  2  Afront face  Abottom face  2  Aside face  2  Atop white rectangle  2  Atop green rectangle
 2  2546.8  8160  2  4800  2  2880  2  2400
 33 414 cm 2
1 4 3
 πr
2 3
1 4
   π  33
2 3
 56.548... cm3
8 Vhemisphere 
1
Vcone  πr 2 h
3
1
  π  32  7
3
 65.973... cm3
Vtop  Vhemisphere  Vcone
 56.548...  65.973...
 122.521...
 123 cm3
Exercise 3-10: Error in area and volume calculations
1 a Largest possible length  14.8  0.05
 14.85 m
Largest possible area  14.852
 220.5225 m2
b Smallest possible length  14.8  0.05
 14.75 m
Smallest possible area  14.752
 217.5625 m2
c The length 14.8 m is measured to three significant figures, so the area should be given
to three significant figures.
Area  14.82
 219.04
 219 m2
2 a Largest possible area  10.35  6.75
 69.8625 m2
Smallest possible area  10.25  6.65
 68.1625 m 2
The actual area lies between 68.1625 m2 and 69.8625 m2.
b 10.3 m has three significant figures. 6.7 m has two significant figures. So 6.7 m is the
measurement with fewer significant figures.
c The length 6.7 m is measured to two significant figures, so the area should be given to
two significant figures.
Area  10.3  6.7
 69.01
 69 m 2
8.45  9.25
2
 39.081 25 m 2
3 a Largest possible area 
8.35  9.15
2
 38.201 25 m2
Smallest possible area 
b Each length is measured to two significant figures, so the area should be given to two
significant figures.
8.4  9.2
2
 38.64
Area 
 39 m 2
4 a Largest possible diameter  12 750 km
4 3
πr
3
4
  π  63753
3
 1.085  1012 km3
Largest possible volume 
Smallest possible diameter  12 650 km
4 3
πr
3
4
  π  63253
3
 1.060  1012 km3
Smallest possible volume 
b
4 3
πr
3
4
  π  63503
3
 1.07  1012 km3
Volume 
Rounded to 3 significant figures because the measurement 12 700 km is to 3 significant
figures
5 a Largest possible length  3.4  0.05
 3.45 cm
Largest possible surface area  6  3.452
 71.415 cm2
Smallest possible length  3.4  0.05
 3.35 cm
Smallest possible surface area  6  3.352
 67.335 cm2
b The side length is measured to two significant figures, so the area should be given to
two significant figures.
Surface area  6  3.42
 69.36
 69 cm 2
c Largest possible volume  3.453
 41.063 625 cm3
Smallest possible volume  3.353
 37.595 375 cm3
d The side length is measured to two significant figures, so the volume should be given to
two significant figures.
Volume  3.43
 39.304
 39 cm3
6 a Largest possible surface area  4πr 2
 4  π  17452
 38 264 913 km 2
Smallest possible surface area  4πr 2
 4  π  17352
 37 827 603 km 2
b The diameter is measured to three significant figures, so the surface area should be
given to three significant figures.
Surface area  4πr 2
 4  π  17402
 3.8  107 km 2
7 a 32 cm and 28 cm each have two significant figures, so the volume should be given to
two significant figures.
1
V  πr 2 h
3
1
  π  142  32
3
 6600 cm3
b 12.3 cm and 64.9 cm each have three significant figures, so the volume should be given
to three significant figures.
V  πr 2 h
 π  12.32  64.9
 30 800 cm3
c The measurement with the least number of significant figures is 1.4 m, with two
significant figures. The volume should, therefore, be given to two significant figures.
V  Ah
 11.5  1.4
 16 m3
8 12.3 cm and 64.9 cm each have three significant figures, so the surface should be given to
three significant figures.
Surface area  2πr 2  2πrh
 2  π  12.32  2  π  12.3  64.9
 5970 cm 2
Sample HSC problem
a Surface area  πr 2  2πrh
 π  0.62  2  π  0.6  3.5
 14 m2
b Capacity  πr 2 h
 π  0.62  3.5
 3.958 m 2
 3.958  1000 L
 4000 L
1 2
πr h  3.958
3
c
1
 π  r 2  3.5  3.958
3
π  r 2  3.5  3  3.958
π  r 2  3.5  11.874
11.874
r2 
π  3.5
r 2  1.079 88...
r  1.079 88...
r  1.3091...
d  2r
 2  1.3091...
 2.1 m
Revision
1 a Absolute error   0.1  2
  0.05 m
b Percentage error 
0.05
 100%
5.8
  0.86%
c Smallest possible area  5.752
 33.0625 m2
Largest possible area  5.852
 34.2225 m2
d The length has two significant figures, so the area must be given correct to two
significant figures.
A  5.82
 34 m2
e V = Ah
 34  0.05
 1.7 m3
2 a Area  area of rectangle  area of semicircle
1
 lw  πr 2
2
1
 8  15   π  42
2
2
 150 cm
b Area  area of rectangle  area of rectangle  area of triangle
1
 lw  lw  bh
2
1
 100  20  180  20   20  20
2
2
 5400 m
θ
 πr 2
360
35

 π  462
360
 646.3 cm 2
3 a A
θ
 2πr
360
35

 2  π  46
360
 28.1 cm
b l
4 a A  πr 2
 π  502
 7853.98 m 2
Cost  7853.98  0.95
 $7461.28
b A  π(R 2  r 2 )
 π  652  502 
 5400 m 2
h
 d f  4d m  d l 
3
11
  6  4  12  4 
3
 212.7 m 2
5 A
6 a Find the height of the trapezium.
0.8
h  1.12  0.82
 0.75 m
h
1.1
1
a  b h
2
1
   2.9  1.3  0.75
2
 1.575 m 2
Atop face 
Afront rectangle  1.3  0.9
 1.17 cm2
Aleft rectangle  1.1  0.9
 0.99 m2
Aback rectangle  2.9  0.9
 2.61 m2
Surface area  2  Atop face  Afront rectangle  2  Aleft rectangle  Aback rectangle
 2  1.575  1.17  2  0.99  2.61
 8.9 m 2
b V = Ah
 1.575  0.9
 1.4 m3
7 V  πr 2 h
 π  1.42  7
 43.1026... m3
 43.1026... 1000 L
 43102.65 L
Number of bottles  43102.65  0.75
 57 470
8 a Surface area  4πr 2
 4  π  202
 5026.55 cm2
4 3
πr
3
4
  π  203
3
 33 510.32 cm3
b V
1
Ah
3
1
  18  15  12
3
 1100 m3
9 a V
1
Ah
3
1
  π  42  10
3
 170 cm3
b V

10 a V 

1 4 3 4 3
  πR  πr 
2 3
3


1 4
4

   π  113   π  83 
2 3
3

 1715 cm3
1 4 3
 πr
2 3
1 4
   π  83
2 3
 1072 cm3
b Vinternal 
The bowl can hold 1072 mL, which is approximately 1.1 L.
1
11 Afront face  lw  bh
2
1
 4.8  5.2   4.8  2.7
2
2
 31.44 m
V  Ah
 31.44  6.5
 204 m3
1
12 a Afront face  lw  πr 2
2
1
 10  13   π  52
2
 169.27 cm 2
V  Ah
 169.27  28
 4700 cm3
b Length of straight edges  10  13  13
 36 cm
1
 2πr
2
1
  2 π5
2
 15.7 cm
Length of curved edge 
Crust length  36  15.7
 52 cm
© Cengage Learning Australia Pty Ltd 2013 MAT12MMWS10023 Measurement: Further
applications of area and volume
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