Chapter 9 – Geometry and measures:
Length, area and volume
4
a  b . This is the
The area of the parallelogram is
h
same as two trapezia.
Exercise 9A
1
2
3
2
2
5
Two of 20 cm and two of 16 cm
2
2
2
8 cm, 25.1 cm, 50.3 cm
2
5.2 m, 16.3 m, 21.2 m
2
6 cm, 37.7 cm, 113 cm
2
1.6 m, 10.1 m, 8.04 m
6
a
7
trapezium = 56, square area = 9, shaded area = 56
2
– 9 = 47 cm
a
c
5 cm
18 m
8
4, because the total area doubled is about 32 m
9
80.2%
a
c
b 8 cm
d 12 cm
2
2
25 cm
2
100 cm
b 36 cm
2
d 0.25 m
57 m
b 702.5 cm
10 1 100 000 km
11 160 cm
c
5
4 complete revolutions
6
1p : 3.1 cm , 2p : 5.3 cm2, 5p : 2.3 cm2,
2
10p : 4.5 cm
7
0.83 m
8
38.6 cm
Exercise 9D
9
Claim is correct (ratio of the areas is just over 1.5 :
1)
1
ai
bi
ci
di
2
2 cm, 6 cm
3
a 73.8 cm
4
area of sector = 1 × π × 82 = 16π,
12 a
many possible correct answers,
eg base 6 cm, top 4cm, height 1.6π. Shaded
area is 8π, trapezium must be the the same
b the dimensions cannot be exact due to the
value of π in the area of the circle
2
18 cm
11 9 cm
2
b 4 cm
2
2
12 Divide 31.3 by π to get about 10 m. This is the
diameter of the tree. Is your classroom smaller than
10 m × 10 m? It probably isn’t, but you need to
check.
14 a 2π
b I 8π
ii
18π
2
2
a 96 cm
2
d 125 cm
2
No, she has used the sloping side instead of the
perpendicular height. It should be 6 × 4 = 24 cm²
4
b 70 cm
2
e 10 cm
c
f
a 500 cm²
43.6 cm
20 cm
2
112 m
Each parallelogram has an area of 30 cm². The
height of each is 5 cm so the length of each must
be 6 cm. x = 6 + 4 + 6 = 16 cm so Freya is
incorrect.
b 3 × 5 = 15
2
2
6
2
1
2
22.3 cm
2
20.7 cm
2
98.0 cm
2
55.6 cm
4
a 107 cm
2
b 173 cm
7
Exercise 9B
ii
ii
ii
ii
b 20.3 cm
5
iii 32π
A = 2π × r2
5.59 cm
8.29 cm
16.3 cm
15.9 cm
area of circle = π × 42 = 16π
13 45 complete revolutions
3
8
180
a

b If arc length is 10 cm, distance along chord
joining the two points of the sector on the
circumference will be less than 10 cm, so angle
at centre will be less than 60°
2
a 66.8°
b 10 cm
9
Let sector have radius R and arc length C, the
angle of the sector is found by
θ = 360 x C
2xπxR
2
and so the area will be 360 x C x π x R
2 x π x R x 360
= CR
2
Exercise 9C
1
2
2
8.80 m
10 a
84 m
2
4
c
2
a
b
c
d
2
a 30 cm
2
d 42 cm
g 3 cm
b 77 cm
2
e 40 m
h 10 cm
2
c
f
24 cm
6 cm
2
10 (36 − 72) cm
11 36.5 cm
2
2
12 16 cm (15.7)
2
2
Area = 15 cm
3
a 27.5 cm, 36.25 cm
2
b 33.4 cm, 61.2 cm
2
c 38.5 m, 90 m
In square Y, the four quarter circles will join
2
13 Each square has side length of r
Shaded part of square X = r2 – 1 πr2
4
= r2( 1 – 1 π )
4
together to give an area of radius 1 r, so
not exactly cylindrical. One possible answer is height of
6.6 cm and diameter of 8 cm.
2
shaded area
2
in Y = r2 – π  r  = r2 – 1 πr2 = r2(1 – 1 π),
2
4
4
which is the same as square X.
a i
b i
c i
2
a
3
a
21 cm2
48 cm2
36 m2
432 m3
ii
ii
ii
63 cm3
432 cm3
324 m3
b 225 m3
c
9
About 127 cm
Exercise 9G
1332 m3
A cross-section parallel to the side of the pool
always has the same shape.
b About 3 21 hours
1
2
7.78 g/cm3
10 A diameter of 10 cm and a length of 5 cm give a
volume close to 400 cm3 (0.4 litres).
Exercise 9E
1
8
4
V=
5
27 = 3 × 3 × 3, 27 + 37 = 64 = 43, 4 – 3 = 1. Hence
the side length is 1 small cube longer, hence 2 cm
longer
1 a 56 cm3
d 84 cm3
2
1
3
3
a
(1.5 + 3) × 1.7 × 2 = 7.65 m3
21 cm3
54 cm2
ii
ii
b 168 cm3
e 160 cm3
c 1040 cm3
1
base area × h = 3 × 9 × 9 × 10 = 270 cm
3
Put the apexes of the pyramids together. The 6
square bases will then form a cube.
b If the side of the base is a then the height will
1
be 2 a.
Total volume of the 6 pyramids is a3.
1
Volume of one pyramid is 6 a3 =
1
1
1
3 × 2 × a × a2 = 3 × height × base area
6
a i
b i
210 cm3
270 cm2
7
146 cm3
8
78 m3 (78.3 m3)
4
6.9 m ( 3 height of pyramid)
9
327 litres
5
a
10 10.2 tonnes
6
208 g
11 She was silly because 188160 is simply all the
numbers multiplied together. The volume is
672 cm2
7
1.5 g
8
1
1
3
73.3 m3
b 45 m3
c
3250 cm3
× 6.4 × 6.4 × H = 81.3
3 x 81.3
So H = 6.4 x 6.4 = 5.954 = 6.0 (2sf)
Exercise 9F
1 a
b
c
d
i
i
i
i
226 cm3
14.9 cm3
346 cm3
1060 cm3
ii
ii
ii
ii
207 cm2
61.3 cm2
275 cm2
636 cm2
2 a
b
c
d
i
i
i
i
72 cm3
112 cm3
180 cm3
600 m3
ii
ii
ii
ii
48 cm2
56 cm2
60 cm2
120 m2
3
Volume = π × (0.3)2 × 4.2 = 0.378π
Cost = 0.378 × π × £67.50 = £80.16 which is £80 to
2sf
4
1.23 tonnes
5
Label should be less than 10.5 cm wide so that it
fits the can and does not overlap the rim and more
than 23.3 cm long to allow an overlap.
6
Volume = π × 32.52 × 100 = 331830.7cm3
1 litre = 1000 cm3
volume = 331830.7 ÷ 1000 = 331.8307 litres =
332 litres (3sf)
7
There is no right answer. Students could start with
the dimensions of a real can. Often drinks cans are
9
14.4 cm
10 Volume of pyramid =
1
3
× 6 × 9 × 15 = 270 cm3
Volume of part cut off top =
1
3
× 3 × 2 × 5 = 10 cm3
So frustum = 270 – 10 = 260 cm3
volume of frustrum
260
Hence volume of pyramid = 270 =
26
27
Exercise 9H
1
a i
b i
c i
2
935 g
3
3560 cm
3
314 cm
3
1020 cm
ii
ii
ii
1430 cm
2
283 cm
2
679 cm
2
3 Total area = πrl + πr2 = π × 3 × l + π × 32
= (3l + 9)π = 24π
So 3l + 9 = 24, so 3l = 24 – 9 = 15
l = 15
3 =5
4 a 816π cm3
5 24π cm2
× 100 = 57%
b 720π mm3
6 a 4 cm
b 6 cm
c Various answers, e.g. 60° gives 2 cm, 240°
gives 8 cm
7 If radius of base is r, slant height is 2r.
Area of curved surface = πr × 2r = 2πr2, area
of base = πr2
8 2.7 g/cm3
9 2.81 cm
10 252π cm2
Exercise 9I
1 a 36π cm3
b 288π cm3 c 1330π cm3
2 a 36π cm2
b 100π cm2 c 196π cm2
3 65 400 cm3, 7850 cm2
4 a 1960 cm2 (to 3sf) b 7444 cm3 (to nearest
unit)
5
125 cm
6
6232
7
7.8 cm
8
a
9
48%
The surface area, because this is the amount of
material (leather or plastic) needed to make the
ball
b Surface area can vary from about 1470 cm2 to
1560 cm2, difference of about 90 cm2. This
seems surprisingly large.
10 Radius of sphere = base radius of cylinder = r,
height of cylinder = 2r Curved surface area of
cylinder = circumference × height = 2πr × 2r = 4πr2
= surface area of sphere
Review Questions
1
29.4 cm2
2
721 cm2
3
5740 cm3(to 3sf)
4
610 g (2sf)
5
17.5 cm
6
360 g
7
56.5 cm
8
Call length of square 2x, so that radius of arcs is x.
2
Then area of square = 4x
Area of each semicircle =
1
2
2
πx so area of 4 semi
2
circles is 2πx
Area of shaded part is: area of 4 semicircles – area
2
of square = (2π – 4)x
So percentage shaded =
2π  4
2π  4
x2 × 100 =
4
4 x2
9
1
3