Volumes of common solids 253
4.0 cm
P
E
A
D
R
B
Q
C
2.0 cm
3.6 cm
1.0 cm
3.0 cm
6.0 cm
Figure 27.22
The height of the large cone = 3.6 + 7.2
= 10.8 cm
Volume of frustum of cone
= volume of large cone
− volume of small cone cut off
=
1
3
π(3.0)
2
(10.8) −
1
3
π(2.0)
2
(7.2)
= 101.79 − 30.16 = 71.6 cm
3
(ii) Method 2
From above, volume of the frustum of a cone
=
1
3
πh(R
2
+ Rr + r
2
)
where
R = 3.0 cm,
r = 2.0 cm and h = 3.6 cm
Hence, volume of frustum
=
1
3
π(3.6)
(3.0)
2
+ (3.0)(2.0) + (2.0)
2
=
1
3
π(3.6)(19.0) = 71.6 cm
3
Problem 25. Find the total surface area of the
frustum of the cone in Problem 24.
(i) Method 1
Curved surface area of frustum = curved surface
area of large cone − curved surface area of small
cone cut off.
From Figure 27.22, using Pythagoras’ theorem,
AB
2
= AQ
2
+ BQ
2
from which AB =
10.8 2 + 3.0 2
= 11.21 cm
and
AD
2
= AP
2
+ DP
2
from which AD =
7.2 2 + 2.0 2
= 7.47 cm
Curved surface area of large cone = πrl
= π(BQ)(AB) = π(3.0)(11.21)
= 105.65 cm
2
and curved surface area of small cone
= π(DP)(AD) = π(2.0)(7.47) = 46.94 cm
2
Hence, curved surface area of frustum
= 105.65 − 46.94
= 58.71 cm
2
Total surface area of frustum
= curved surface area
+ area of two circular ends
= 58.71 + π(2.0)
2
+ π(3.0)
2
= 58.71 + 12.57 + 28.27 = 99.6 cm
2
(ii) Method 2
From page 252, total surface area of frustum
= πl(R + r) + πr
2
+ π R
2
where l = BD = 11.21 − 7.47 = 3.74 cm,
R = 3.0 cm and r = 2.0 cm. Hence,
total surface area of frustum
= π(3.74)(3.0 + 2.0) + π(2.0)
2
+ π(3.0)
2
= 99.6 cm
2