Frustum of a pyramid
If the top of a pyramid is cut away by a plane which is
parallel to the base of the pyramid, the remaining part is
called a frustum of the pyramid.
The removed part is also a pyramid of smaller volume.
In this case,
volume of the
volume of the
Volume of the frustum
 volume of

volume of
volume of
of a pyramid
 larger pyramid  smaller pyramid
pyramid VPQRS
frustum ABCDSPQR pyramid VABCD
Refer to the figure on the right.
Volume of pyramid VPQRS
 1  30  8 cm 3
3
 80 cm 3
If the volume of pyramid VABCD
is 200 cm3, then
V
base area
= 30 cm2
S
P
D
A
8 cm
R
Q
C
B
volume of frustum ABCDSPQR
 volume of pyramid VABCD  volume of pyramid VPQRS
 (200  80) cm3
 120 cm3
Follow-up question
The figure shows a frustum ABCDHEFG.
Its lower base is a square of side 7 cm. The
volume of pyramid VEFGH is 75 cm3 and
the height of pyramid VABCD is 12 cm.
Find the volumes of
A
(a) pyramid VABCD,
(b) frustum ABCDHEFG.
V
F
7 cm
Solution
1 2
 7  12 cm3
3
 196 cm 3
(a) Vol ume of pyramid VABCD 
H
E
12 cm
G
D
C
B
Follow-up question (cont’d)
The figure shows a frustum ABCDHEFG.
Its lower base is a square of side 7 cm. The
volume of pyramid VEFGH is 75 cm3 and
the height of pyramid VABCD is 12 cm.
Find the volumes of
A
(a) pyramid VABCD,
(b) frustum ABCDHEFG.
V
H
E
F
7 cm
Solution
(b) Volume of frustum ABCDHEFG  (196  75) cm 3
 121 cm 3
12 cm
G
D
C
B
Example 6
The figure shows a solid ABCDHEFGI which
is formed by drilling out a pyramid IEFGH
from the frustum ABCDHEFG. The upper
base and the lower base of the frustum are
squares of sides 4 cm and 12 cm respectively.
It is known that the height of pyramid IEFGH
is
4
that of pyramid VEFGH. Find the
5
volumes of
(a)
(b)
frustum ABCDHEFG,
solid ABCDHEFGI.
Solution
(a)
1
3
Volume of pyramid VABCD   (12  12)  (15  7.5) cm
3
 1080 cm 3
1

 (4  4)  7.5 cm 3
Volume of pyramid VEFGH
3
 40 cm 3
Volume of frustum ABCDHEFG
 volume of pyramid VABCD  volume of pyramid VEFGH
 (1080  40) cm3
 1040 cm3
(b)
4
∵ Height of pyramid IEFGH   height of pyramid VEFGH
5
4
∴ Height of pyramid IEFGH   7.5 cm
5
 6 cm
1
3
Volume of pyramid IEFGH   (4  4)  6 cm
3
 32 cm 3
Volume of solid ABCDHEFGI
 volume of frustum ABCDHEFG  volume of pyramid IEFGH
 (1040  32) cm3
 1008 cm3
CP (P.154)
ID03 (P.151)
380 cm3
Ex.10A (P.159)
h=5
420 cm3
6 cm
304 cm3
Total Surface Areas of Pyramids
The figure shows a pyramid VABCDE and its net.
The pyramid has a base (orange region) and a number of
lateral faces (green regions).
Its total surface area can be found by summing its base area
and the areas of all the lateral faces.
Total surface area  total area of  base area
of a pyramid
lateral faces
V
Refer to the figure on the right.
area of △VBC
= 40 cm2
Total surface area of right pyramid VABCD
 total area of lateral faces + base area
 4  area of △VBC + area of ABCD
 (4  40 +
 224 cm2
82)
cm2
D
C
A
8 cm
8 cm
B
The base is a square.
Follow-up question
Find the total surface area of the
right pyramid in the figure.
area = 65 cm2
area = 40 cm2
Solution
Total surface area of the right pyramid
 total area of lateral faces + base area
 [(2  40 + 2  65) + 14  8] cm2
 (80 + 130 + 112) cm2
 322 cm2
14 cm
8 cm
The base is a rectangle.
Example 4
In the figure, VABCD is a right pyramid
with a rectangular base ABCD. Find
(a)
(b)
the lengths of AB and BC,
the total surface area of pyramid VABCD.
Solution
(a)
Consider right-angled triangle VBE.
Consider right-angled triangle VBF.
BF  VB 2  VF 2
BE  VB 2  VE 2
 25 2  24 2 cm
 25 2  20 2 cm
 49 cm
 225 cm
 15 cm
 7 cm
AB  2 BE
 14 cm
∴
BC  2 BF
 30 cm
(Pyth. theo
(b)
1
 AB  VE
2
  1  14  24 cm2
2
 168 cm2
Area of △VAB 
1
 BC  VF
2
1
  30  20 cm 2
2
 300 cm 2
Area of △VBC 
∴
Total area of lateral faces  2  (168  300) cm 2
 936 cm 2
∴
Total surface area of pyramid VABCD
 total area of lateral faces  base area
 (936  14  30) cm 2
 1356 cm 2
Example 5
The figure shows a regular pyramid VABCD
with a square base ABCD of side 6 cm. Its
volume is 48 cm3.
(a)
(b)
(c)
(a)
Find the height VO of the pyramid.
Find the length of VE.
Find the total surface area of the pyramid.
Let h cm be the height VO of the pyramid.
1
48   (6  6)  h
3
h4
∴
The height VO of the pyramid is 4 cm.
(b)
Consider right-angled triangle VOE.
1
OE  AB  3 cm
2
VE  VO 2  OE 2
 4 2  32 cm
 5 cm
(Pyth. theorem)
(c)
1
Area of △VBC   BC  VE
2
1
  6  5 cm 2
2
 15 cm 2
∵
All the lateral faces of pyramid VABCD are congruent
∴
triangles.
Total surface area of pyramid VABCD
 total area of lateral faces  base area
 (4  15  6  6) cm 2
 96 cm 2
P.153
ID04 (P.152)
ID05 (P.153)
85 cm2
Ex.10A (P.156)
43.3 cm2
173 cm2
340 cm2
168 cm2
Ex.10A (P.159)
P.155
384 cm2
736 cm2