SURFACE AREA AND VOLUME OF A PRISM
Prism-
Lateral Edges-
A prism is a solid, whose side faces are
parallelograms and whose ends (or bases) are
congruent parallel rectilinear figures.
In this figure given below, there is a prism whose
ends are rectilinear figures PQRST and P'Q'R'S'T'.
The lines of intersection of the lateral faces of a
prism are called the lateral edges of the prism.
In the above figure PP', QQ', RR', SS' and TT' are
the lateral edges of the prism.
Regular PrismIf ends are regular figures then prism is called a
regular prism.
S'
Right Prism-
T'
X'
R'
P'
Q'
S
T
X
P
R
Z
A prism is called a right prism if its lateral edges are
perpendicular to its ends (bases). Otherwise it is
said to be an oblique prism.
The prism shown in figure (I) is an oblique prism
whereas the prism shown in figure (II) is a right
prism.
Q
Base of Prism-
T'
R'
P'
Q'
The end on which a prism may be supposed to
stand is called the base of the prism.
In the above figure, PQRST and P'Q'R'S'T' are
the bases of the prism. Every prism has two bases.
T
In the above figure, Q' Z is the perpendicular
distance between the ends PQRST and P'Q'R'S'T'.
So, it is the height of the prism shown in the above
figure.
Axis of a PrismThe straight line joining the centres of the ends of a
prism is called the axis of the prism.
In the above figure, a straight line passing through
X' and X' is the axis of the prism.
Lateral FacesAll faces other than the bases of a prism are known
as its lateral faces.
In the above figure, PQQ'P', QRR'Q', RSS'R' etc
are lateral faces.
S
P
Height of a prismThe perpendicular distance between the ends of a
prism is called the height of the prism.
S'
R
Q
In a right prism, length of the prism is same as its
height. Also, all lateral edges are of the same length
equal to the height of the prism. It is also evident
from the definition of a right prism that its all lateral
faces are rectangles. The number of lateral edges
and lateral faces of a prism is same as the number
of sides in the base of the prism.
Triangular PrismA prism is called a triangular prism if its ends are
triangles.
Right Triangular PrismA triangular prism is called a right triangular prism
if its lateral edges are prependicular to its ends.
The prism shown in figure (III) is a triangular prism
whereas the prism shown in figure (IV) is a right
triangular prism.
R'
P'
 Total surface area = 3a × h +
R'
X
Q'
3 2
a
2
X
P'
Q'
Volume =

3 2
a h
4
Some Examples
X'
Q.1 Find the area of the base of a right triangular prism
having volume of 1476 cm2 and height 18 cm.
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X'
A prism is said to be quadrilateral prism or a pentagonal
prism or a hexagonal prism etc according as the number
of sides in the rectilinear figure forming the ends (base)
is four or five or six etc.
If the ends of a quadrilateral prism are parallelograms,
then it is also known as a parallelopiped.
A quadrilateral prism with its ends as squares is called a
rectangular solid or a cubiod.
Figure (V) shows a right pentagonal prism and figure
(VI) shows a rectangular solid.
T'
R'
T
Axis
X'
P'
S
 Area of the base =
1476
cm2 = 82 cm2
18
Q.2 The base of a right prism is an equilateral triangle
with a side 10 cm and its height is 25 cm. Find its
volume, lateral surface area and total surface area.
Sol. Volume = Area of the base × Height.
R'
X'
P'
Area of the base =
Q'
 3
3
2
  side 
4

2
=  4  10  cm2 = 25 3 cm2


Q'
X
P
Volume
 Area of the base = Height
Base is an equilateral triangle.
S'
T'
Sol. Volume = Area of the base × Height


V = 25 3  25 cm3

R
Q
= 625 3 cm3
Volume and Surface Area of a Right Prism-
Lateral surface area
(i) Volume of a right prism-
= Perimeter of the base × Height
V= Area of the Base × Height
= (10 + 10 + 10) × 25cm2 = 750 cm2
(ii) Lateral Surface area of a right prism-
Total Surface area
L.S.A. = Perimeter of the Base × Height
= Lateral Surface area + Area of ends
(iii) Total Surface area of a right prism-
T.S.A. = Lateral Surface area + Area of Ends
= Lateral Surface area + 2 (Area of the Base)
For Example : If the base of a right prism is an
equilateral triangle of side a and height h, then,

 3

2
2
=  750  2  4  10   cm




2
=  750  50 3  cm
Q.4 A right prism of height 15 cm stands on a triangular
base whose sides are 13 cm, 14 cm and 15 cm,
find its lateral surface area, total surface area and
volume.
 Lateral surface area = 3 a × h
2
Sol. If a, b, c are the length of the sides of a triangle
and s is the semi-perimeter, then its area
A  s s  a s  b s  c  where s =
1
a  b  c
2
s=
Sol. Lateral surface area
= Perimeter of the base × Height
Here, a = 13 cm, b = 14 cm and c = 15 cm.

Q.6 The base of a right prism is an equilateral triangle
of side 8 cm. If the lateral surface area of the prism
is 960 cm2. Find its volume.
 960 = (8 + 8 + 8) × Height
1
13  14  15 = 21 cm.
2
 Height =
960
cm  40 cm
24
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Perimeter of the base = 2 s = 42 cm.
Area of the base
Area of the base =
 21  21  13   21  14    21  15

 21  8  7  6  7  3  8  7  3  2

2
2
 72  32  42 = 7 × 3 × 4 cm = 84 cm

Lateral surface
Volume of the prism


 16 3  40  640 3 cm3
= (42 × 15) cm2 = 630 cm2
Total surface area
= Lateral surface area + 2 (Area of the base)
= (630 + 2 × 84) cm2 = 798 cm2
Q.7 A right triangular prism of height 18 cm and of
base sides 5 cm, 12 cm and 13 cm is transformed
into another right triangular prism on a base of
sides 9 cm, 12 cm and 15 cm. Find the height of
new prism and the change in the whole surfae area.
Sol. Perimeter of the base = (5 + 12 + 13 ) cm = 30 cm
Volume = Area of the base × Height
 Semi-perimeter s = 15 cm.
= (84 × 15 )cm3 = 1260 cm3
Here, a = 5 cm, b = 12 cm and c = 13 cm
Q.5 A right prism stands on a triangular base. The
volume of the prism is 606 cm3 and the sides of
the base are 5 cm, 5 cm and 8 cm. Find the height
of the prism.
Sol. Area of triangle  s s  a s  b s  c 

Area of the base  s s  a s  b s  c 
= 15  10  3  2 = 30cm2
Let V1 be the volume of the prism.
V1 = Area of the base × Height
Here, a = 5 cm, b = 5 cm and c = 8 cm.
 V1 = (30 × 18) = 540 cm3
1
s   5  5  8  9cm
2
 A=
3
 (8) 2  16 3 cm 2
4
= Area of the base × Height
= Perimeter of the base × Height

3
2
  side 
4
Let S be the total surface area of the prism.
S = Lateral surface area + 2 (Area of the base)
9  9  5   9  5   9  8
= Perimeter of the base × height
=
9  4  4  1  12 cm
2
+ 2 (Area of the base)
= (30 × 18 + 2 × 3) = 600 cm2
Volume = Area of the base × Height
Height =
Let h be the height of the new prism.
Volume
606

 50.5cm
Area of the base 12
Semi-perimeter s1 = (9 + 12 + 15)/2 = 18
The sides of the base:-
3
a = 9 cm, b = 12 cm and c = 15 cm
of the prism.
a : b : c = 5 :12 : 13
Area of base  s1 s1  a s1  bs1  c 
 a = 5k, b = 12k and c = 13k
 18(18 – 9)(18 –12)(18 –15)  54cm

2
 a + b + c = 30k  60 = 30k
[  a + b + c = 60cm (Given]
Volume = Area of the base × Height
 540 = 54 × h 
 k=2
h = 10 cm

a = 10, b = 24, and c = 26
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[Volume of the two prisms are same]
The height of new prism is 10 cm.
semi-perimeter s = 60/2 = 30.
Let S1 be the total surface area of the new prism.

S1 = Perimeter of the base × Height
= (36 × 10 + 2 × 54 ) = 468 cm2

30  30  10  30  24  30  26 
=
+ 2 (Area of the base)
s  s  a  s  b  s  c 
Area of the base =
= 30  20  6  4 =
Change in the whole surface area = S – S1
= (600 – 468) = 132 cm2
56 5 4 6 4
= 5 × 6 × 4 = 120 cm2
Volume of the prism = Area of the base × Height
Q.8 The base of a right triangular prism is an equilateral
triangle. If its height is halved and each side of the
base is doubled, find the ratio of the volume of the
two prisms.
Sol. Let a be the length of each side of the base of the
given prism and h be its height. then volume
= 120 × 40 = 4800 cm3
Total surface ara = Perimeter of the base × Height
+ 2 (Area of the base)
= (60 × 40 + 2 × 120) = 2640 cm2
Q.10 The total surface area of a right triangular prism
of the height 4 cm is 72 3 cm2. If the base of the
prism is an equilateral triangle, find its volume.
 3 2

 V1   4 a  h 
Sol. Let each side of the base of the prism be a cm.
Let a1 be the length of each side of the base of the
new prism and h1 be its height. Then,
a1 = 2a
and h1 =
total surface area = 72 3 cm2
 Perimeter of the base × height
h
2
+ 2 (Area of the base) = 72 3
If V2 is the volume of the new prism

3 2
3
h  3
V2 
a1  h1 
 2 a 2    a 2h 
4
4
2  4


3 2
a h
V1
1
4


V2
2
3 2
a h
2
Q.9 The perimeter of the base of right triangular prism
is 60 cm and sides of the base are in the ratio 5 :
12 : 13. Find its volume and total surface area, if
its height is 40 cm.
Sol. Let a, b, c be the lengths of the sides of the base
4

 3 2
3a  4  2 
a   72 3
 4


3a 2  24a  144 3  0

a 2  8 3  144  0

a 2  12 3a  4 3a  144  0

a a  12 3  4 3 a  12 3  0

a a  4 3 a  12 3  0







as a base and the other regular hexagon. If both
of the prisms have equal heights and volumes, then
find the ratio between the length of each side at
their bases.
[  a + 12 3  0 as a >0]
 a4 3

a4 3

Volume of the prism
= Area of the base × Height
3
 4 3
4
(b) 300 cm3
(c) 1200 cm3
(d) 600 cm3
prism is (use
2
 4  48 3cm3
The base of a prism is a right-angles triangle and
the two sides containing the right angle are 8 cm
and 15 cm. If its height is 20 cm, then the volume
of the prism is
The base of a prism is a regular hexagon. If every
edge of the prism measures 1 metre, then the
volume of the prism is
(a)
(c)
3.
3 2
cu m
2
(b)
3 3
cu m
2
6 2
cu metre
5
(d)
5 3
cu metre
2
The base of a right prism is a pentagon whose
sides are in the raio 1: 2 : 2 : 1: 2 and its height
is 10 cm. If the longest side of the base be 6 cm,
the volume of the prism is
4.
5.
8.
3 = 1.73)
2
(a) 1200 cm
(b) 2400 cm2
(c) 3600 cm2
(d) 4380 cm2
9. The base of a right prism is a tra-pezium. The
lengths of the parallel sides are 8 cm and 14 cm
and the distance between the parallel sides is 8
cm. If the volume of the prism is 1056 cm3, then
the height of the prism is
(a) 44 cm
(b) 16.5 cm
(c) 12 cm
(d) 10.56 cm
10. If the altitude of a right prism is 10 cm and its
base is an equilateral triangle of side 12 cm, then
its total surface area (in cm2) is
(a) 270 cm3
(b) 360 cm3
(a) (5 + 3 3 )
(b) 36 3
(c) 540 cm3
(d) None of these
(c) 360
(d) 72(5+ 3 )
What is the total surface area of a triangular prism
whose height is 30 m and the sides of whose base
are 21 m, 20 m and 13 m respectively?
(a) 1872 sq m
(b) 1725 sq m
(c) 1652 sq m
(d) 1542 sq m
The base of a prism is a triangle of which the
sides of the prism is 4200 cubic cm. What is the
height? Find its lateral area also.
(a) 20 cm, 1400 sq cm
(b) 25 cm, 700 sq cm
11. The base of a right prism is a right- angled triangle
whose sides are 5 cm, 12 cm and 13 cm. If the
area of the total surface of the prism is 360 cm2,
then its height (in cm) is
(a) 10
(b) 12
(c) 9
(d) 11
12.
If the base is a right rectangular prism is left
unchanged and the measure of the lateral edges
are doubled, then its volume will be
(a) unchanged
(b) tripled
(c) doubled
(d) quadrupled
13. The height of a right prism with a square base is 15 cm.
If the area of the total surface of the prism is 608 sq.
cm, its volume is:
(c) 20 cm, 700 sq cm
(d) 10 cm, 1400 sq cm
6.
(d) 2 : 3
3:2
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2.
(c)
6 :1
(a) 160 cm3
 
Exercise on Prism
1.
(b)
The base of a solid right prism is a triangle whose
sides are 9 cm, 12 cm and 15 cm. The height of
the prism is 5 cm. Then, the total surface area of
the prism is
(a) 180 cm2
(b) 234 cm2
(c) 288 cm2
(d) 270 cm2
The base of a right prism is an equilateral triangle
of area 173 cm2 and the volume of the prism is
10380 cm3. The area of the lateral surface of the
=
7.
(a) 1: 6
There are two prism, one has equilateral triangle
5
(a) 910 cm3
(b) 920 cm3
(c) 960 cm3
(d) 980 cm3
Explanations
14. The base of a right prism is an equilateral triangle of
side 8 cm and height of the prism is 10 cm. Then the
volume of the prism is:
(a) 320 3 cubic cm
(b) 160 3 cubic cm
(c) 150 3 cubic cm
(d) 300 3 cubic cm
1.
Volume of the prism = Area of the base × Height
1
 15  8  20  1200 cm3
2
2. Given prism is a solid with regular hexagonal base
 Its volume = Area of the base × Height
15. A right prism stands on a base 6 cm equilateral triangle
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(a) 9
3 3
3 3
1 
cu m
2
2

and its volume is 81 3 cm3. The height (in cms) of
the prims is:
Since, the area of regular hexagon with 


3 3 2

side1m 
m


2
(b) 10
(c) 12
(d) 15
16. A prism has as the base a right angled triangle whose
sides adjacent to the right angles are 10cm and 12cm
long. The height of the prism is 20cm. The density of
the material of the prism is 6gm.cubic cm. The weight
of the prism is
(1) 6.4 kg
(2) 72 kg
(3) 3.4 kg
(4) 4.8 kg
A
17. The perimeter of the triangular base of a right prism
is 15 cm and radius of the incircle of the triangular
base is 3 cm. If the volume of the prism be 270 cm3,
then the height of the prism is–
(a) 6 cm
(b) 7.5 cm
(c) 10 cm
(d) 12 cm
18. The base of a prism is a right angled triangle with two
sides 5 cm and 12 cm. The height of the prism is 10
cm. The total surface area of the prism is–
(a) 360 sq. cm
(b) 300 sq. cm
(c) 330 sq. cm
(d) 325 sq. cm
19. The base of a right prism is an equilateral triangle. If
the lateral surface area and volume is 120 cm2, 40 3
cm3 respectively then the side of base the prism is.
(a) 4 cm
(b) 5 cm
(c) 7 cm
(d) 40 cm
20. The base of a right prism is an equilateral triangle. If
its height is one-fourth and each side of the base is
tripled, then the ratio of the volumes of the old to the
new prism is (a) 4 : 3
(b) 1 : 4
(c)
1:2
(d)
3
90°
B
E
O
3
3.
C
3
6 cm
3
AB  3 2
AE  3 2
ABE  45
AEB  45
D
Area of BCDE = 18 cm2
Area of OAE =
1
 3  3  4.5cm2
2
 Area of ABE = 9 cm2
 Area of ABCDE (base) = (18 + 9) = 27 cm2
Volume of prism = Area of the base × Height
= (27 × 10) = 270 cm3
4.
Total surface area of prism
= Lateral surface area + 2 ×(Area of base)
Here, s =
a  b  c 20  20  13
 27
=
2
2
 Required area = (21 + 20 + 13) × 30 + 2 ×
27  27  21 27  20  27  13
= 54 × 30 + 2 27  6  7  14
4:9
= 1620 + 2 × 126 = 1872 sq m
5.
Answer Key
1.(c)
2.(b) 3.(a) 4.(a) 5.(a) 6.(b) 7.(c)
8.(c)
9.(c) 10.(d) 11.(a) 12.(c) 13.(c) 14.(b)
Let the side a = 17 cm, b = 25 cm, c = 28 cm
 a  b  c   17  25  28 
  
  35cm
2
2
Then, s  
(s – a) = (35 – 17) = 18 cm
15.(a) 16.(b) 17.(d) 18.(a) 19.(a) 20.(d)
(s – b) = (35 – 25) = 10 cm
6
(s – c) = (35 – 28) = 7 cm
a = 20 cm
Perimeter of triangle = 3a = 3 × 20 = 60 cm
Lateral surface area = Perimeter of triangle × height
= 60 × 60 = 3600 cm2
s  s  a s  b s  c 
Area of a triangle =
Hence, area of the base = 35  18  10  7 sq. cm
Volume of the prism = Area of the base × Height
9.
Area of traperium =
 4200 
 Height of the prism =  210   20cm
=
Lateral Area = Perimeter of the base × Height
1
× 8 × (8 + 14)
2
= 88 cm2
 Volume = area of traperium × height
 1056 = 88 × h
 h = 12 cm
10. Total surface area = Permeter base × height + 2 × area
of base
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= (17 + 25 + 28) × 20
= 1400 sq cm
6.
1
× height × (sum of perallel sides)
2
Let the height of each prism be h units and the
length of each side of equilateral triangle at thebase
of first prism be a units and that the second prism
having regular hexagon as base be b units.
(See the figures given below)
= 36 × 10 + 2 ×

11.
b
I
Total surface area = 360
(perimeter base) × height + 2 × area of base = 360
1

(5 + 12 + 13) × h + 2 ×   5  12  = 360
2
II
According to the question,
30 h = 300
h = 10 cm

12. The base of the prism is rectangular and we are not
3 2
3 2
changing the base so length & breadth will be remain
a h 
b h
4
2
same.If we double the lateral edges it means we are
doing double its height so Volume of the prism will
1 2 3 2
be doubled.
a  b

 a 2  6b2
4
2
13. Perimeter of square base of side a = 4a
Lateral surface area = 4a × h
a
6

 a = 6b

Total surcace area = Lateral surface area + 2 × area
b
1
of base
 a:b= 6 :1
608 = 4ah + 2a2

Base of prism is a right angle triangle
608 = 60a + 2a2

a2 + 30a – 304 = 0

1
2
× 9 × 12 = 54 cm
 area of base =
a2 + 38a – 8a – 304 = 0

2
a(a + 38) – 8 (a + 38) = 0

Potal surface area = Perimeter of box x height + 2 ×
(a + 38) (a – 8) = 0

area of base
a = 8 cm.

= (9 + 12 + 5) × 5 + 2 × 54
2
Volume of prism = area of base × height
= 288 cm
= 64 × 15
Volume = area of base × height
= 960 cm3
 10380 = 173 × h
14. Volume = area of base × height
 h = 60 cm
Volume of first prism = Volume of second prism
8.

= 360 + 72 3 = 72 5  3 cm2
a
7.
3
× 12 × 12
4
 area of equilateral triangle =
3 2
a = 173
4
=
7
3
(8)2 × 10
4
= 160 3 cm.
15. area of base of prism =
40 3
3a
=
120
12
3
×6×6
4
a = 4 cm
= 9 3 cm2
20.
Volume = area of base × height
81 3 = 9 3 × height
V1
=
V2
3 2
2
a1 .h1
 a1 
h1
4
=  . h
a


3 2 2
2
2
a1 .h
4
h = 9 cm
2
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om
 a1   h1  4
.

=
 3a1   h2 / 4  9
1
16. area of base =
× base × height
2
=
1
× 10 × 12 = 60 cm2
2
 Volume of prism = Area of base × height
= 60 × 20 = 1200 cm3
Material used for 1 cubic cm. = 6 gm
Material used for 12 00 cm3 = 1200 × 6 = 7200 gm
= 7.2 kg
17. Perimeter of triangle = 15cm
15
 semiperimeters = 2 cm
inradius r = 3 cm
 = r.s =
15
45
×3=
cm
2
2
Volume of prism = area of base × height
270 =
45
×h
2
h = 12 cm
18. Third side of right angle triangle will be 13 cm. Tolal
surface area.
= (Perimeter of base) × h + 2 × area of base
= (5 + 12 + 13) × 10 + 2 ×
1
× 5 × 12
2
= 300 + 60 = 360 sq. cm
19. Volume of prism V = area of base × height
=
3 2
a ×h
4
Lateral surface area of prism (L.S.A)
= Perimeter of base × height = 3a × h
3 2
a h
V
4
=
=
l.S.A
3a  h
8
 a2 = 3a1 & h2 =
h1
2
SURFACE AREA AND VOLUME OF A PYRAMID
Pyramid-
X
A pyramid is a solid whose base is a plane rectilinear
figure and whose side-faces are triangles having a
common vertex outside the plane of the base.
D
E
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Figure of this below shows a pyramid XABCDE. The
base of this pyramid is the pentagon ABCDE and
triangles XAB, XBC, XCD, XDE and XEA are five
faces. If the base of a pyramid is a triangle, a
quadrilateral and a square, then it is called triangular
pyramid, quadrilateral pyramid and square pyramid
respectively. Similarly, a pyramid is called a
pentagonal, hexagonal, septagonal and octagonal
according as the number of sides of the base is 5, 6,
7 or 8.
X
F
O
C
M
A
B
Also, in right-angled triangle XOM, we have XM2
= OM2 + XO2 (By pythangoras theorem)
 XM =
XO 2  OM 2
Right Pyramid-
A
P
B
Vertex-
D
E
O
C
The common vertex of the triangular faces of a
pyramid is called the vertex of the pyramid. In
the above figure, 'X' is the vertex of the pyramid
XABCDE.
Height-
Axis-
A pyramid is said to be right pyramid of the
perpendicular dropped from the vertex on the base
meets the base at its cenral point, ie the centre of the
inscribed or cirumscribed circle. In other words, the
vertex of the pyramid lies on the perpendicular to
the base drawn through its centre. Otherwise, the
pyramid is called an oblique pyramid.
The pyramid shown in the above figue I is an oblique
pyramid whereas figure II given above shows a right
pyramid.
Regular Pyramid-
The height of a pyramid is the length of the
perpendicular from the vertex to the base.
A pyramid is said to be a regular if its base is a regular
figure. ie all sides of its base are equal.
In the above figure, XP is the height of the
pyramid XABCDE.
In case of a right regular pyramid the lateral edges
are equal and the lateral faces are congruent triangles.
Volume and Surface Area of a Pyramid-
The axis of a pyramid is the straight line joining
the vertex to the central point of the base.
In the above figure, XO is the axis of the pyramid
XABCDE.
(i) Volume of a pyramid
=
1
× Area of the Base × Height
3
(ii) Lateral surface area of a pyramid
Lateral Edges-
The edgews through the vertex of a pyramid are
known as its lateral edges.
= Sum of areas of all the lateral triangular faces.
L.S.A. =
Slant HeightThe slant height of a regular right-pyramid is the
line-segment joining the vertex to the mid-point
of anyone of the sides of the base.
The figure given below shows a right regular
pyramid, in which O is the centre of the base and
XM is the slant height.
9
1
× perimeter of the base × slant height.
2
(iii) Total surface area of a pyramid
= Sum of areas of all lateral faces + Area of the
base.
T.S.A. =
1
× perimeter of the base × slant height
2
+ Area of the base.

=
For a right pyramid with an equilateral triangle of side
'a' as base and height 'h'.
(v) Total surface area of the regular tetrahedron
(i) Lateral edge or Lateral height
=
= 4
a2
h 
3
2
h2 
Q. 1 Find the volume of the right pyramid the area of whose
base is 60 cm2 and height 15 cm.
a2
12
Sol.
Volume of a right pyramid.
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(iii)
Lateral surface area
=
1
=
(Perimeter of the base × slant height)
2
slant height) +
(v) Volume =
3 2
a
4
1

=   60  15  = 200 cm3
3
Q. 2 Find the height of the right pyramid whose volume is
750 cm3 and area of whose base is 250 cm2.
Sol.
1
3
×
× a2 × h2
3
4
Volume of a right pyramid
=
=
3 2
a h.
12
1
(Area of the base × Height)
3
Let h be the height of the given right pyramid.
Then,
1
(Length of an edge
2
of the base × Slant height)
(vi)Area of lateral face =
750 =
Tetrahedron and Regular Tetrahedron
A tetrahedron is a pyramid whose base is a triangle.
It has six edges and four triangular faces.
 h=
When the length of each edge of a regular
tetrahedron is given 'a' and 'h' is height.
1
× 250 × h
3
750  3
= 9 cm
250
Q. 3. A right pyramid has its base as an equilateral triangle
A tetrahedron whose all the edges are of equal length
is called a regular tetrahedron. In a regular
tetrahedron all the four faces are congruent
equilateral triangles.
of side 40 cm and its height is 24 3 cm. Find
the volume of the pyramid.
Sol.
(i) Height of the regural tetrahedron
h=
1
(Area of the base) × (Height)
3
 Volume of the given pyramid
1
(iv)Total surface area = (Perimeter of the base ×
2

3 2
a = 3 × a 2.
4
Some Examples
(ii) Slant height =
6.
3 3 2
3 3 2
a =
h
4
2
 3 2 
Volume of the pyramid =  12 a  h 


Here, a = 40 h = 24 3
2
×a
3
 3

2
 Volume of the pyramid =  12  40  24 3 


(ii) Slant height of the regular tetrahedron
= 9600 cm3
=
3
× a.
2
Q. 4 Find the volume of a regular tetrahedron whose each
edge is of 12 2 cm.
(iii) Volume of the regular tetrahedron
=
2
3 3
× a3 =
h
12
8
Sol.


 a  3 h 

2 

Volume of a regular tetrahedron =
 volume of the given tetrahedron
(iv) Lateral surface area of the regular
tetrahedron
10
=
2
× 12 2
12

3

2
(edge)3
12
cm.
2
=
× 1728 × 2 2 = 72 cm3
12
Sol.
a = 9 cm, b = 12 cm, c = 15 cm.
Q. 5 Find the lateral surface area and total surface area of
a right pyramid in which the base is an equilateral
 s=
 Area of the base
triangle of area 16 3 cm3 and length of each
lateral edge is 5 cm.
Sol.
Let the length of each side of the base is a and h
is the height of the pyramid.
3 2
a = 16 3
4

s(s - a)(s -b)(s -c)
=
18(18-9)(18-12)(18-15)
=
18  9  6  3 = 54 cm2
Volume of the tetrahedron
 a2 = 64
 a = 8 cm
Lateral edge = 5 cm
h2 
a2
= 5 or,
3
h2 
64
=5
3
1
× 54 × 15 = 270 cm3
3
Area of the base = 16 3 cm2

a2
11 64
h2 


 9  3 cm
12
3 12
1
(Perimeter of the base × Slant height)
2
l=
=
3 2
(a ) = 16 3  a2 = 64
4
 a = 8 cm
Let h be the height of the pyramid and l be its
slant height.
Lateral Surface Area
=
=
Let the length of each side of the base be a cm.
11
cm
3
Slant height =
1
(Area of the base × Height)
3
base of area 16 3 cm2. If the area of one of its
lateral faces is 40 cm2, find the volume of the
pyramid.
Sol.
 h=
=
Q. 8 A right pyramid stands on an equilateral triangular
64
75  64
= 25  h2 =
3
3
 h2 +

=
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Area of the base = 16 3 cm2

a b c
9  12  15
 18
=
2
2
h2 
1
(8 + 8 + 8) × 3 = 36 cm2
2
Total surface area
a2
12
 l 2 = h2 +
64
a2
 l 2 = h2 +
12
12
 l 2 = h2 +
64
12
= Lateral surface area + Area of the base


= 36  16 3 cm
2
Q. 6 If 'a' be the length of the perpendicular drawn from a
vertex of a regular tetrahedron to its opposite face
and each edge of length 2b, show that 3a2 = 8b2 .
Sol.
a = height of the tetrahedron
=
 a=
2
× 2b
3
 a2 =
Area of one lateral surface = 40 cm2

1
(a × l) = 40
2
 a × l = 80
 l = 10
2
× (Length of an edge)
3
............(i)
 8 × l = 80
Put l = 10 in eqn (i)
8b 2
3
100 = h2 +
 3a2 = 8b2
Q. 7 Find the volume of a tetrahedron the sides of whose
base are 9 cm, 12 cm and 15 cm and height 15
11
 h=
284
3
16
16
 h2 = 100 –
3
3
Volume of the pyramid
A
=
1
(Area of the base × Height)
3
=
1
× 16 3 ×
3
284
16
=
3
3
B
F
X'
284 cm3
G
C
E
D
Let X' be the centre of regular hexagon.
Sol.
 XX'G = 90° and XX'G is the right-angled triangle.
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Q. 9 Area of a regular hexagon is 216 3 cm2. A pyramid
of the height 6 cm is formed upon the regular
hexagon. Find the slant surface area of the
pyramid.
Let each side of the regular hexagon be a units.
 Area of the regular hexagon =
 XG =
3 3 2
a
2
(XX ')2  (X ' G) 2
XX' is given, now we have to find X'G. Six equal
triangles can be drawn in a regular hexagon and
area of each triangle is
 3 3 a2 = 216 3  a2 = 144
2
1

=   216 3  = 36 3 cm2
6
 a = 12 cm
 each side of regular hexagon = 12 cm
x
In  X'EF,,
1
 2 × EF × X'G = 36 3
1
 2 × 12 × X'G = 36 3
 X'G = 6 3 cm
A
B
Slant height of the pyramid
F
X'
C
G
=
(XX ') 2  (X ' G) 2
=
(6) 2  (6 3) 2 =
Slant surface area
E
D
Slant surface area
=
144 = 12 cm
1
(perimeter of base) × slant height
2
XG is the line joining the mid-point of any side of
regular hexagon to point X of the pyramid, i.e.
slant height of pyramid.
=
1
× perimeter of base × slant height
2
=
1
× (12 × 6) × 12 = 432 cm2
2
Q. 10
Find the lateral surface area, total surface area
and volume of a right pyramid with equilateral
triangle as a base in which the length of each side
of the base is 4 cm and slant height is 5 cm.
Sol.
length of each side of the base a = 4 cm and slant
height = 5 cm.
Let h be the height of the pyramid,
Slant height =
12
h2 
a2
12
 5=
 h2 = 25 –
height of the pyramid h = 5 cm
4
 25 = h +
3
16
h 
12
2
2
4
71
=
3
3
 h=
 Slant height =
h2 
a2
=
12
=
25 
10
25
=
cm
3
3
71
cm
3
lateral surface area
=
=
Total surface area
= Lateral surface area + area of the base
(30 + 4 3 ) cm2
=
1
× (Lateral surface area)
3
=
150
50
1
×
=
cm2
3
3
3
Volume of the pyramid
Q. 13
The base of a right pyramid is an equilateral
triangle each side of which is 2 cm long. Every
slant edge is 3 cm long. Find the lateral surface
area and the volume of the pyramid.
Sol.
a = 2 cm and salnt edge = 3 cm.
1
=
(Area of the base × height)
3
Sol.
=
 3 2
1
×  4 4  ×


3
=
4
3
71
3
Let h be the height of the pyramid
71 cm3
h2 
Slant edge =
Find the volume, lateral surface area and tatal
surface area of a right trianglular pyramid the
lenght of whose edge is 12 cm.
The pyramid is a tetrahedron whose edge is of
length 12 cm.
Volume of the pyramid =
10
150
1
(10 + 10 + 10) ×
=
cm2
3
3
2
 Area of one side face

3 2
=  30  4  4 


Q. 11
1
(Perimeter of the base × Slant height)
2
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1
(4 + 4 + 4) × 5 = 30 cm2
2
=
h2 
 3=
4
23
 h2 = 9 – 3 = 3  h =
2
× (edge)2
12
2
× (12)3 = 144 2 cm3
12
=
3 3
Lateral surface area =
× (edge)2
4
=
Q. 12
Sol.
h2 
23
cm
3
a2
12
23 4

=
3 12
8 = 2 2 cm
Lateral surface area
3 3
× (12)2 = 108 3 cm2
4
=
1
(Perimeter of the base × Slant height)
2
3 × (edge)2
=
1
( 2 + 2 + 2) × 2 2 = 6 2 cm2
2
Total surface area =
=
a2
3
4
4
 9 = h2 + 3
3
 Slant height =
=
100
12
Lateral surface area
1
(Perimeter of the base × Slant height)
2
=
=
25 
3 ×(12)2 = 144 3 cm2
Volume of the pyramid
The base of a right pyramid is an equilateral
triangle of side 10 cm and its vertical height is 5
cm find its slant height and area of one side face.
lenght of each side of the base a = 10 cm.
13
=
1
(Area of the base × height)
3
1
3
=
×
× (2)2 ×
3
4
Q. 14
Sol.
23
=
3
23
cm3
3
The base of a right pyramid is an equilateral
triangle of side 4 cm. The height of the pyramid
is half of its slant height. Find the volume and the
length of a slant edge of the pyramid.
length of each side of the base a = 4 m.
4.
5.
l
h=
2
1
2
 h=
6.
h2 
a2
12
 3h2 =
a2
a2
 4h2 = h2 +
12
12
3h2 =

16
12
7.
[  a = 4 m]
4
 h = 9
2
 l=
2
h=
m
3

4
3
=
a2
h 
=
3
2
8.
4 16

9 3
52
2 13
=
m
9
3
9.
Volume of the pyramid
=
1
(Area of the base × Height)
3
If a regular square pyramid has a base of side 8 cm
and height 45 cm, then its volume is
(a) 480 cm3
(b) 900 cm3
(c) 640 cm3
(d) 960 cm3
The base of a pyramid is an equilateral triangle of
side 1 cm. If the height of the pyramid is 4 cm, then
the volume is
(a) 0.550 cm3
(b) 0.577 cm3
(c) 0.678 cm3
(d) 0.750 cm3
=
8 3 3
m
9
The volume of the pyramid on a square base of side
15 cm and height 10 cm is
(a) 750 cm3
(c) 2250 cm
(b) 700 cm3
3
A right pyramid is on a regular hexagonal base. Each
side of the base is 10 m and the height is 30 m . The
volume of the pyramid is
(a) 2500 m3
(b) 2550 m3
(c) 2598 m3
(d) 5196 m3
There is a pyramid on a base which is a regular
hexagon of side 2a. If every slant edge of this pyramid
5a
, then the volume of this pyramid is
2
(a) 3a 3
(b) 3a 3 2
(c) 3a 3 3
(d) 6a 3
The area of the square base of a right pyramid is 36
cm2. If the area of each triangle forming the slant
surface is 15 cm3, then the volume of the pyramid is
(a) 64 cm3
(b) 48 cm3
(c) 24 cm3
(d) 36 cm3
If the area of the base of a regular hexagonal pyramid
32 3 m2, then the volume of the pyramid is :
Exercise on Pyramid
(d) 1125 cm
3
The volume of a pyramid whose base is an equilateral
trianlge is 12 cm3. If the height of the pyramid is
3 3 cm metres, then each side of the base is
3.
(d) 72 cm
is 96 3 m2 and the area of one of its side faces is
1
2
3
=
×
× (4)2 ×
3
3
4
2.
(c) 24 cm
is of length
[  l = 2h]
Lateral edge =
1.
(b) 18 cm
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Let h be the height of the pyramid and l be slant
height.
(a) 36 cm
(a) 2 cm
(b) 3 cm
(c) 4 cm
(d) 6 cm
If the base of a right pyranmid is a triangle with sides
5 cm. 12 cm and 13 cm respectively and the volume
of the pyramid is 360 cubic cm, then the height of
the pyramid is
14
(a) 380 3 m 3
(b) 382 3 m 3
(c) 384 3 m 3
(d) 386 3 m 3
10. A right pyramid has an equilateral triangular base of
side 4 cm. If the numerical value of its total surface
area is three times the numerical value of of its
volume, then height is
(a) 8 cm
(b) 6 cm
(c) 10 cm
(d) 12 cm
11. The base of a right pyramid is a square of side 40
cm long. If the volume of the pyramid is 8000
cm3, then its height is:
(a) 5 cm
(b) 10 cm
(c) 15 cm
(d) 20 cm
12. The base of a right pyramid is a square of side 16
cm long. If its height be 15 cm, then the area of
the lateral surface in square centimetre is:
(a) 136
(b) 544
(c) 800
(d) 1280
13. The base of a right pyramid is an equilateral triangle of side 10 3 cm. If the total surface area
of the pyramid is 270 3 sq. cm., its height is
21. A right pyramid stands on a rectangular base 32 cm
long and 10 cm in width. If the height of the pyramid
is 12 cm. Find its whole surface area.
(a) 933 cm3
(b) 936 cm3
(c) 934 cm3
(d) 935 cm2
Answer Key
(a) 12 3 cm
(b) 10 cm
1.(a) 2.(c) 3.(c) 4.(d) 5.(b) 6.(c) 7.(c)
(c) 10 3 cm
(d) 12 cm
8.(b)
15.(d) 16.(a) 17.(d) 18.(c) 19.(a) 20.(b) 21.(b)
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14. If the base of a right pyramid is triangle of sides 5
cm, 12 cm, 13 cm and its volume is 330 cm3, then
its height (in cm) will be
(a) 33
(b) 32
(c) 11
(d) 22
15. A right pyramid stands on a square base of diagonal 10 2 cm. If the height of the pyramid is 12
cm, the area (in cm2) of its slant surface is
(a) 520
(b) 420
(c) 360
(d) 260
Explanations
1.
2.
16. The length of each edge of a regular tetrahedron
is 12 cm. The area (in sq. cm) of the total surface
of the tetrahedron is
(a) 288 3
(b) 144 2
(c) 108 3
(d) 144 3
(a) 33
(b) 32
(c) 11
(d) 22
8
2 cm 2
9
(b)
=
1
× Area of the base × Height
3
=
1
× 15 × 15 × 10 = 750 cm3
3
Let a = side of the equilateral triangle
1  3 a2 

 3 3 = 12

3  4
3.
 a2 = 16  a = 4 cm
Volume of a pyramid
=
1
× Area of the base × Height
3
Since, 52 + 122 = 132, the base of the pyramid is right
triangle , Now, let the height be h,
Now, according to the question,
7
3cm 2
9
 240 =
(c)
8
3cm 2
9
(d)
7
2 cm 2
9
9 2
c.c
4
(3) 9 3 c.c
(2)
1

1
×   5 12  × h
2
3
 h = 24 cm.
19. Each edge of the regular tetrahedron is 3 cm.then its
volume is
(1)
3 a2
4
Now, according to the question,
The base of right pyramid is a equilateral triangle
of side 4 cm. The height of the pyramid is half of its
slant height. Its volume is
(a)
Volume of a pyramid
 Area of the equilateral triangle =
17. If the base of a right pyramid is triangle of sides 5 cm,
12 cm, 13 cm and its volume is 330 cm3, then its height
(in cm) will be
18.
9.(c) 10.(a) 11.(c) 12.(b) 13.(d) 14.(c)
4.
4 2
c.c
9
Volume of a pyramid =
=
(4) 27 3 c.c
5.
20. A pyramid on a square base has four equilateral triangles on its four other faces, each edges being 10m.
Find its volume.
(a) 235.7 m3
(b) 288.7 m3
3
(c) 532.7 m
(d) 352.7 m3
15
1
× Area of the base × Height
3
1
× 8 × 8 × 45 = 960 cm3
3
Volume of the pyramid =
1
× Area of the base × Height
3
AB = BC = CD = EF = FA = 2a
A
PE =
1
5a
and OE = 2a
2
1
2
x
 h = OP =
1
2
B
1
× Area of the base ×
3
Volume of the pyramid =
C
D
3a
 5a 
2
   4a =
2
2
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Height
=
2
3
1
x2 = (1)2 –   =
2
4
 x=
8.
1
3a
1
×
×6×
× 2a × a 3 = 3a 3 3
3
2
2
Side of the square base =
Let G be the centroid of ABCD and M be the
mid point of AD.
3
2
 Area of  ABC =
O
3
4
Volume of the pyramid
6.
=
1
3
3
×
×4=
3
4
3
=
1.732
= 0.577 cm3
3
D
h
M
G
A
Slant height OM =
Volume of a pyramid
=
36 = 6 cm
C
B
OG 2  MG 2 =
h 2  32
 Area of triangle forming the slant surface
1
× Area of the base × Height
3
=
1
3 3
×
× (10)2 × 30 = 2598 m3
3
2
[
Area of the regular hexagon of side
1
×6×
2
h2  9
=
= 3 h2  9
Given : 3 h 2  9 = 15
3 3 2
a=
a sq units]
2
7.
 h = 4 cm
 Volume of the pyramid =
P
1
× Area of the base ×
3
Height
h
E
1
× 36 × 4 = 48 cm3
3
D
a a 3
F
2a
A
9.
Area of regular hexagon of side a =
3 3 2
a
2
C
O

3 3 2
a = 96 3  a = 8 m
2
Let h be the height of the pyramid. Then area of one
B
side face of the pyramid =
16
1
a × l , where l is the
2
slant height of the face.

1
a × l = 32 3
2
 h = 15 cm
 l=8 3
3a 2
+ h2 = l 2

4

E
12.
3  64
+ h2 = 64 × 3
4
G
A
 1
 h2 = 64 × 3 1 4  = 196  h = 12 m


Volume of the pyramid =
=
F
1
× Area of the base × h
3
m
at
hs
wi
th
ab
hi
na
y.c
om

D
 slant height =
 l=
16
h2 
12
h2 
Slant height =
Lateral surface area =
a2
h 
12
2
a2
 152  8 2  17 cm
4
1
× Perimeter of base ×slant height
2
1
= × 64 × 17 = 544 cm2
2
4
h2 
3
 l=
C
Perimeter of square = 4a = 4 × 16 = 64 cm.
1
× 96 3 × 12 = 384 3 m3
3
10. Let a be the length of each side of the base, h be the
height and l be the slant height of the pyramid.
13.
D
It is given that the numerical value of its total surface
area is three times the numerical value of of its
volume.

B
A
A
O
Lateral surface area + Area of the base = 3 × (Volume)

1
(4 + 4 + 4) ×
2
h2 
4
+
3
3
× 42
4
B
 6
4
h  = 4 3 (h – 1)
3
C
BE = 5 3
4
+ 4 3= 4 3h
3
h2 
E
AB = 10 3 cm
1  3  42  h 
=3×



3  4
 6
B
C
AE =
OE =
2
2
10 3   5 3 
2
= 15 cm
1
× 15 = 5 cm
3
If height of pyramid = h cm
 2 4
 36  h  3  = 48 (h – 1)2
Slant height =
Total surface area =
 2 4
 3  h  3  = 4 (h – 1)2
1
× area of base × slant height +
2
area of base
 3h2 + 4 = 4 (h2 – 2h + 1)
 8h = h2  h = 8 cm
11. Volume =
h 2  25
270 3 =
1
× area of base × height
3
1
× 30
2
3 ×
h 2  25 +
h2 + 25 = 169
h=
1
 8000 = 3 × 40 × 40 × h
144 = 12 cm
14. area of base =
17
1
× 5 × 12 = 30 cm2
2
3
× 10 3
4


2
 Volume = area of base × hiegh
330 = 30 × h

h = 11 cm
15.
=
19. Volume of tetrahedron =
52 12 2 = 13 cm
2
×3×3×3
12
=
9 2
c.c
4
m
at
hs
wi
th
ab
hi
na
y.c
om
height
=
1
× 40 × 13 = 260 cm2
2
20.
=
17. Volume of pyromid =
Volume =
3 (12)2 = 144 3 cm2
=
1
× area of base × height
3
1 1
× × 5 × 12 × h
3 2
h = 33 cm
4
h
O
l
12
D
C
4 D
O
D
C
10
F
A
l
h=
2
 l = 2h
2 3
1
× (10)2 × 5 3
3
P
4
4
1
× area of base × slant height
3
E
A
OD =
a 3
=5 3
2
= 288.67m2
21. Total surface area of pyramid = lateral surface area +
area of base.
slant height for base side AB and CD.
330 =
B
Slant edge = a = 10 m
slant height =
3 2
a
4
16. Total surface area of terahedron = 4 ×
18.
=
1
× perimetar of base × slant
2
area of slant surface =
3 cm2
2 3
a
12
(  a = side of tetrahedron)
1
side of square base =
× 10 2 = 10 cm
2
slant height =
8
9
B
32
l1 =
=
2
(inradius)
3
similarls slant height for base side AD and CD
l2 =
In  POD
l2 = h2 + OD2
4h2 = h2 +
(5) 2  (12) 2 = 13 cm
(16) 2  (12) 2 = 20 cm.
Area of triangles n side AB and CD
4
3
=2×
1
× 32 × 13
2
= 416 cm2
Area of triangles on side AD and BC
4
2
3h2 =
 h=
3
3
1

= 2 ×   20 10  = 200 cm2
2
1
Volume = × area of base × height
3
lateral surface area = 416 + 200 = 616 cm2
area of base = 32 × 10 = 320 cm2
Total surface area = 616 + 320
= 936 cm2
1
2
3
= ×
× (4)2 ×
3
3
4
18