Volume and Surface Area of Solids

Surface Area of Solids

● Surface Area of a Prism
● Surface Area of a Cylinder
● Surface Area of a Pyramid
● Surface Area of a Cone
● Surface area of a Frustum
● Surface Area of the Sphere

Surface Area of a Prism

A prism is a solid with uniform cross- section. The surface area of a prism is the sum of the area
of its faces.

Surface Area of a Cylinder

For the surface area of a cylinder, it is best to visualise the net of the solid: it is made up of one
rectangular piece and two circular pieces

Area of circular pieces = 2 × πr2

Area of rectangular piece = 2πr × h
Total surface area = 2πr2 + 2πrh
= 2πr(r + h)

Worked Example
If the height of a cylinder is 7 cm and the radius of its circular top is 3 cm, calculate its surface
area.

Total surface area = 2πr(r + h)

= 2π × 3 × (3 + 7)
= 6π × 10
= 60π
= 188 cm2 (3 s.f.)
The total surface area is 188 cm2.
Work To Do
1. Calculate the surface area of each of the following cylinders:
a) r = 2 cm, h = 6 cm
Total surface area = 2π2 + πDH✔
22 22
2x 7
x 22 + 7
x4x6=
25.14+75.43=100.57
=100.57cm2 ✔

b) r = 4 cm, h = 7 cm
T.S.A = 2 πr2+π𝐷𝐻✔
22 22
2x 7
x 42 + 7
x8x7
100.57 + 176=
=276.57cm2✔

c) r = 3.5 cm, h = 9.2 cm

T.S.A = 2 πr2 + π𝐷𝐻
2 x 22/7 x 3.52 + 22/7 x 7 x 9.2
77 + 202.4
=279.4cm2✔

d) r = 0.8 cm, h = 4.3 cm

T.S.A =2 π2 + π𝐷𝐻✔
2 x 22/7 x 0.82 + 22/7 x 1.6 x 4.3=
4.023+21.623
=25.646cm2✔

2. Calculate the height of each of the following cylinders. Give your answers to 1 d.p.
a) r = 2.0 cm, surface area = 40 cm2
40=2 x 22/7 x 2.02 + 22/7 x 4 x h✔
40=25.14 + 12.57h
40-25.14=14.86
14.86=12.57h ✔
h= 1.18cm✔
 b) r = 3.5 cm, surface area = 88 cm2
88=2x 22/7 x 3.52 + 22/7 x 7 x h✔
88 = 77+22h
88-77=22h
11=22h✔
h=0.5cm✔

c) r = 5.5 cm, surface area = 250 cm2

250=2 x 22/7 x 5.52 + 22/7 x 11 x h
250 = 190.14 + 34.57h
250-190.14=
59.86=34.57h
h=1.73cm

d) r = 3.0 cm, surface area = 189 cm2

189 = 2 x 22/7 x 3.02 + 22/7 x 6 x h
189 = 56.57 + 18.86h
189-56.57
132.43 =18.85h
h = 7.03cm

Surface Area of a Pyramid

The surface area of a pyramid is the sum of the area of the slanting faces and the area of the
base.
Surface area = base area + area of the four triangular faces (take the slanting height marked
green below)

Worked example
Find the surface area of the following pyramid

Solution
Surface area = base area + area of the four triangular faces
= (14 x 14) + (½ x 14 x 9)4
= 196 + 252
= 448 mm2

Work To Do
1. Calculate the surface area of a regular tetrahedron with edge length 2 cm.
(𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
𝑎+𝑏+𝑐
s= 2

(3(3 − 2)(3 − 2)(3 − 2))

3
=1.73 x 4 = 6.92
=6.92cm2

2. The rectangular-based pyramid shown has a sloping edge length of 12 cm. Calculate its
surface area.

12+12+5
2
= 14.5
12+12+8
2
= 16
Area = 16(16 − 12)(16 − 12)(16 − 8)
Area = 45.25 x 2 =90.5
Area = 14. 5(14. 5 − 12)(14. 5 − 12)(14. 5 − 5)
Area = 29.34 x 2 = 58.68
Surface area =90.5 +58.68 + 40 = 189.18cm2
 3. Two square-based pyramids are glued together as shown (right). Given that all the
triangular faces are congruent, calculate the surface area of the whole shape.

5+5+4
2
=7
Area = 7(7 − 5)(7 − 5)(7 − 4)
Area = 9.17 x 8
Surface area = 73.32cm2

Surface Area of a Cone

Total surface area of a cone= πr2 + πrl
Curved surface area of a cone =πrl

The surface area of a cone comprises the area of the circular base and the area of the curved
face. The area of the curved face is equal to the area of the sector from which it is formed.

Worked example
Find the surface area of the cone below
Solution
= (3.14 x 4 x 4) + (3.14 x 4 x 5)
= 50.24 +62.8
=113.04 cm2
Note;
Always use slanting height, if it’s not given find it using Pythagoras theorem

1. Calculate the surface area of each of the following cones:

No radius

Surface area = πr2 + πrl

(3.142 x 152) +(3.142 x 15 x 20)
706.95 + 942.6
=1649.55cm2

1. Two cones with the same base radius are stuck together as shown. Calculate the
surface area of the shape.
 Surface area = πr2 + πrl
Surface area = (3.142 x 8 x 15 ) + (3.142 x 8 x 30)
Surface area = 1131.12cm2

Surface Area of a Frustum

The bottom part of a cut pyramid or cone is called a frustum. Example of frustums are bucket, a
lampshade and a hopper.

Example
Find the surface area of a fabric required to make a lampshade in the form of a frustum whose
top and bottom diameters are 20 cm and 30 cm respectively and height 12 cm.
Solution
Complete the cone from which the frustum is made, by adding a smaller cone of height x cm.

Surface Area of the Sphere

A sphere is solid that it’s entirely round with every point on the surface at equal distance from
the Centre.

Surface area is = 4πr2

Example
Find the surface area of a sphere whose diameter is equal
to 21 cm
Solution
Surface area = 4πr2
= 4 x 3.14 x 10.5 x 10.5
= 1386 cm2

Volume of Solids
● Volume of a Prism
 ● Volume of a Pyramid
● Volume of a Sphere
● Volume of a Cone
● Volume of a Frustum
.

Volume is the amount of space occupied by an object. It’s measured in cubic units.
Generally volume of objects is base area x height

Volume of a Prism
A prism is a solid with uniform cross section. The volume V of a prism with cross section area A
and length L is given by V = AL

Example

Solution
Volume of the prism = area of the cross-section x length
= ½ × 6 × 3 × 10
= 90cm2

Example
Find the volume of a regular hexagonal prism with edge length 8 ft and height 12ft
Solution

h = 4 tan 60 = 6.9 ft
Area of 1 triangle = ½ x 8 x 6.9 = 27.6 ft2
Area of 6 triangles = 27.6 x 6 = 165.6 ft2

Volume = cross section area x length

V = 165.6 x 12 = 1987.2 ft2

Volume of a Pyramid
Volume of a pyramid = ⅓ Ah
Where A = area of the base and h = vertical height
Example
Find the volume of a pyramid with the vertical height of 8 cm and width 4 cm length 12 cm.
Solution.

Volume = ⅓ x 12 x 4 x 8
= 128cm3

Volume of a Sphere
4
V= 3
πr2

Volume of a Cone
Volume = ⅓ area of base × height
1
V= 3 πr2

Example
Calculate the volume of a cone whose height is 12 cm and length of the slant height is 13 cm
Solution
Volume
Radius = base ?
Vertical height = triangle height
Slant height = the hypotenuse
r2 +h2 = H2
132 - 122 = 25
25 = 5 cm
V = ⅓ πr2
⅓ x 3.142 x 52 x 12 =314.2 cm3
Volume of a Frustum
Volume = volume of large cone – volume of smaller cone
Example
A frustum has base radius 2 cm and height 3.6 cm. if the height of the cone from which it was
cut was 6 cm, calculate
a. The radius of the top surface
6 2
3.6
= ?
7.2 = 6 x r
1.2cm = r✔

b. The volume of the frustum

volume of frustum = volume of original pyramid - volume of cut-off pyramid
Volume of original pyramid= ⅓πR2H
⅓ x 3.142 x 22 x 3.6 = 15.08cm✔
Volume of cut-off pyramid
⅓ πr2h
⅓ x 3.142 x 1.22 x 6 = 9.05
Volume of the frustum = 15.08 - 9.05=
= 6.03 cm3
Solution

Work to do
1. Metal cube of side 4.4cm was melted and the molten material used to make a sphere.
Find to 3 significant figures the radius of the sphere (3mks)
Volume of the cube = 4.43
=85.184cm3
4
Volume of sphere = 3
πr3
4
85.184 = 3
x 3.142 x r3
85.184 = 4.19r3
20.33 = r3
r=2.73cm
 2. Two metal spheres of diameter 2.3cm and 3.86cm are melted. The molten material is
used to cast equal cylindrical slabs of radius 8mm and length 70mm. If 1/20 of the metal
is lost during casting. Calculate the number of complete slabs casted. (4mks)
4
Volume of sphere = 3
πr3
4 4
( 3 x 3.142 x 1.153) + ( 3 x 3.142 x 1.933)
1
6.37 + 30.12 = 36.49cm3 x 20
= 1.82cm3
36.49 - 1.82 = 34.67cm3
Volume of a cylinder = πr2l
3.142 x 0.82 x 7= 14.08cm3
34.67 ÷ 14.08 = 2.5
= 2 slabs

3. The volume of a rectangular tank is 256cm3. The dimensions are as in the figure. Find
the value of x (3 marks)

Volume =A.C.S X L
256 cm3 =(¼ x)x(x-8) x 16
256 = ¼ x2 x 2x x 16
256 = ¼ x2 x 32x
256 = 8x3
x3 = 32
3
x = 32
x = 3.17

¼ x(x-8) x.16
¼ x2-2x x 16 = 256