IGCSE Surface Area and Volume 2_u5_s1.ppt

Volume of a pyramid
Volume of a pyramid =
1
3
 base area  perpendicular height
h

Calculate the volume of the rectangular-based pyramid.
6
cm
5 cm
4
cm
Volume =
1
3
 base area  height

1
3
(5 4) 6
40 cm3
A B
C
D
E

Surface area of a pyramid
Surface area = sum of the areas of all the faces of the pyramid
h

Calculate the surface area of the rectangular-based pyramid.
6
cm
5 cm
4
c
m
A B
C
D
E
First find the length of EX and EY.
EX2
2.52
 62
EX2
42.25
EX 6.5
EY2
22
 62
EY2
40
EY 6.325
Use Pythagoras on triangle EOX.
Use Pythagoras on triangle EOY.
O X
Y

6.5
6.5
6.325
6.325
5 cm
4
cm
A B
C
D
E
E
E E
Surface area = sum of areas of faces
Area of rectangle ABCD = 4 × 5
= 20 cm2
Area of triangle BCE = ½ × 4 × 6.5
= 13 cm2
Area of triangle CDE = ½ × 5 × 6.325
= 15.81 cm2
= 20 + 13 + 13 + 15.81 + 15.81
= 77.6 cm2
NET OF
PYRAMID

Volume of a cone
Volume of a cone =
1
3
 base area  perpendicular height

1
3
r 2
h
h
r

Calculate the volume of the cone.
Volume =
1
3
 base area  height

1
3
( 42
) 7
117 cm3
7
cm
4 cm

Surface area of a cone
l
h
r
+
The surface of a cone is made from a flat circular base and a curved surface.
The curved surface is made from a sector of a circle.
FLAT
BASE
CURVED SURFACE
=
l l
Curved surface area of a cone = , where is the slant height
rl l
Total surface area of a cone = r2
rl

Calculate
a the curved surface area of the cone,
b the total surface area of the cone.
12
cm
5 cm
a First calculate the slant height using Pythagoras.
l
l
l2
l2
l
Curved surface area rl
 5 13
204 cm2
b Total surface area r2
rl
 52
 65
25  65
65
90
283 cm2
52
122
169
13

The straight edges of the sector are joined together to make a cone.
Calculate a the curved surface area of the cone,
b the radius of the base of the cone,
c the height of the cone. 280o
4
c
m
4
c
m
a Curved surface area = area of sector

280
360
 42
39.1 cm2

112
9

b Curved surface area rl
39.1 r 4
r 
39.1
4
r 3.11 cm
c Using Pythagoras
3.11
4
h
h2
 3.112
42
h2
6.321
h 2.51 cm

When you make a cut parallel to the base of a cone and remove the
top part, the part that is left is called a frustum.
FRUSTUM
Volume of frustum = volume of large cone – volume of smaller cone

Calculate the volume of the frustum.
All lengths are in cm.
3
6
8
h
You must first find the height
of the smaller cone using
similar triangles.
h
3

h  8
6
6h 3h  24
3h 24
h 8
Volume of large cone 
1
3
r 2
h

1
3
 62
16
192
Volume of small cone 
1
3
r 2
h

1
3
 32
8
24
Volume of frustum 192  24
528 cm3
3
6
8

Volume and surface area of a sphere
Volume of a sphere 
4
3
r 3
Surface area of a sphere 4r 2
Volume and surface area of a hemisphere
Volume of a hemisphere
2
3
r3
Curved surface area of a hemisphere 2r 2
A hemisphere is half a
sphere.

The sphere has radius 10 cm.
Calculate a the volume of the sphere,
b the surface area of the sphere.
a Volume 
4
3
r3

4
3
 103
4189 cm3
b Surface area 4r 2
4  102
1257 cm2

The solid hemisphere has radius 6 cm.
Calculate
a the volume of the hemisphere,
b the curved surface area of the hemisphere,
c the total surface area of the hemisphere.
6 cm
a Volume 
2
3
r3

2
3
 63
452 cm3
b Curved surface area 2r 2
2  62
226 cm2
c Total surface area = area of base circle + curved surface area
 62
 226
339 cm2

The solid is made from a cylinder and a hemisphere.
The cylinder has a height of 8 cm and a radius of 3 cm.
Calculate the volume of the solid.
Volume of cylinder r 2
h
 32
8
72
Volume of hemisphere 
2
3
r3

2
3
 33
18
Total volume 72 18
90
283 cm3

IGCSE Surface Area and Volume 2_u5_s1.ppt

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