Title : Cone and Pyramid
Subject :
Mathematics
Target Audience : Form 3
Usage :
Lecturing
Prerequisite knowledge :
1. Pythagoras’ Theorem.
2. Area of some plane figures
e.g. square, rectangle, triangle, circle, sector
3. Ratio and proportion
Objectives :
Let the students know and apply the mensuration
concepts
Egyptian Pyramid
These are pyramids
V
Slant edges



Vertex



height

D
A




B
C
x
x
x
cube
Three congruent pyramids
x
x
1
x3
3
Volume of the pyramid =
x
=
=
1 2
x x
3
1
 base area  height
3
For any pyramid,
Volume of pyramid =
1
 base area  height
3
Example 1
The figure shows a pyramid with a rectangular base ABCD of
area 192 cm2, VE = 15 cm and EF = 9 cm, find the volume of
the pyramid.
V
Solution :
VF2 = (152 - 92 ) cm2
V
VF =
15 cm
E
D
A
F
E 9 cm B
15 cm
C
F
15 - 9 cm
2
= 12 cm
9 cm
Volume of the pyramid
=
1
 base area  height
3
1
 192  12) cm3
=(
3
= 768 cm3
2
B
 Pyramid B
 A frustum
Pyramid A
=
-
Volume of the frustum = Volume of Pyramid A - Volume of Pyramid B
Example 2
The base ABCD and upper face EFGH of the frustum are squares of side 16
cm and 8 cm respectively. Find the volume of the frustum ABCDEFGH.
Solution :
V
6 cm
H
E
Volume of VEFGH = (
G
1
 ( 8  8 )  6) cm3
3
= 128 cm3
F
1
12 cm
Volume of VABCD = (  ( 16  16 )  12) cm3
3
D
A
C
= 1024 cm3
Volume of frustum ABCDEFGH
B
= (1024 - 128 ) cm3
= 896 cm3
C
V
V
C
D
D
A
C
B
V
V
A
B
V
Total surface area of pyramid VABCD =
+
+
lateral faces
+
+
Base
Total surface area of a pyramid
= Base area + The sum of of the area of all lateral faces
Example 3
The figure shows a pyramid with a rectangular base ABCD of area 48
cm2. Given that area of  VAB = 40 cm2 , area of  VBC = 30 cm2,
find the total surface area of the pyramid.
Solution :
V
Total surface area of pyramid VABCD
D
A
C
B
= Area of ABCD + (Area VAB + Area VDC +
Area VBC + Area VAD )
= Area of ABCD + (Area VAB  2) +
(Area VBC  2)
= 48 cm2 + ( (40  2) + (30  2)) cm2
= 188 cm2
How to generate a cone?
…...
…...
How to calculate the curved
surface area ?
l
l
2πr
r
Cut here
Curved surface area
l
θ
After cutting the cone,
r
Curved surface area = Area of the sector
Curved surface area = 1/2 ( l ) ( 2π r )
=πrl
Curved surface area = πr l
Volume of a cone
r
1
h
3
r
Volume of a cone = 1 πr2 h
3
h
How to calculate
total surface area of a cone?
l
l
+
r
Total surface area =πr2 + πr l
r
Examples
1 a)
If h = 12cm, r= 5 cm, what
is the volume?
Answer:
1
Volume = 3 πr2h
= 1 π (52) ( 12)
3
= 314 cm3
b) what is the total surface area?
2
=
π5
Based Area
= 25πcm2
Slant height = 122 + 5 2
= 13 cm
Curved surface area = π(5) ( 13)
= 65π cm2
Total surface area = based area + curved surface area
= 25π+65π= 90π
= 282.6cm2 (corr.to 1 dec.place)
Volume of Frustum
Volume of Frustum

=
R
r
Volume of frustum = volume of big cone - volume of small cone
=
1 3
πR
3
=
1
3
-
π1 r3
3
π( R3 - r3 )
Start
Now
Exit
Q1
The volume of a pyramid of
square base is 96 cm3. If its
height is 8 cm, what is the
length of a side of the base?
A. 2 cm
B. 2 3cm
Answer
is C
Answer
C. 6cm
D. 12cm
E. 36cm
Help
To Q2
Q2
A
X
In the figure, the volumes of
the cone AXY and ABC are
16 cm3 and 54 cm3
respectively, AX : XB =
Y
A. 2 : 1
B
C
Answer
is A
B. 2 : 3
C. 8 : 19
D. 8 :27
E.
3
16 : 3 38
Answer
Help
To Q3
V
Q3
D
M
A
B
C
In the figure, VABCD
is a right pyramid with
a rectangular base. If
AB=18cm, BC=24cm
and CV=25cm, find
a) the height (VM) of
the pyramid,
Answer
b) volume of the
a) 20cm
b) 2880cm3
pyramid.
Help
To Q4
Q4
A
50cm
B
C
The figures shows a right
circular cone ABC. If AD=
48cm and AC= 50cm, find
(a) the base radius (r) of the cone,
(b) the volume of the cone.
(Take  = 22 )
7
Answer
Help
Let V is the volume of the pyramid and y be the length of
a side of base
1
V = 3  base area  height
what is the length of
1
96 = 3  y2  8
side of the base?
a
288 = 8y2
36 = y2
Back to Q1
y=6
Therefore, the length of a
side of base is 6 cm
To Q2
AX : XB = ?
Hints: Using the concept
of RATIOS
A
AX
16
( AB )3 = 54
8
3=
( AX
)
AB
27
AX
2
=
AB
3
AB = AX + XB and AX = 2, AB = 3
3 = 2 + XB
X
B
C
Back to Q2
XB = 1
Therefore, AX : XB = 2 : 1
Y
To Q3
a) the height (VM) of the
pyramid
AC2 =182 + 242
AC2 = 900
AC = 30cm
MC = 1 AC =15cm
2
252 = VM2 + MC2
625 = VM2 + 152
625 - 225 = VM2
VM2 = 400
VM = 20cm
Therefore, the height
(VM) of the pyramid
is 20 cm
b) volume of the
pyramid.
Volume of the pyramid is:
1 ×base area ×height
3
= 1 ×18 ×24 ×20
3
= 2880cm3
Therefore, the volume of
the pyramid is 2880cm3
Back to Q3
To Q4
(a) the base radius (r)
(b) the volume of the cone
The volume (V) of cone is:
The radius is r, therefore:
502
=
482
+
1
V = 3  r2 h
1
22
= 3  7  142  48
r2
2500 = 2304 + r2
196 = r2
= 704 cm3
The volume is 704 cm3
A
(Take  = 22/7)
r = 14
The radius is 14cm.
Back to Q4
50cm
B
C