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```					Volume of a pyramid
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
           Vertex

       

Slant edges   
       height
D

   C


A                             
B
x
x

x

cube

Three congruent pyramids
x                               1
Volume of the pyramid =   x3
x                                     3
1 2
x                          =       x x
3
1
=       base area  height
3

For any pyramid,

1
Volume of pyramid =        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 :
V                                           VF2 = (152 - 92 ) cm2
15 cm                VF =    15 - 9 cm
2    2

E           F          = 12 cm
D              C                 9 cm
F
A                                Volume of the pyramid
E 9 cm B
15 cm                                    1
=         base area  height
3
1
=(    192  12) cm3
3
= 768 cm3
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.

V                  Solution :
1
Volume of VEFGH = (       ( 8  8 )  6) cm3
6 cm                          3
H           G
= 128 cm3
E           F
12 cm                                                        1
Volume of VABCD = (  ( 16  16 )  12) cm3
3
D
C
= 1024 cm3

Volume of frustum ABCDEFGH
A                   B
= (1024 - 128 ) cm3

= 896 cm3
C

V
V

D       C
D           C   V               V
A           B           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.

V              Solution :
Total surface area of pyramid VABCD

= Area of ABCD + (Area VAB + Area VDC +
D            C                       Area VBC + Area VAD )

A            B         = 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
Remark :   Area of sector = 1/2r2 (θ/2)= 1/2 r2 θ or 1/2 r l
θ

r
After cutting the cone,
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

h
1
h
3
r

Volume of a cone = 1 πr2 h
3
How to calculate
total surface area of a cone?

l
l   +         r
r

Total surface area =πr2 + πr l
Examples
1 a)   If h = 12cm, r= 5 cm, what
is the volume?

1
Volume = 3 πr2h

= 1 π (52) ( 12)
3

= 314 cm3
b) what is the total surface area?
Based Area = π52
= 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            1
=   3
πR3     -   π r3
3
1
=   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
C. 6cm
is C                           Help
D. 12cm
E. 36cm               To Q2
Q2
A
In the figure, the volumes of
the cone AXY and ABC are
16 cm3 and 54 cm3
X         Y       respectively, AX : XB =
A. 2 : 1
B                 C
B. 2 : 3             Answer
C. 8 : 19
D. 8 :27
is A
E.   3   16 : 3 38    To Q3
Q3            V              In the figure, VABCD
is a right pyramid with
a rectangular base. If
D
AB=18cm, BC=24cm
and CV=25cm, find
C
M
A
B
a) the height (VM) of
the pyramid,
b) volume of the
a) 20cm                                   Help
pyramid.
b) 2880cm3
To Q4
Q4
A
The figures shows a right
50cm
circular cone ABC. If AD=
48cm and AC= 50cm, find
B              C    (a) the base radius (r) of the cone,
(b) the volume of the cone.
(Take  = 22 )
7

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       a
1
96 = 3  y2  8                  side of the base?
288 = 8y2

Back to Q1
36 = y2
y=6
Therefore, the length of a                        To Q2
side of base is 6 cm
Hints: Using the concept
AX : XB = ?                 of RATIOS
AX       16                                A
( AB )3 = 54

( AX )3 = 278
AB                                   X            Y
AX      2
AB  = 3                             B                C
AB = AX + XB and AX = 2, AB = 3
Back to Q2
3 = 2 + XB
XB = 1
Therefore, AX : XB = 2 : 1                  To Q3
a) the height (VM) of the     b) volume of the
pyramid                       pyramid.
AC2 =182 + 242             Volume of the pyramid is:
AC2 = 900
1 ×base area ×height
AC = 30cm                   3
MC = 1 AC =15cm         = 1 ×18 ×24 ×20
2                   3
252 = VM2 + MC2         = 2880cm3
625 = VM2 + 152         Therefore, the volume of
625 - 225 = VM2         the pyramid is 2880cm3
VM2 = 400
Back to Q3
VM = 20cm
Therefore, the height
(VM) of the pyramid
is 20 cm                                   To Q4
(a) the base radius (r)          (b) the volume of the cone
The volume (V) of cone is:
The radius is r, therefore:
1
502   =   482   +   r2              V = 3  r2 h
1    22
2500 = 2304 + r2                      = 3  7  142  48
196 = r2                              = 704 cm3
r = 14                             The volume is 704 cm3
The radius is 14cm.                  A
(Take  = 22/7)
Back to Q4                           50cm

B               C

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