form 3 cone and pyramid mathematics usage : title : subject : target audience : lecturing
TRANSCRIPT
Form 3
Cone and Pyramid
Mathematics
Usage :
Title :
Subject :
Target Audience :
Lecturing
Prerequisite knowledge :
1. Pythagoras’ Theorem.
3. Ratio and proportion
2. Area of some plane figures
e.g. square, rectangle, triangle, circle, sector
Let the students know and apply the mensuration concepts
Objectives :
Egyptian Pyramid
These are pyramids
Vertex
Slant edges height
V
A
B
CD
xx
x
cube
Three congruent pyramids
31
x2 x =
= 31
base area height
Volume of the pyramid = 3
1x3
xx
x
For any pyramid,
Volume of pyramid =31 base area height
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
A
CD
B
15 cm9 cmE
F
Solution : VF2 = (152 - 92 ) cm2
VF = 22 9 - 15 cm
= 12 cm
Volume of the pyramid
= 3
1 base area height
= ( 3
1 192 12) cm3
= 768 cm3
V
E F
15 cm
9 cm
Pyramid B
A frustum
Pyramid A
= Volume of Pyramid A -Volume of the frustum Volume of Pyramid B
B
= -
Example 2The 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
A
E
DC
B
H G
F
6 cm
12 cm
Solution :
Volume of VEFGH = ( 31
( 8 8 ) 6) cm3
= 128 cm3
Volume of VABCD = ( 31
( 16 16 ) 12) cm3
= 1024 cm3
Volume of frustum ABCDEFGH
= (1024 - 128 ) cm3
= 896 cm3
V
D C
BA
V
BA
D C
VV
V
C
Base area The sum of of the area of all lateral faces+=
Total surface area of a pyramid
Total surface area of pyramid VABCD =
+ + + +
lateral faces
Base
V
D C
BA
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 :
Total surface area of pyramid VABCD
+ (Area VAB + Area VDC +
Area VBC + Area VAD )
+ (Area VAB 2) +
(Area VBC 2)
+ ( (40 2) + (30 2)) cm2
= 188 cm2
= Area of ABCD
= Area of ABCD
= 48 cm2
How to generate a cone?
…...
…...
How to calculate the curved surface area ?
Cut here
l
r
l
2πr
Curved surface area = πr lCurved surface area = πr l
Curved surface area
Curved surface area = Area of the sectorCurved surface area = 1/2 ( l ) ( 2π r )
= π r l
After cutting the cone, θ
r
l
Volume of a cone
r
h
r
h 31
Volume of a cone = πr2 h
Volume of a cone = πr2 h
13
How to calculate total surface area of a cone?
Total surface area =πr2 + πr l Total surface area =πr2 + πr l
+r
ll r
Examples1 a) If h = 12cm, r= 5 cm, what
is the volume?
Answer:
Volume = πr2h13
13
= π (52) ( 12)
= 314 cm3
b) what is the total surface area?Based Area = π52
= 25πcm2
Slant height
= 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)
= 122 + 5 2
Volume of Frustum
Volume of Frustum
= -
R r
= πR3 - π r3
31
31
31
π( R3 - r3 ) =
Volume of frustum = volume of big cone - volume of small cone
Start Now Exit
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?
Q1
Answer is C
8cm
A. 2 cm
B. 2 3cm
C. 6cm
D. 12cm
E. 36cm
Help
Answer
To Q2
In the figure, the volumes of the cone AXY and ABC are 16 cm3 and 54 cm3 respectively, AX : XB =
Q2
Answer is A
A
X Y
B CA. 2 : 1
B. 2 : 3
C. 8 : 19
D. 8 :27
E. 3 16 : 3 38
Help
Answer
To Q3
V
D
C
A
B
M
Q3 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,
b) volume of the
pyramid. Help
Answer
To Q4
a) 20cm
b) 2880cm3
A
CB
50cm
48cm
Q4
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 = )227
Help
Answera) 14cm
b) 704cm3
Let V is the volume of the pyramid and y be the length of a side of base
V = base area height13
96 = y2 813
288 = 8y2
36 = y2
y = 6
Therefore, the length of a side of base is 6 cm
Back to Q1
To Q28c
m
what is the length of a side of the base?
( )3 = ABAX 16
54
ABAX ( )3 = 8
27
AXAB = 2
3
AB = AX + XB and AX = 2, AB = 3
3 = 2 + XB
XB = 1
Therefore, AX : XB = 2 : 1
Hints: Using the concept of RATIOS
Back to Q2
To Q3
A
X Y
B C
AX : XB = ?
AC2 =182 + 242
AC2 = 900
AC = 30cm
252 = VM2 + MC2
625 = VM2 + 152
625 - 225 = VM2
VM2 = 400
VM = 20cm
MC = AC =15cm21
Therefore, the height (VM) of the pyramid is 20 cm
Volume of the pyramid is:
= ×18 ×24 ×2013
×base area ×height31
= 2880cm3
Therefore, the volume of the pyramid is 2880cm3
Back to Q3
To Q4
a) the height (VM) of the pyramid
b) volume of the pyramid.
The radius is r, therefore:
502 = 482 + r2
2500 = 2304 + r2
196 = r2
r = 14
The radius is 14cm.
The volume (V) of cone is:
V = r2 h31
= 142 4831 22
7
= 704 cm3
The volume is 704 cm3
Back to Q4
A
CB
50cm48
cm
(a) the base radius (r) (b) the volume of the cone
(Take = 22/7)