egyptian pyramidverulam.s3.amazonaws.com/resources/ks4/maths/year 11 maths grade a... · 3 = 1 x2 x...
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3
1x2 x =
= 3
1 base area height
Volume of the pyramid =
3
1
x3 x
x
x
For any pyramid,
Volume of pyramid = 3
1 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
C D
B
15 cm 9 cm E
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 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
A
E
D C
B
H G
F
6 cm
12 cm
Solution :
Volume of VEFGH = ( 3
1 ( 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
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
B A
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
Curved surface area = πr l
Curved surface area Remark : Area of sector = 1/2r2 (θ/2)= 1/2 r2 θ or 1/2 r l
Curved surface area = Area of the sector
Curved surface area = 1/2 ( l ) ( 2π r )
= π r l
After cutting the cone,
θ
r
l
Examples
1 a) If h = 12cm, r= 5 cm, what
is the volume?
Answer:
Volume = πr2h 1 3
1 3
= 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
= -
R r
= πR3 - π r3
3
1
3
1
3
1
π( R3 - r3 ) =
Volume of frustum = volume of big cone - volume of small cone
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
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 C A. 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
C B
50cm
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 = ) 22 7
Help
Answer
Let V is the volume of the pyramid and y be the length of
a side of base
V = base area height 1 3
96 = y2 8 1 3
288 = 8y2
36 = y2
y = 6
Therefore, the length of a
side of base is 6 cm
Back to Q1
To Q2
what is the length of a
side of the base?
( )3 = AB AX 16
54
AB AX
( )3 = 8 27
AX AB
= 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 =15cm 2 1
Therefore, the height
(VM) of the pyramid
is 20 cm
Volume of the pyramid is:
= ×18 ×24 ×20 1 3
×base area ×height 3 1
= 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.