Cones and Pyramids
Egyptian Pyramid
These are pyramids
V
Slant edges
height
D
A
Vertex
B
C
x x x
cube Three congruent pyramids
x x
1
Volume of the pyramid =
3
=
x
1
x3 x2 x
3 1
=
base area height
3
For any pyramid, Volume of pyramid =
1 3
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
Solution :
VF2 = (152 - 92 ) cm2
V
VF =
15 cm
E
D F
A
E 9 cm B 15 cm
C
15
2
-9
= 12 cm
F 9 cm
Volume of the pyramid = =(
1
base area height
3 1
192 12) cm3
3
= 768 cm3
2
cm
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
1
6 cm
H E
Volume of VEFGH = ( ( 8 8 ) 6) cm3 3
G
= 128 cm3
F
1
12 cm
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 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 Remark :
Area of sector = 1/2r2 (θ/2)= 1/2 r2 θ or 1/2 r l
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 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
1 96 = 3 y2 8
what is the length of a side of the base?
288 = 8y2 36 = y2
Back to Q1
y=6 Therefore, the length of a side of base is 6 cm
To Q2
AX : XB = ? 3 = 16 ( AX ) AB 54 AX 8 ( AB )3 = 27
Hints: Using the concept of RATIOS A
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 = 12 AC =15cm 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 = 704 cm3
r2
2500 = 2304 + r2 196 = r2
The volume is 704 cm3 A (Take = 22/7)
r = 14 The radius is 14cm. Back to Q4
50cm
B
C
These are pyramids
V
Slant edges
height
D
A
Vertex
B
C
x x x
cube Three congruent pyramids
x x
1
Volume of the pyramid =
3
=
x
1
x3 x2 x
3 1
=
base area height
3
For any pyramid, Volume of pyramid =
1 3
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
Solution :
VF2 = (152 - 92 ) cm2
V
VF =
15 cm
E
D F
A
E 9 cm B 15 cm
C
15
2
-9
= 12 cm
F 9 cm
Volume of the pyramid = =(
1
base area height
3 1
192 12) cm3
3
= 768 cm3
2
cm
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
1
6 cm
H E
Volume of VEFGH = ( ( 8 8 ) 6) cm3 3
G
= 128 cm3
F
1
12 cm
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 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 Remark :
Area of sector = 1/2r2 (θ/2)= 1/2 r2 θ or 1/2 r l
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 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
1 96 = 3 y2 8
what is the length of a side of the base?
288 = 8y2 36 = y2
Back to Q1
y=6 Therefore, the length of a side of base is 6 cm
To Q2
AX : XB = ? 3 = 16 ( AX ) AB 54 AX 8 ( AB )3 = 27
Hints: Using the concept of RATIOS A
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 = 12 AC =15cm 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 = 704 cm3
r2
2500 = 2304 + r2 196 = r2
The volume is 704 cm3 A (Take = 22/7)
r = 14 The radius is 14cm. Back to Q4
50cm
B
C
