012
and
Find the volume of frustum of cone,
the area of whose ends are 40 sq metures and 10 sq metres and height is 9 metres
Answers
Answer:
Answer:
The Volume of the frustum is 213.6456 cubic meter.
Step-by-step explanation:
Given : Frustum of cone areas of whose ends are 40 square meter and 10 square meter and height is 9 meter.
To find : The volume of the frustum ?
Solution :
The area of the one end of cone is A_1=40\ m^2A
1
=40 m
2
The radius of the one end of frustum is A=\pi {r_1}^2A=πr
1
2
40=\pi {r_1}^240=πr
1
2
r_1=\sqrt{\frac{40}{\pi}}r
1
=
π
40
r_1=3.56\approx 3.6r
1
=3.56≈3.6
The area of the other end of cone is A_2=10\ m^2A
2
=10 m
2
The radius of the other end of frustum is A=\pi {r_2}^2A=πr
2
2
10=\pi {r_2}^210=πr
2
2
r_2=\sqrt{\frac{10}{\pi}}r
2
=
π
10
r_2=1.78\approx 1.8r
2
=1.78≈1.8
The height is h=9 meter.
The volume of the frustum is
V = \frac{1}{3}\pi h(r_1^2 + r_2^2+ (r_1\times r_2))V=
3
1
πh(r
1
2
+r
2
2
+(r
1
×r
2
))
Substitute the value,
V = \frac{1}{3}\times 3.14\times 9((3.6)^2 +(1.8)^2+ (3.6\times1.8))V=
3
1
×3.14×9((3.6)
2
+(1.8)
2
+(3.6×1.8))
V = \frac{1}{3}\times 3.14\times 9(12.96 +3.24+6.48)V=
3
1
×3.14×9(12.96+3.24+6.48)
V = \frac{1}{3}\times 3.14\times 9\times 22.68V=
3
1
×3.14×9×22.68
V = 213.6456\ m^3V=213.6456 m
3
Therefore, The Volume of the frustum is 213.6456 cubic meter.