Find the Volume of a frustum of a regular hexagonal pyramid the sides of whose bases are 5m and 10m respectively and whose altitude is 15m
A. 2325 m^3
B. 2452 m^3
C. 2244 m^3
D. 2273 m^3
Answers
Given : A frustum of a regular hexagonal pyramid the sides of whose bases are 5m and 10m respectively
altitude is 15m
To Find : Volume
A. 2325 m^3
B. 2452 m^3
C. 2244 m^3
D. 2273 m^3
Solution:
Volume of the frustum of a pyramid = (h/3) (A₁ + A₂ + √A₁A₂)
A₁ = Area of Lower base
A₂ = Area of upper base
h = altitude (height ) = 15 m
A₁ = 6 * (√3 / 4) * 10²
A₂ = 6 * (√3 / 4) * 5²
√A₁A₂ = 6 * (√3 / 4)*50
Volume of the frustum of a pyramid = (15/3) (6 * (√3 / 4) * 10² + 6 * (√3 / 4) * 5²+ 6 * (√3 / 4)*50)
= 5 (6 * (√3 / 4)) ( 100 + 25 + 50)
= 5 (3 * (√3 / 2)) ( 175)
= 2,625√3 /2
= 2,273.25
= 2,273 m³
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