# Question #204b9

Aug 12, 2016

$366 f {t}^{3}$

#### Explanation:

On observation of the given figure it appears that the solid is a combination of two pyramids, So the problem can be solved by using the following formula

$\text{Volume of a pyramid"=1/3xx"Area of its base"xx "its height}$

The base of each pyramid is a trapezium and its area is determined by the following formula

$\text{Area of Trapezium}$
$= \frac{1}{2} \times \text{sum of its two parallel sides"xx "distance between them}$

Pyramid - I
Its base is a Trapezium ABGC having parallel sides $B G = 8 '$ and $A C = 5 '$ and the distance between these sides is $A B = 9 '$
Again the height of this Pyramid $B F = 7 '$ (Since each face angle at B is ${90}^{\circ}$)

So volume of this pyramid-I

${V}_{1} = \frac{1}{3} \times \text{Area of its Trapezoidal base ABGC"xx "Its height } B F$

$\text{ } = \frac{1}{3} \times \frac{1}{2} \left(B G + A C\right) \times A B \times B F$

$\text{ } = \frac{1}{3} \times \frac{1}{2} \left(8 + 5\right) \times 9 \times 7 f {t}^{3} = 136.5 f {t}^{3}$

Pyramid - II

Its base is a Trapezium ADEC having parallel sides $A C = 5 '$ and $D E = 12 '$ and the distance between these sides is $A D = 9 '$
Again the height of this Pyramid $E K = B A = 9 '$ (Since each face angle at A is ${90}^{\circ}$)

So volume of this pyramid-II

${V}_{2} = \frac{1}{3} \times \text{Area of its Trapezoidal base ADEC"xx "Its height } B A$

$\text{ } = \frac{1}{3} \times \frac{1}{2} \left(D E + A C\right) \times A D \times B A$

$\text{ } = \frac{1}{3} \times \frac{1}{2} \left(12 + 5\right) \times 9 \times 9 f {t}^{3} = 229.5 f {t}^{3}$

So total volume of the solid

$V = {V}_{1} + {V}_{2} = \left(136.5 + 229.5\right) f {t}^{3} = 366 f {t}^{3}$