Question #204b9

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

`"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

`"Area of Trapezium"`
`=1/2xx"sum of its two parallel sides"xx "distance between them"`

Pyramid - I
Its base is a Trapezium ABGC having parallel sides `BG=8'` and `AC=5'` and the distance between these sides is `AB=9'`
Again the height of this Pyramid `BF=7'` (Since each face angle at B is `90^@`)

So volume of this pyramid-I

`V_1=1/3xx"Area of its Trapezoidal base ABGC"xx "Its height "BF`

`" "=1/3xx1/2(BG+AC)xxABxxBF`

`" "=1/3xx1/2(8+5)xx9xx7ft^3=136.5ft^3`

Pyramid - II

Its base is a Trapezium ADEC having parallel sides `AC=5'` and `DE=12'` and the distance between these sides is `AD=9'`
Again the height of this Pyramid `EK=BA=9'` (Since each face angle at A is `90^@`)

So volume of this pyramid-II

`V_2=1/3xx"Area of its Trapezoidal base ADEC"xx "Its height "BA`

`" "=1/3xx1/2(DE+AC)xxADxxBA`

`" "=1/3xx1/2(12+5)xx9xx9ft^3=229.5ft^3`

So total volume of the solid

`V=V_1+V_2=(136.5+229.5)ft^3=366ft^3`