Question

A corporation building is in the form as shown alongside. The vertical cross section parallel to the width side of the building is a rectangle of size 5 m x 2 m mounted by a semicircle of radius 1.5 m. The inner measurements of a cuboid are 6 x 5 x 2 m. Find the volume of the corporation and the total internal surface area excluding the floor.

Answer 1

Volume:
As the corporation building is of uniform cross-section: its volume $=$ Its area of cross section $\times$ its length
$=[5\times 2+\frac{1}{2}\times \frac{22}{7}\times (1.5{)}^2]\times 6$
$=13.535\times 6$
$=81.21m^3$
For total internal surface area excluding the floor:
Perimeter of cross section excluding the floor $=2+\frac{2\pi r}{2}+2=4+\pi \times 1.5=8.71m$
Required internal surface area $=$ Perimeter of cross section $\times$ length $+2\times$ area of cross section
$=8.71\times 6+2\times 13.53=79.32m^2$