Question

A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of cylinder is 24m. The height of of the cylindrical portion is 11m while the vertex of the cone is 16m above the ground. Find the area of the canvas required for the tent.

Answer 1

We know,
$l^2=r^2+h^2\Rightarrow l^2={12}^2+5^2$
=144 +25=169
l=13m
Now, Area of the canvas required
=C.S.A of the cylinder +C.S.A of the cone
$=2\pi rh+\pi rl=\frac{22}{7}[2\times 12\times 11+12\times 13]=\frac{22}{7}\times (264+156)=\frac{22}{7}\times 420=1320m^2$