Question

The interior of a building is in the form of a right circular cylinder of diameter 4.2 m and height 4 m surmounted by a cone of same diameter. The height of the cone is 2.8 m. Find the outer surface area of the building.

Answer 1

We have,

Radius of the cylinder = Radius of the cone = r = $\frac{4.2}{2}$ = 2.1 m,

Height of the cylinder, H = 4 m and

Height of the cone, h = 2.8 m

Also,

The slant height of the cone,
l = $\sqrt{r^2+h^2}$

= $\sqrt{{2.1}^2+{2.8}^2}$

= $\sqrt{4.41+7.84}$

= $\sqrt{1}2.25$

= 3.5 m

Now,

The outer surface area of the building = CSA of the cylinder + CSA of the cone

= 2πrH+πrl
= πr(2H+l)

= $\frac{22}{7}$ ×2.1×(2×4+3.5)

= 6.6 × 11.5

= 75.9 $m^2$

So, the outer surface area of the building is 75.9 $m^2$.