A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of the cylinder is 32 m, and the height of the cylinder is 14 m, while the vertex of the cone is 26 m above the ground. What is the area of the canvas required for this tent? (Where π = 3.14)

Correct option is D

Given:

Diameter of cylinder = 32 m

Height of cylinder =14 m

Vertex of cone above ground = 26 m

Formula Used:

Lateral surface area of cone = $$\pi r l$$​​

Slant height $$l= \sqrt{r^{2}+h^{2}}$$​​

Curved Surface Area  = $$2 \pi rh$$​​

Solution:

Radius of cylinder = Radius of Cone = $$\frac{32}{2}=16\ m$$​​

Height of cone = 26 - 14 = 12 m

Slant height(l) = $$\sqrt{16^{2}+12^{2}}=\sqrt{256+144} = \sqrt{400} = 20\ m$$​​

The area of canvas required is lateral surface area of conical and cylindrical tent

= LSA of Cone + CSA of Cylinder

= $$\pi \times 16 \times 20 + 2\times \pi \times16 \times 14$$​​

= $$\pi \times 16(20 +2 \times 14)$$​​

= $$3.14 \times 16(48)$$​​

= $$2411.52\ m^{2}$$​​