A heap of wheat is in the form of a cone whose base diameter is 8.4 m and height is 1.75 m. The heap is to be covered by canvass. What is the area (in m$^2$) of the canvas required? (Use $\pi = \frac{22}{7}$)

60.06

We know that,

H² = B² + P²

The curved surface area of the cone = πrl

We have,

Diameter of cone = 8.4m

Height of cone = 1.75m

Radius of cone = \(\frac{8.4}{2}\) = 4.2 m

Using Pythagoras Theorem,

= l = \(\sqrt {4.2^2 + 1.75^2}\) = 4.55 m

The curved surface area of the cone = \(\sqrt {3}\) × 4.2 × 4.55 = 60.06 m²