radius of heap of wheat r = 10.5/2 = 5.25 m and height h = 3 m
Volume of heap of wheat = 1/3πr²h = 1/3 × 22/7 × 5.25 × 5.25 × 3 = 22 × 0.75 × 5.25 = 86.625 m³.
Hence, the volume of heap of wheat is 86.625 m³.
Let, the slant height of heap of wheat = l m
We know that, l² = h² + r² ⇒ l² = 3² + (5.25)² ⇒ l² = 9 + 27.5625 ⇒ l² = 36.5625
⇒ l = √36.5625 = 6.05 m (approx.)
Required area of canvas = πrl
= 22/7 × 5.25 × 6.05 = 22 × 0.75 × 6.05 = 99.825 m²
Hence, the required area of canvas to protect wheat is 99.825 m².