Given,
Diameter of the conical heap (d) = 10.5 m
Radius of the conical heap (r) = Diameter 2 = 10.5 2 \dfrac{\text{Diameter}}{2} = \dfrac{10.5}{2} 2 Diameter = 2 10.5
Height of the conical heap (h) = 3 m
By formula,
Volume of the conical heap (V) = 1 3 π r 2 h \dfrac{1}{3}πr^2h 3 1 π r 2 h
Substituting values we get :
V = 1 3 × 22 7 × ( 5.25 ) 2 × 3 = 1 3 × 22 7 × 27.5625 × 3 = 1 3 × 22 7 × 82.6875 = 1819.125 21 = 86.625 m 3 V = \dfrac{1}{3} \times \dfrac{22}{7} \times (5.25)^2 \times 3 \\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 27.5625 \times 3 \\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 82.6875 \\[1em] = \dfrac{1819.125}{21} \\[1em] = \text{86.625 m}^3 V = 3 1 × 7 22 × ( 5.25 ) 2 × 3 = 3 1 × 7 22 × 27.5625 × 3 = 3 1 × 7 22 × 82.6875 = 21 1819.125 = 86.625 m 3
By formula,
Slant height (l) = r 2 + h 2 \sqrt{r^2 + h^2} r 2 + h 2
Substituting values we get :
l = ( 5.25 ) 2 + ( 3 ) 2 = 27.5625 + 9 = 36.5625 = 6.047 m l = \sqrt{(5.25)^2 + (3)^2} \\[1em] = \sqrt{27.5625 + 9} \\[1em] = \sqrt{36.5625} \\[1em] = 6.047 \text{ m} l = ( 5.25 ) 2 + ( 3 ) 2 = 27.5625 + 9 = 36.5625 = 6.047 m
The area of the canvas required to cover the heap of wheat (A) = Curved surface area of conical heap = πrl
Substituting values we get :
A = 22 7 × 5.25 × 6.047 = 698.4285 7 = 99.775 m 2 . A = \dfrac{22}{7} \times 5.25 \times 6.047 \\[1em] = \dfrac{698.4285}{7} \\[1em] = 99.775 \text{ m}^2. A = 7 22 × 5.25 × 6.047 = 7 698.4285 = 99.775 m 2 .
Hence, the volume of the conical heap is 86.625 m3 and the area of the canvas required is 99.775 m2 .
