Question

plz someone solve this question

Answer 1

volume of cone = pi. r^2 .h /3
=22/7 x (10.5/2)^2 x 3/3
= 22/7 x 441/16
=11 x 63 /8
= 693/8 = 86.625 m^3

surface area of canvas = pi.r .l
where l is the slant height
l =root {r^2 +h^2}
=root {441/16 + 9}
= 6 ( approx )
now surface area of canvas = 22/7 x 21/4 x 6 = 22 x 3 x 6/4 = 11 x 3 x 3 =99 m^2

Answer 2

Hi there,

given: diameter = 10.5m, height = 3m

therefore, radius = 10.5 ÷ 2
= 105 ÷ 20 [ (10.5 x 10) / (2 x 10) ]
= 21/4m


volume of a conem
= 1/3 x π x r² x h
= 1/3 x 22/7 x 21/4 x 21/4 x 3
= 11 x 63/8
= 693/8
= 86.625m³

therefore the volume of conical heap of wheat is 86.625m³

to find the area of canvas we need curved surface area of the conical heap of wheat, and for that we need slant height (l)

l² = r² + h²
l² = (21/4)² + 3²
l² = 441/16 + 9
l² = 441/16 + 144/16 ( making denominator common)
l² = 585/ 16
l² = 36.5626
l² ≈ 36
l ≈ 6m

Now,

curved surface area of the conical heap of wheat,
= π x r x l
= 22/7 x 21/4 x 6
= 11 x 3 x 3
= 99m²

therefore the canvas required for covering the conical heap of wheat is 99m².


Hope it helps you out!