Question

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Answer 1

Answer:

(1) given diameter = 24m => radius = 12m , height = 3.5m

volume of rice = volume of cone

volume of rice = πr^2h/3

= 22×144× 3.5/21 = 528 cubic metres

the canvas i.e required to cover is nothing but its curved surface area [ CSA]

so canvas required to cover the heap = CSA of cone

= π × radius(r) × slant height (l)

l^2 = r^2 + h^2

l^2 = 144 + 12.25

l^2 = 156.25 => l = 12.5m

so CSA = 22×12×12.5/7 = 471.4 sq.m

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(2) given dimensions of cylinder and cone are equal

r = 6cm , h = 8cm

volume of the reamining solid = volume of cylinder - volume of cone

[ volume of cylinder = πr^2h, volume of cone = πr^2h/3 ]

so volume of remaining solid

= πr^2h - πr^2h/3

= πr^2h[ 1 - 1/3 ]

= 3.14×36×8[2/3]

= 904.32×2/3 = 602.88 cubic.cm

l^2 = r^2 + h^2

=> l = √[36+64] = √[100] = 10cm

in the same way TSA of remaining solid = TSA of cylinder -

TSA of cone

= 2πrh + 2πr^2 - ( πrl + πr^2)

= πr [ 2h + 2r - l - r ]

= 3.14 *6 [ 16 + 12 - 10 - 6]

= 18.84 [ 12] = 226.08 sq.cm

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