Question

a heap of wheat is in the form of a cone whose diameter is 8m and height is 3m .Find it's volume .The heap is to convered by canvas to protect it from rain.find the area of canvas required​

Answer 1

Answer:

Step-by-step explanation:

Dimensions of the conical shaped heap

Diameter = d = 10.5 m

Radius = r = d / 2

r = 10.5 / 2 = 5.25 m

i ) volume of the heap = volume of the cone

V = ( pi × r ^2 × h ) /3

= (22 / 7 ) × ( 5.25 )^2 × 3 / 3

= 86.625 cubic cm

ii ) let the slant height of the cone = l cm

l^2 = r^2 + h^2

= 3^2 + ( 5.25 )^2

= 9 + 27.5625

= 36.5625

Therefore ,

l = 6.046 ( approx)

The area of the canvas required to from rain

= curved surface area of the cone

= pi × r × l

= (22/ 7 ) × 5.25 × 6.046

= 697.62 / 7

= 99.66 cm^2

Answer 2

Volume of the heap is 50.29 m^3 and the area of the canvas required is 62.86 m^2.

Given :

Diameter of the heap = 8m

Height of the heap = 3m

To find :

Volume of the heap and the area of canvas required

Solution :

Radius = Diameter/2

= 8/2

= 4m

Volume of the heap = volume of the cone

We know,

Volume of heap = π r^2(h/3)

= 22/7 × 4^2 (3/3)

= 22/7 × 16 × 1

= 50.29 m^3

Let the slant height be l m

We know,

l^2 = r^2 + h^2

=> l^2 = 4^2 + 3^2

=> l^2 = 16 + 9

=> l^2 = 25

=> l = √25

=> l = 5 m

We also know,

Curved surface area = πrl

= 22/7 × 4 × 5

= 62.86 m^2

Hence, Volume of the heap is 50.29 m^3 and the area of the canvas required is 62.86 m^2.

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