A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.

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Solution

Radius (r) of heap

Height (h) of heap = 3 m

Volume of heap

Therefore, the volume of the heap of wheat is 86.625 m³.

Area of canvas required = CSA of cone

Therefore, 99.825 m² canvas will be required to protect the heap from rain.

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A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume.
The heap is to be covered by canvas to protect it from the rain. Find the area of the canvas required.
$[A s s u m e π = \frac{22}{7}]$

Q. Question 9
A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume.
The heap is to be covered by canvas to protect it from the rain. Find the area of the canvas required.

[A s s u m e π = \frac{22}{7}]

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and height is

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Q. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3m. And its volume . The heap is to be covered by canvas to protect it from rain . Find the area of the canvas required.

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A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.

Radius (r) of heap Height (h) of heap = 3 m Volume of heap Therefore, the volume of the heap of wheat is 86.625 m3. Area of canvas required = CSA of cone Therefore, 99.825 m2 canvas will be required to protect the heap from rain.

Radius (r) of heap

Height (h) of heap = 3 m

Volume of heap

Therefore, the volume of the heap of wheat is 86.625 m³.

Area of canvas required = CSA of cone

Therefore, 99.825 m² canvas will be required to protect the heap from rain.