Question

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,
​

Answer 1

Given :

• Diameter of a cone = 10.5m

• Height of a cone = 3m

To Find :

• Volume of a cone

• Slant height of a cone

• Area of the canvas required

Solution :

As we know that,

$\sf{Radius = \dfrac{Diameter}{2} }$

$\sf{Radius = \dfrac{10.5}{2} }$

$\sf{Radius = 5.25m}$

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Now,

⠀$\boxed{\sf{ Volume \: o f \: a \: cone (heap) = \dfrac{1}{3}\pi {r}^{2}h }}$

Where,

• r = Radius of a cone ( 5.25m )

• h = Height of a cone ( 3m )

By substituting values according to the formula we get,

$\sf{⟹ \: \dfrac{1}{3} \times \dfrac{22}{7} \times (5.25)^{2} \times 3 }$

$\sf{⟹ \: \dfrac{1}{\cancel{3}} \times \dfrac{22}{7} \times 5.25 \times 5 . 25 \times\cancel{ 3 }}$

$\sf{⟹ \dfrac{22}{7} \times 27.5625}$

$\sf{⟹ \dfrac{606.375}{7} }$

$\sf\pink{⟹ 86.625 {m}^{3} }$

Hence, The Volume of a cone is 86.625m³

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Now, we will find slant height by using Pythagoras theorem.

AC² = AB² + BC²

Now, AC = l (slant height) , AB = h (height of a cone) and BC = r (radius of a base)

⠀⠀⠀$\boxed{\sf{ {l}^{2} = {h}^{2} + {r}^{2} }}$

Where,

• h = Height of a cone

• r = Radius of a cone

$\sf{⟹ {l}^{2} = {3}^{2} + {5.25}^{2} }$

$\sf{⟹ {l}^{2} = 9 + 27.5625}$

$\sf{⟹ {l}^{2} = 36.5625 }$

$\sf{⟹ l = \sqrt{36.5625 } }$

$\sf\pink{⟹ l = 6.0467m }$

Hence, The Slant height of a cone is 6.0467m

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⠀$\boxed{\sf{ Curved \: surface \: area \: of \: a \: cone \: = \pi rl }}$

$\sf{⟹ \dfrac{22}{7} \times 5.25 \times 6.0467 }$

$\sf{⟹ \dfrac{115.5 \times 6.0467}{7} }$

$\sf{⟹ \dfrac{698.39}{7} }$

$\sf\pink{⟹ 99.77 {m}^{2}(approx) }$

Hence, The Area of a canvas required is 99.77m²(approx)

Answer 2

Given :-

Diameter = 10.5 m

Radius = 10.5/2 = 5.25 m

Height = 3 m

To Find :-

Heap is to be covered by canvas to protect it from rain. Find the area of the

canvas required.

Solution :-

Volume of a cone = $\dfrac{ 1}{3} π r ² h$

At first we need to find slant height

l² = r² + h²

l² = 5.25² + 3²

l² = 27.5625 + 9

l² = 36.5625

l = 6.046 = 6 m

Now,

Volume = $\dfrac{ 1}{3 }\:× \:\dfrac{22}{7}\: × \:5.25\: × \:5.25 \: ×\: 3$

Volume =$\dfrac{ 22}{7}\: × \:27.5625$

Volume = 86.6 m

Now,

Area of canvas = πrl

Area = 3.14 × 5.25 × 6

Area = 3.14 × 31.5

Area = 98.91 m

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