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 rain. Find the
area of the canvas required.​

Answer 1

Answer:

• Volume of the heap is 86.625 m³
• The area of the canvas required is 99.73 cm²

Step-by-step explanation:

Given :

• Diameter of the heap is 10.5 m
• Height of the heap is 3 m

To Find :

• Volume of the heap.
• Area of the canvas required.

Calculation :

Firstly let's find the value of radius. To calculate the radius, we will divide diameter by 2.

• Radius = Diameter/2

$\longrightarrow$ Radius = $\rm \dfrac{10.5}{2}$ m

$\longrightarrow$ Radius = 5.25 m.

Henceforth, radius is 5.25 m.

⠀⠀⠀________________

~ Calculating the volume of heap,

To calculate the volume of heap, we'll us this formula :-

• $\boxed{ \rm{ \gray{Volume (V) = \dfrac{1}{3} \times \pi \times \: r^2 \times h}}}$

$\longrightarrow$ Volume of heap = $\rm \dfrac{1}{3} \times \dfrac{22}{7} \times (5.25)^2 \times 3$

$\longrightarrow$ Volume of heap = $\bf = 86.625 \: m^3$

Therefore, volume of heap is 86.625 m³.

⠀⠀⠀_________________

~ Calculating the slant height of the canvas,

Let's assume the slant height of canvas as 'x'

$\longrightarrow$ Slant height = $\rm \sqrt{r^2 + h^2}$

Where,

• Radius is 5.25 m
• Height is 3 m

$\longrightarrow$ Slant height = $\rm \sqrt{5.25^2 + 3^2}$

$\longrightarrow$ Slant height = $\sqrt{(5.25 \times 5.25) + (3 + 3)}$

$\longrightarrow$ Slant height = $\sqrt{27.5625 + 9}$

$\longrightarrow$ Slant height = $\bf \: 6.05$

Hence, slant height of the canvas is 6.05 m.

⠀⠀________________

~ Calculating the area of canvas,

To calculate the area of canvas, we'll be using this formula

• $\boxed{ \rm{ \gray{Area = \pi \times r \times l}}}$

Putting the values,

$\longrightarrow$ $\rm{ \dfrac{22}{7} \times 5.25 \times 6.05}$

$\longrightarrow$ $\bf{99.73 \: cm^2}$

Therefore, area of the canvas required is 99.73 cm².

Answer 2

$\huge\bf{Answer :}$

• Volume of the heap = 86.625 m³ ✓
• Area of the canvas = 99.73 m² ✓

$\huge\bf{Concept :}$

• Volume refers to a particular space which is occupied by a shape or a substance. By finding the radius of the heap, we can calculate the required volume of the heap.
• For finding the area of canvas, we'll calculate the lateral or curved surface area of the heap in which base would not be included. For that, we'll also find the slant height.

Let's solve the problem.

Given :-

• Shape of the heap is conical.
• Diameter of the heap = 10.5 m
• Height ( h ) of the heap = 3 m

[ Note - All the calculations would be done in metres, and no conversion will take place. ]

To find :-

• Volume of the heap, and
• Area of canvas used up in covering the heap.

$\huge\bf{Explanation :}$

Here, we can observe that radius is not given but diameter is provided. So, we'll find out the radius from the relationship between radius and diameter.

We know that,

☆ Diameter = 2 × Radius

• Inserting the values.

10.5 = 2 × Radius

∴ Radius ( r ) = 10.5/2 = 5.25 m

Now, we'll find the volume since we've both radius and height.

As we know that,

Volume of the conical heap = 1/3 × π × r² × h

• Putting the values of radius and height and π.

⇒ 1/3 × 22/7 × 5.25 × 5.25 × 3

⇒ 86.625 m³ ✔️

Now, we're asked to find the area of canvas. So, we'll calculate the lateral surface area. But, for that we need to find out the slant height ( l ) of the heap.

We know that,

l = $\sqrt{ {radius}^{2} + {height}^{2} }$

⇒ $\sqrt{5.25 {}^{2} + 3 ^{2} }$

⇒ $\sqrt{27.5625 + 9}$

⇒ 6.05 m

Now, we've founded all the necessary dimensions and can find the lateral surface area.

Lateral surface area = π × r × l

⇒ 22/7 × 6.05 × 5.25

⇒ 99.73 m² ✔️