Question

a tent is in the form of a cone mounted on a cylinder. The radius and height of the cylinder and 16m and 30m respectively. The height of the cone is 12m . Find the area of the canvas required.​

Answer 1

Given :-

• A tent in the form of a cone mounted on a cylinder.
• The radius, r and height, H of the cylinder are 16 m and 30 m respectively.
• The radius and height of the cone are 16 m and 12 m respectively.

To Find :-

• Area of the canvas required.

Solution :-

The canvas will be required to cover the cone and cylinder, which means we will have to find the curved surface of cone and cylinder and then add them to get the required canvas.

Curved Surface Area of Cone,

We have,

• Radius, r = 16 m
• Height, h = 12 m

So,

⇒ C.S.A of Cone = πrl

⇒ C.S.A of Cone = πr√(h² + r²)

⇒ C.S.A of Cone = π × 16 × √( 144 + 256 )

⇒ C.S.A of Cone = 16π√400

⇒ C.S.A of Cone = 320π m²

Similarly, Let us find the Curved Surface Area of cylinder,

We have,

• Radius, r = 16 m
• Height, H = 30 m

So,

⇒ C.S.A of Cylinder = 2πrH

⇒ C.S.A of Cylinder = 2×π×16×30

⇒ C.S.A of Cylinder = 960π m²

Further, The area of canvas required is,

⇒ CSA of Cone + CSA of Cylinder

⇒ 320π + 960π

⇒ 1280π m²

Hence, The area of canvas required

is 1280π m²

Answer 2

$\huge\bold{\mathbb{QUESTION}}$

A tent is in the form of a cone mounted on a cylinder. The radius and height of the cylinder are $\sf 16\:m$ and $\sf 30\:m$ respectively. The height of the cone is $\sf 12\:m .$ Find the area of the canvas required.

$\huge\bold{\mathbb{GIVEN}}$

A tent is in the form of a cone mounted on a cylinder.

For the cylinder,

• $\sf r$ (radius) $\sf = 16\:m$

• $\sf h$ (height) $\sf = 30\:m$

For the cone,

• $\sf r$ (radius) $\sf = 16\:m$

• $\sf h$ (height) $\sf= 12\:m$

$\huge\bold{\mathbb{TO\:FIND}}$

The area of the canvas required.

$\huge\bold{\mathbb{SOLUTION}}$

First, we have to find the Curved Surface Area (C.S.A) of the cylinder.

We know that:

$\:\:\:\,\,\:★[[\boxed{\sf{C.S.A\:of\:cylinder=2πrh\:unit²}}]]★$

Here,

• $\sf r = 16\:m$

• $\sf h = 30\:m$

Putting the values of $\sf r$ and $\sf h.$

C.S.A of cylinder

$\sf \longrightarrow 2πrh\:m²$

$\sf \longrightarrow (2 × π × 16 × 30)\:m²$

$\sf \longrightarrow 960π\:m²$

Now, we have to find the Curved Surface Area (C.S.A) of the cone.

We know that:

$\:\:\:\:\:\:\:\:\:★[[\boxed{\sf{C.S.A\:of\:cone=πrl\:unit²}}]]★$

Here,

• $\sf r = 16\:m$

• $\sf h = 12\:m$

Putting the values of $\sf r$ and $\sf h.$

C.S.A of the cone

$\sf \longrightarrow πrl\:m²$

$\sf \longrightarrow \{π × r × \sqrt{(h² + r²)}\}\:m²$

$\sf \longrightarrow \{π × 16 × \sqrt{(12² + 16²)}\}\:m²$

$\sf \longrightarrow \{π × 16 × \sqrt{(144 + 256)}\}\:m²$

$\sf \longrightarrow (π × 16 × \sqrt{400})\:m²$

$\sf \longrightarrow (π × 16 × 20)\:m²$

$\sf \longrightarrow 320π\:m²$

So,

• C.S.A of the cylinder $\sf = 960π\:m²$

• C.S.A of the cone $\sf = 320π\:m²$

Total area of the canvas

$=$ C.S.A of cylinder + C.S.A of cone

$\sf = (960π + 320π)\:m²$

$\sf = 1280π\:m²$

$\huge\bold{\mathbb{HENCE}}$

Total area of the canvas $\sf = 1280π\:m²$

$\huge\bold{\mathbb{THEREFORE}}$

The area of the canvas required is $\sf1280π\:m².$

$\huge\bold{\mathbb{DONE}}$