Question

Curved surface area of a cone is 251.2 cm² and
radius of its base is 8 cm. Find its slant height
and perpendicular height. ( te = 3.14) (3 marks)​

Answer 1

• The slant height of the cone = 10 cm.
• The perpendicular height of the cone = 6 cm.

Given :

• Curved surface area of a cone (C.S.A.) = 251.2 cm²
• The radius of the base of a cone = 8 cm.
• Value of π = 3.14

To Find :

• The slant height of the cone.
• The perpendicular height of the cone.

Solution :

1) The slant height of the cone.

Let,

The slant height be x.

Given,

• Curved surface area (C.S.A.) of the cone = 251.2 cm²

We know that,

• Curved surface area (C.S.A.) of the cone = πrl

That means,

• πrl = 251.2 cm²

Where,

• r = radius = 8 cm.
• l = slant height.
• π = 3.14.

$: \implies \rm 3.14 \times 8 \: cm \times l = 251.2 \: {cm}^{2} \\ \\ : \implies \rm 25.12 \: cm\times l = 251.2 \: {cm}^{2} \\ \\ : \implies \rm l = \frac{251.2 \: {cm}^{2} }{25.12 \: cm} \\ \\ : \implies \rm l = 10 \: cm \\ \\ \: \: \therefore \: \rm \: l = 10 \: cm$

Hence,

The slant height of the cone is 10 cm.

_______________________________

2) The perpendicular height of the cone.

Let,

The perpendicular height be x.

We know that,

• l² = r² + h²

Where,

• l = slant height = 10 cm.
• r = radius = 8 cm.

$: \implies \rm {l}^{2} = {r}^{2} + {h}^{2} \\ \\ : \implies \rm {(10 \: cm)}^{2} = {(8 \: cm)}^{2} + {(x)}^{2} \\ \\ : \implies \rm 100 \: {cm}^{2} = 64 \: {cm}^{2} + {x}^{2} \\ \\ : \implies \rm 100 \: {cm}^{2} - 64 \: {cm}^{2} = {x}^{2} \\ \\ : \implies \rm 36 \: {cm}^{2} = {x}^{2} \\ \\ : \implies \rm \sqrt{36 \: {cm}^{2} } = x \\ \\ : \implies \rm 6 \: cm = x \\ \\ \: \: \rm \therefore \: \: x = 6 \: cm$

Hence,

The perpendicular height of the cone is 6 cm.