Question

The entire surface of a solid cone of base radius 3 cm and height 4 cm is equal the entire surface of a solid right circular cylinder of diameter 4 cm. Find the ratio of their curved surfaces.​

Answer 1

The ratio of the curved surfaces of the cone and cylinder is 15/16

Step-by-step explanation:

Given: Radius of the cone base ($R_{cone}$) = 3cm

Height of the cone ($H_{cone}$) = 4cm

Diameter of the cylinder $D_{cyl}$ = 4cm

Total Surface Area of the cone = Total Surface Area of the cylinder

To Find: The ratio of the curved surfaces of the cone and cylinder

Solution:

• Finding ratio of the curved surfaces of the cone and cylinder

T.S.A. of the cone is

$T.S.A._{cone} = \pi R_{cone} (R_{cone} + \sqrt{R_{cone}^{2} + H_{cone}^{2} } )$

$\Rightarrow T.S.A._{cone} = \pi \times 3 \times (3 + \sqrt{3^{2} + 4^{2} } )$

$\Rightarrow T.S.A._{cone} = 24\pi \ cm^{2}$

Given that $T.S.A._{cone} = T.S.A._{cylinder}$, therefore,

$T.S.A._{cylinder} = 2 \pi R_{cyl} (R_{cyl} + H_{cyl} ) = 24 \pi \ cm^{2}$

$\Rightarrow T.S.A._{cylinder} = 2 \pi \times 2 (2 + H_{cyl} ) = 24 \pi \ cm^{2}$

$\Rightarrow H_{cyl} = 4 \ cm$

Now, for the ratio of the curved surface area,

$C.S.A._{cone} = \pi R_{cone} \sqrt{ R_{cone}^{2} + H_{cone}^{2} }$

$\Rightarrow C.S.A._{cone} = \pi \times 3 \times \sqrt{ 3^{2} + 4^{2}} = 15 \pi \ cm^2$

And,  $C.S.A._{cyl} = 2 \pi R_{cyl} H_{cyl}$

$\Rightarrow C.S.A._{cyl} = 2 \times \pi \times 2 \times 4 = 16 \pi \cm^{2}$

Therefore, the ratio is

$\frac{C.S.A._{cone}}{C.S.A._{cyl}} = \frac{15 \pi}{16 \pi} = \frac{15}{16}$

Hence, the ratio of the curved surfaces of the cone and cylinder is 15/16