# Ex.13.1 Q3 Surface Areas and Volumes Solution - NCERT Maths Class 10

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## Question

A toy is in the form of a cone of radius $$3.5\; \rm{cm}$$ mounted on a hemisphere of same radius. The total height of the toy is $$15.5 \;\rm{cm}$$. Find the total surface area of the toy.

## Text Solution

What is known?

The toy is in the form of a cone of radius $$=3.5\, \rm{cm}$$ mounted on a hemisphere with the same radius. The total height of the toy is $$15.5 \,\rm{cm}$$.

What is unknown?

The total surface area of the toy.

Reasoning:

We can create the figure of the toy as per given information

From the figure it’s clear that total surface area of the toy includes CSA of the cone and hemisphere.
Total surface area of the toy $$=$$ CSA of the hemisphere $$+$$ CSA of the cone

We will find the total area of the toy by using formulae;
CSA of the hemisphere$$= 2\pi {r^2}$$
where $$r$$ is the radius of the hemisphere

CSA of the cone $$= \pi rl$$
where $$r$$ and $$l$$ are the radius and slant height of the cone respectively.

Slant height of the cone, $$l = \sqrt {{r^2} + {h^2}}$$

Height of the cone, $$h =$$ total height of the toy $$–$$ height of the hemisphere

Steps:

Radius of the hemisphere, $$r = 3.5cm$$
Height of the hemisphere = radius of the hemisphere,  $$r = 3.5cm$$

Radius of the cone,  $$r = 3.5cm$$

Height of the cone $$=$$ Total height of the toy $$–$$ height of the hemisphere
$h = 15.5cm - 3.5cm = 12cm$

Slant height of the cone,  $$l = \sqrt {{r^2} + {h^2}}$$

\begin{align} l &= \sqrt {{r^2} + {h^2}} \\l &= \sqrt {{{\left( {3.5{cm}} \right)}^2} + {{\left( {12{cm}} \right)}^2}} \\l &= \sqrt {12.25{c{m^2}} + 144{c{m^2}}} \\l &= \sqrt {156.25{c{m^2}}} \\l &= 12.5{cm}\end{align}

Total surface area of the toy $$=$$ CSA of the hemisphere $$+$$ CSA of the cone

\begin{align}&= 2\pi {r^2} + \pi rl\\&= \pi r\left( {2r + l} \right)\\&= \frac{{22}}{7} \times 3.5{cm} \times \left( {2 \times 3.5{cm} + 12.5{cm}} \right)\\&= \frac{{22}}{7} \times \frac{7}{2}{cm} \times \left( {7{cm} + 12.5{cm}} \right)\\&= 11{cm} \times 19.5{cm}\\&= 214.5{c{m^2}}\end{align}

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