# Ex.13.4 Q3 Surface Areas and Volumes Solution - NCERT Maths Class 10

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## Question

A fez, the cap used by the Turks, is shaped like the frustum of a cone (see Fig. 13.24). If its radius on the open side is $$10\,\rm{cm}$$, radius at the upper base is $$4\,\rm{cm}$$ and its slant height is $$15\,\rm{cm, }$$ find the area of material used for making it.

## Text Solution

What is Known?

A fez is shaped like the frustum of a cone with radius of open side $$10\,\rm{cm},$$

radius at the upper base $$4\,\rm{cm}$$ and Slant Height $$15 \,\rm{cm}$$

What is Unknown?

Area of the material used for making Fez

Reasoning: Since the fez is in the shape of frustum of a cone and is open at the bottom.

Therefore,

Area of material used for making fez $$=$$ Curved Surface Area of the frustum $$+$$ Area of the upper circular end

We will find the Area of material by using formulae;

CSA of frustum of a cone $$= \pi \left( {{r_1} + {r_2}} \right)l$$

where $$r1, r2$$ and $$l$$ are the radii and slant height of the frustum of the cone respectively.

Area of the circle $$= \pi {r^2}$$

where $$r$$ is the radius of the circle.

Steps:

Slant height, $$l =15\,\rm{cm}$$

Radius of open side $$r_1=10\,\rm{cm}$$

Radius of upper base $${r_2} =4\,\rm{cm}$$

Area of material used for making fez $$=$$ Curved Surface area of the frustum $$+$$ area of the upper circular end

\begin{align}&= \pi \left( {{r_1} + {r_2}} \right)l + \pi {r^2}\\&= \pi \left[ {\left( {{r_1} + {r_2}} \right)l + r_2^2} \right]\end{align}

\begin{align}&= \frac{{22}}{7}\left[ {\left( {10cm + 4cm} \right)15cm + {{\left( {4cm} \right)}^2}} \right]\\&= \frac{{22}}{7}\left[ {14cm \times 15cm + 16c{m^2}} \right]\\&= \frac{{22}}{7}\left[ {210c{m^2} + 16c{m^2}} \right]\\&= \frac{{22}}{7} \times 226c{m^2}\\&= \frac{{4972}}{7}c{m^2}\\&= 710\frac{2}{7}c{m^2}\end{align}

\begin{align} 710\frac{2}{7}c{m^2}\end{align} of the material used for making Fez.

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