# Ex.13.3 Q7 Surface Areas and Volumes Solution - NCERT Maths Class 9

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## Question

A joker’s cap is in the form of a right circular cone of base radius $$7\rm\, cm$$ and height $$24\,\rm cm.$$ Find the area of the sheet required to make $$10$$ such caps.

Video Solution
Surface-Areas-And-Volumes
Ex exercise-13-3 | Question 7

## Text Solution

Reasoning:

Curved surface area of a right circular cone of base radius $$r$$ and slant height $$l$$ is $$= \pi rl$$ and $$l = \sqrt {{r^2} + {h^2}}$$

What is known?

(i) Caps box radius and height.

(ii) Number of caps.

What is unknown?

Area of the sheet required to $$10$$ caps.

Steps:

Base Radius $$(2r) = 7\rm\,cm$$

Height $$h = 24\,\rm cm$$

Slant height \begin{align}l = \sqrt {{r^2} + {h^2}} \end{align}

\begin{align} &= \sqrt {{7^2} + 24} \\& = \sqrt {49 + 576} \\ &= \sqrt {625} \\ &= 25\rm\,\,cm \end{align}

Curved surface area of cone $$=$$ Surface area of the sheet required for $$1$$ cap\begin{align}\,\pi rl. \end{align}

\begin{align} &= \frac{{22}}{7} \times 7 \times 25\\ &= 550\rm\,\,{m^2} \end{align}

Curved surface area for $$10$$ caps \begin{align} = 550 \times 10 = 5500\rm\,\,{m^2} \end{align}

The area of the sheet required to make $$10$$ such caps is \begin{align}5500\rm\,\,{m^2}. \end{align}

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