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

## 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.

## 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 \(\begin{align} = 550 \times 10 = 5500\rm\,\,{m^2} \end{align}\)

**Answer:**

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