# Ex.13.4 Q2 Surface Areas and Volumes Solution - NCERT Maths Class 10

## Question

The slant height of a frustum of a cone is \(4\,\rm{cm}\) and the perimeters (circumference) of its circular ends are \(18 \,\rm{cm}\) and \(6\, \rm{cm.}\) Find the curved surface area of the frustum.

## Text Solution

**What is ****Known?**

Slant height of a frustum of a cone is \(4\,\rm{cm}\) and the circumference of its circular ends are \(18\, \rm{cm}\) and \(6 \,\rm{cm.}\)

**What is unknown?**

The curved surface area of the frustum.

**Reasoning:**

Draw a figure to visualize the shape better

Using the circumferences of the circular ends of the frustum to find the radii of its circular ends

Circumference of the circle \( = 2\pi r\)

where *\(r\)* is the radius of the circle.

We will find the CSA of the frustum by using formula;

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.

**Steps:**

Slant height of frustum of a cone, \(l = 4 \rm cm\)

Circumference of the larger circular end, \({C_1} = 18 \rm cm\)

Radius of the larger circular end,\(\begin{align} {r_1} = \frac{{{C_1}}}{{2\pi }} = \frac{{18 \rm cm}}{{2\pi }} = \frac{9}{\pi } \rm cm\end{align} \)

Circumference of the smaller circular end, \({C_2} = 6 \rm cm\)

Radius of the smaller circular end,\(\begin{align} {r_2} = \frac{{{C_2}}}{{2\pi }} = \frac{{6 \rm cm}}{{2\pi }} = \frac{3}{\pi } \rm cm\end{align} \)

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

\[\begin{align}&= \pi \left( {\frac{9}{\pi } \rm cm + \frac{3}{\pi } \rm cm} \right) \times 4 \rm cm\\&= \pi \times \frac{{12}}{\pi } \rm cm \times 4cm\\&= 48 \rm c{m^2}\end{align}\]

The curved surface area of the frustum is \(48 \;\rm cm^2.\)