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

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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, \(l = 4cm\)

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

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

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

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

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

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

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

  
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