Question

A table lamp is in the form of a frustum of a cone whose area of the top and the bottom circle bases are $212\,{{\rm{m}}^2}$ and $103\,{{\rm{m}}^2}$. Find the total surface area, if the slant height is $10\,{\rm{m}}$ and top and bottom radii are $2\,{\rm{m}}$ and $4\,{\rm{m}}$.(Use $\pi = 3$)
A) $595\,{{\rm{m}}^{\rm{2}}}$
B) $550\,{{\rm{m}}^{\rm{2}}}$
C) $510\,{{\rm{m}}^{\rm{2}}}$
D) $505\,{{\rm{m}}^{\rm{2}}}$

Answer 1

Hint: In the given problem, first we need to find the total surface area of the frustum. The smaller value of area is the area of the top circle and the bigger value of area is the area of the bottom circle. We need to use the formula $\pi \left( {r + R} \right) \times l + \pi {r^2} + \pi {R^2}$, here $r,R$ represent the radius of the circle, and $l$ is the length. After substituting the respective values in the formula, we can calculate the total surface area.

Complete step by step answer:
In this given problem, let us take a smaller radius as ‘$r$’; bigger radius as ‘$R$’; slant height as ‘$l$’.
Given,
Area of the smaller base $\left( {\pi {r^2}} \right) = 103\,{{\rm{m}}^{\rm{2}}}$
Area of the bigger base $\left( {\pi {R^2}} \right) = 212\,{{\rm{m}}^{\rm{2}}}$
Slant height, $l = 10\,{\rm{m}}$
Smaller radius, $r = 2\,{\rm{m}}$
Bigger radius, $R = 4\,{\rm{m}}$

The total surface area (TSA) of the frustum is:
$\begin{array}{c} = \pi \left( {r + R} \right) \times l + \pi {r^2} + \pi {R^2}\\ = 3\left( {2 + 4} \right) \times 10 + 103 + 212\\ = 3 \times 6 \times 10 + 315\\ = 180 + 315\\ = 495\,{{\rm{m}}^{\rm{2}}}\end{array}$

Hence, the total surface area of the frustum is $495\,{{\rm{m}}^{\rm{2}}}$.

Note: A frustum (plural: frusta or frustums) is in geometry the portion of a solid (normally a cone or pyramid) that lies between one or two parallel planes that cut it. A right frustum is a parallel truncation of the right cone or pyramid. Here we have to find the total surface area of the frustum for the given data. Since the radius and the length are given, so by substituting the value in the formula, we get our required surface area.