Question

The curved surface area of a right circular cone with height 24 m and radius 7 m is?
(a) $500\text{ }{{\text{m}}^{2}}$
(b) $550\text{ }{{\text{m}}^{2}}$
(c) $\text{607 }{{\text{m}}^{2}}$
(d) $650\text{ }{{\text{m}}^{2}}$

Answer 1

Hint: Calculate the slant height of the right circular cone by using Pythagoras theorem. Then apply the formula for the curved surface area of the cone $=\pi rl$, where $l=$ slant height, $r=$ radius of the cone.

Complete step-by-step answer:

In a right circular cone height is perpendicular to the radius of the cone. Therefore, applying Pythagoras theorem in the right angle triangle formed by height, radius, and slant height, we get, $l=\sqrt{{{r}^{2}}+{{h}^{2}}}$. Here, $l=$ slant height, $r=$ radius of the cone and $h=$ height of cone.

We have been given: \[r=7\text{ m and }h=24\text{ m}\]. Substituting the value of $r\text{ and }h$ in the Pythagorean relation, we get,
\[\begin{align}
  & l=\sqrt{{{7}^{2}}+{{24}^{2}}} \\
 & =\sqrt{49+576} \\
 & =\sqrt{625} \\
 & =25 \\
\end{align}\]
Therefore, slant height of the cone $=l=25\text{ m}$.
Now, we know that curved surface area of cone $=\pi rl$. Therefore, substituting the value of\[\pi ,r\ \text{and }l\], we get,
$\begin{align}
  & \text{C}\text{.S}\text{.A of cone}=\dfrac{22}{7}\times 7\times 25 \\
 & =22\times 25 \\
 & =550\text{ }{{\text{m}}^{2}} \\
\end{align}$
Therefore the curved surface area of the cone is $550\text{ }{{\text{m}}^{2}}$.
Hence, option (b) is the correct answer.

Note: Here, one may note that we have used the value of $\pi =\dfrac{22}{7}$. Since, nothing is provided in the question about the value of $\pi $, so we have assumed it $\dfrac{22}{7}$ to make the calculation easy. Also, we have used Pythagoras theorem in the right circular cone. Here, as you can see in the figure that slant height is the hypotenuse of the right angle triangle.