# Find the ratio of curved surface area of frustum to the cone.

In the figure, there is a cone which is being cut and extracted in three segments having heights $h_1,h_2$ and $h_3$ and the radius of their bases $1$ cm, $2$cm and $3cm$, then The ratio of the curved surface area of the second largest segment to that of the full cone.

$\color{green}{a.)2:9}\\ b.)4:9\\ c.)\text{cannot be determined }\\ d.) \text{none of these}\\$ I found that $h_1=h_2=h_3\\$

and

$\dfrac{A_{\text{2nd segment}}}{A_{\text{full cone}}}=\dfrac{\pi\times (1+2)\times \sqrt{h_1^2+1} }{\pi\times 3\times \sqrt{(3h_1)^2+3^2} }=\dfrac13$

But book is giving option $a.)$

Area is proportional to the square of linear dimension. So the area of the full cone is $k(3^2)$ for some $k$. The area of the second largest segment is $k(2^2) - k(1^2)$, so the ratio is $3:9 = 1:3$. You are right, and the book is wrong.

Let the apex angle of full cone be $2\alpha$. Then from the corresponding right triangles, we have $$h_1+h_2+h_3=3\cot\alpha$$ $$h_2=(2-1)\cot\alpha=\cot\alpha$$ The required ratio of the curved surface areas is $$=\frac{\pi(1+2)\sqrt{(\cot\alpha)^2+(2-1)^2}}{\pi(3)\sqrt{(3\cot\alpha)^2+(3)^2}}=\frac{3\pi\sqrt{1+\cot^2\alpha}}{9\pi\sqrt{1+\cot^2\alpha}}=\frac{1}{3}$$ The ratio obtained is obviously same as you obtained. There is certainly some printing mistake in the options provided in your book. I can't blame the whole book but still some errors/mistakes are common in the books depending on their authors\editors.

The lateral heights for all the segments will be the same and so will the heights.

The curved surface area of a cone $= \pi$(radius)(lateral height)

For whole cone = $\pi(3)(3L)$ (where $L$ is the lateral height of each segment)

For 2nd largest segment = $\pi(2)(L)$ (since segment $2$ has lateral height of $L$)

So the ratio will give = $\dfrac{\pi 2L}{\pi9L} = \dfrac 29$