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 $a.)$
 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.
A: 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.  
A: 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 = $\dfrac{\pi 2L}{\pi9L} = \dfrac 29$ 
