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.
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 = $\dfrac{\pi 2L}{\pi9L} = \dfrac 29$