A right circular cone and a right circular cylinder have equal base and equal height. If radius of the base and the height, are in the ratio 5:12, then the ratio of the total surface area of the cylinder to that of the cone is:
(a) 3:1
(b) 13:9
(c) 17:9
(d) 34:9

Hint: As it is given that the ratio of radius to height is 5:12, using this we can say that radius is 5k and height is 12k. We know that the total surface area of the cylinder is equal to $\left( 2\pi rh+2\pi {{r}^{2}} \right)$ and the total surface area of the cone is $\left( \pi {{r}^{2}}+\pi rl \right)$ , where l is equal to $\sqrt{{{r}^{2}}+{{h}^{2}}}$ . Now substitute the values in the formulas and find the ratio to reach the answer.

Complete step by step solution:
We will start the solution to the above question by drawing the diagram of the situation given in the question.


As it is given that the ratio of radius to height is 5:12, using this we can say that radius is 5k and height is 12k.
Now, we know that the total surface area of the cylinder is equal to $\left( 2\pi rh+2\pi {{r}^{2}} \right)$ .
$\text{TSA of cylinder}=2\pi rh+2\pi {{r}^{2}}=2\pi \times 5k\times 12k+2\pi \times {{5}^{2}}{{k}^{2}}=120\pi {{k}^{2}}+50\pi {{k}^{2}}$
Also, we know that the total surface area of the cone is $\left( \pi {{r}^{2}}+\pi rl \right)$ , where l is equal to $\sqrt{{{r}^{2}}+{{h}^{2}}}$ .
$\text{TSA of cone}=\pi rl+\pi {{r}^{2}}=\pi r\sqrt{{{r}^{2}}+{{h}^{2}}}+\pi {{r}^{2}}=\pi \times 5k\sqrt{25{{k}^{2}}+144{{k}^{2}}}+25\pi {{k}^{2}}$
Now we will find the ratio as asked in the question.
$\dfrac{\text{TSA of cylinder}}{\text{TSA of cone}}=\dfrac{120\pi {{k}^{2}}+50\pi {{k}^{2}}}{\pi \times 5k\sqrt{25{{k}^{2}}+144{{k}^{2}}}+25\pi {{k}^{2}}}=\dfrac{170\pi {{k}^{2}}}{65\pi {{k}^{2}}+25{{k}^{2}}}=\dfrac{170\pi {{k}^{2}}}{90\pi {{k}^{2}}}=\dfrac{17}{9}$
Therefore, the ratio is 17:9. Hence, the answer to the above question is option (c).

Note: Remember that radius and height have the ratio a:b implies r=ak and h=bk, where k is integer and not r=a and h=b, because there is a possibility that when the ratio between them is taken some factors might have been cancelled from numerator and the denominator which is compensated by multiplying k with each. However, in the questions like above, you can solve by taking r=a and h=b as well, because at the end k is cancelled.