MAT Exam 2020 Paper

Concept:The curved surface area of a cone is $\pi r l$, where $r$ is the base radius and $l$ is the slant height. The base area is $\pi r^2$.Explanation:Let the first cone have curved surface area $S_1$, radius $r_1$, and slant height $l_1$.Let the second cone have curved surface area $S_2$, radius $r_2$, and slant height $l_2$.Given: $S_1 = 3S_2$ and $l_2 = 3l_1$.Using $S = \pi r l$, we get:$$\pi r_1 l_1 = 3(\pi r_2 l_2).$$Cancel $\pi$ from both sides:$$r_1 l_1 = 3 r_2 l_2.$$Substitute $l_2 = 3l_1$:$$r_1 l_1 = 3r_2(3l_1) = 9r_2 l_1.$$Cancel $l_1$ from both sides:$$r_1 = 9r_2.$$Now, the ratio of their base areas is:$$\frac{\pi r_1^2}{\pi r_2^2} = \left(\frac{r_1}{r_2}\right)^2 = \left(\frac{9r_2}{r_2}\right)^2 = 9^2 = 81.$$Thus, the required ratio is $81:1$.Answer:$81:1$, which is option B.