Subject: General Aptitude · Chapter: Volume and Surface Area · Exam: · Marks: · Difficulty:
A cone of height 10 cm and radius 5 cm is cut into two parts at half its height. The cut is given parallel to its circular base. What is the ratio of the curved surface area of the original cone and the curved surface area of the frustum?
(a)3 : 1
(b)3 : 2
(c)4 : 1
(d)4 : 3
Answer
Answer (as printed): D
Explanation
We have, $\triangle A O B \sim \triangle C O D$. $$\therefore \frac{A B}{C D}=\frac{O A}{O C} \Rightarrow \frac{5}{C D}=\frac{10}{5} \Rightarrow C D=\frac{5}{2} \mathrm{cm} .$$ Curved surface area of the cone $$\begin{gathered} =\left[\pi \times 5 \times \sqrt{5^{2}+(10)^{2}}\right] \mathrm{cm}^{2} \\ =25 \sqrt{5} \pi \mathrm{cm}^{2} . \end{gathered}$$ Curved surface area of the frustum $$\begin{aligned} & =\pi\left(5+\frac{5}{2}\right) \sqrt{\left(5-\frac{5}{2}\right)^{2}+5^{2}} \\ & =\left(\pi \times \frac{15}{2} \sqrt{\frac{25}{4}+25}\right) \mathrm{cm}^{2}=\left(\pi \times \frac{15}{2} \times \frac{1}{2} \times 5 \sqrt{5}\right) \mathrm{cm}^{2} \\ & =\frac{75 \sqrt{5}}{4} \pi \mathrm{cm}^{2} . \end{aligned}$$ Hence, required ratio $=25 \sqrt{5} \pi: \frac{75 \sqrt{5}}{4} \pi=4: 3$.