Subject: General Aptitude · Chapter: Volume and Surface Area · Exam: 2006 · Marks: · Difficulty:
In a right circular cone, the radius of its base is 7 cm and its height is 24 cm. A cross-section is made through the mid-point of the height parallel to the base. The volume of the upper portion is
(a)$154 \mathrm{cm}^{3}$
(b)$169 \mathrm{cm}^{3}$
(c)$800 \mathrm{cm}^{3}$
(d)$1078 \mathrm{cm}^{3}$
Answer
Answer (as printed): A
Explanation
$r=7 \mathrm{cm}, h=24 \mathrm{cm}$. Now, $\triangle A O B \sim \triangle C O D$. $$\text { So, } \frac{O A}{O C}=\frac{A B}{C D} \Rightarrow \frac{h}{h / 2}=\frac{r}{C D} \Rightarrow C D=\frac{r}{2} \text {. }$$ ∴ Volume of upper portion $$\begin{aligned} & =\frac{1}{3} \pi\left(\frac{r}{2}\right)^{2}\left(\frac{h}{2}\right)=\left(\frac{1}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 12\right) \mathrm{cm}^{3} \\ & \quad=154 \mathrm{cm}^{3} . \end{aligned}$$