Question

A solid right circular cone of height 27 cm is cut into two pieces along a plane parallel to its base at a height of 18 cm from the base. If the difference in volume of the two pieces is 225 cc, the volume, in cc, of the original cone is

• A. 232
• B. 256
• C. 264
• D. 243

Answer 1

The Correct Option is D

As the cone is cut at one-third height from the top (the vertex), the total volume is proportional to the cubes of the heights of the two parts. 

Ratio of volumes two parts = \(\bigg(\frac{1}{3}\bigg)^3:13 = 1:27\)

Hence the bottom part will have volume of \(27 - 1\)i.e., \(26\) parts.

Given \((26 -1)\) i.e., \(25\) parts \(-225\) cc.

Hence the required answer is \(27\) parts = \(\frac{27×225}{25} = 243\) cc.

So, the correct answer is (D): \(243\)

Answer 2

Let the radius of base be \(3r\).
Height of the upper cone is \(9\), so by symmetry radius of the upper cone will be \(r\).
Frustum's volume = \(\frac \pi3(9r^2.27-r^2.9)\)
Volume of the upper cone = \(\frac \pi3.r^2.9\)
Given, the difference in volume of the two pieces is \(225\) cc.
\(\frac \pi3.9.r^2.25 = 225\)
\(r^2=\frac 3\pi\)
Volume of the larger cone,
= \(\frac \pi3.9r^2.27\)

= \(\frac \pi3.9. \frac 3\pi .27\)

\(= 243\)

So, the correct option is (D): 243.