If the triangle is revolved about its any side such that the volume of solid so obtained is $128\pi$, find the radius of the circle.

Question

A triangle with maximum area is cut from a circle. If the triangle is revolved about its any side such that the volume of solid so obtained is $128\pi$ cm$^3$, then find the radius of the circle.

My Attempt:

I think a cone will be obtained.

At first I thought one side of the triangle will become height of the cone and another side will become radius but I think this is not possible as the angle between two sides is not $90$.

Triangle with maximum area will be equilateral, I think.

Let radius of the circle be $r$. And side of the triangle be $a$.

Therefore, $\frac{\sqrt3 a}2=\frac{3r}2\implies a=\sqrt3 r$

Volume of cone is $\frac13\pi R^2H$, here, $R=a=\sqrt3 r$. By taking $H$ as $\frac32r$, I am getting $r^3=128$

Answer given is $\frac{8\sqrt3}3$ cm.

Answer 1

First, you state that the triangle is equilateral, but without proof. So, I would like to quote Arthur's answer from here:

Pick one side of the triangle as base (rotate the circle so that the base is horizontal for clearer view). Then the largest area you can get is by putting the point opposite the base as far away from from the base as possible. This happens to be halfway along the longest arc between the end points of the base. In other words, the opposite point must be exactly in the middle of the two end points of the base for the area to be maximal.

Now, if the triangle is not equilateral, there is at least one point which is not in the middle of the two others, so we can increase the area of the triangle. An equilateral triangle is the only one which cannot be increased this way, so it must be maximal.

Okay, getting to the point now. Call the side of the equilateral triangle $a$, and the radius $r$. The circle is the circumcircle of the equilateral triangle, so the two are related as: $$r = {a \over \sqrt 3} \space\space\space\space\space\space ... \space\space\space (*)$$(which can be proven in many ways). Our equilateral triangle will look like this:
Since all sides are equal, we can rotate it about any side(say $AB$). Before that, we can split the triangle by constructing the altitude (which is also the median) $CD$.

Now, we can rotate the $\triangle ADC$ about $AD$ and $\triangle BDC$ about $BD$. Note that this operation is the same as rotating $\triangle ABC$ about $AB$. Since $CD \perp AB$, after rotating, we will get 2 equal right circular cones, joined base to base (the lower one is inverted). They will look like this (commonly called a bicone):

The radius of both the cones will be $CD = \sqrt3a/2 $, and the height will be $AD = BD = a/2$. Both the cones have equal volume, so the total volume will be: $$2 \times \frac13 \pi \left(\frac{\sqrt3a}{2}\right)^2\frac a2$$Simplifying: $$\frac \pi4 a^3$$Given, $$\frac \pi4 a^3 = 128 \pi$$ So, $$a = 8$$Thus, from $(*)$, we have: $$\color{green}{r = \frac{a}{\sqrt3} = \frac{8}{\sqrt3} = \frac{8\sqrt3}{3}}$$