Question

A right triangle with sides \(3\ \text{cm}\), \(4\ \text{cm}\) and \(5\ \text{cm}\) is rotated about the side of \(3\ \text{cm}\) to form a cone. The volume of the cone so formed is:

• A. \(12\pi\ \text{cm}^3\)
• B. \(15\pi\ \text{cm}^3\)
• C. \(16\pi\ \text{cm}^3\)
• D. \(20\pi\ \text{cm}^3\)

Hint:

In triangle-rotation problems, first fix the axis of rotation — that side becomes the cone’s height.
The other leg becomes the radius, and the hypotenuse becomes the slant height.
Verify with \(\ell^2=r^2+h^2\) before plugging into \(V=\frac{1{3\pi r^2 h\).

Answer 1

The Correct Option is C

Step 1 (Confirm the right triangle and its parts).
Given sides: \(3,4,5\ \text{cm}\).
Check by Pythagoras: \(3^2 + 4^2 = 9 + 16 = 25 = 5^2\) the right angle is between the \(3\ \text{cm}\) and \(4\ \text{cm}\) sides.
Hence, legs (perpendicular sides) are \(3\) and \(4\); hypotenuse is \(5\).
Step 2 (Understand the rotation and map triangle parts to cone parts).
Axis of rotation \(=\) the side about which the triangle is rotated.
Triangle is rotated about the \(3\ \text{cm}\) side this side sweeps out the central axis of the cone \(\boxed{h=3\ \text{cm}}\) (height).
The side perpendicular to this axis at the right angle is \(4\ \text{cm}\); when rotated, its endpoint traces a circle of radius \(4\ \text{cm}\) \(\boxed{r=4\ \text{cm}}\).
The hypotenuse \(5\ \text{cm}\) rotates to form the lateral generator (slant height) \(\boxed{\ell=5\ \text{cm}}\) (not directly used in volume, but good for a check).
Step 3 (Consistency check using the right-cone relation).
For a right circular cone formed from a right triangle: \(\ell^2 = r^2 + h^2\).
Here \(r^2 + h^2 = 4^2 + 3^2 = 16 + 9 = 25 = 5^2 = \ell^2\) dimensions are consistent.
Step 4 (Write the volume formula and substitute).
Volume of a cone: \(V = \dfrac{1}{3}\pi r^2 h\).
Substitute \(r=4,\ h=3\):
\[ V = \frac{1}{3}\pi \times (d)^2 \times 3 = \frac{1}{3}\pi \times 16 \times 3 = 16\pi\ \text{cm}^3. \]
Step 5 (Unit and option check).
All dimensions are in cm, so volume is in \(\text{cm}^3\).
\(\boxed{16\pi\ \text{cm}^3}\) matches Option 3.
Step 6 (Why other options are plausible distractors).
If one mistakenly takes the axis as \(4\ \text{cm}\) and radius \(3\ \text{cm}\), the result would be \(V=\frac{1}{3}\pi\cdot 3^2\cdot 4=12\pi\ \text{cm}^3\) (Option (a) — this corresponds to rotating about the 4 cm side}, not the 3 cm side.
Options \(15\pi\) and \(20\pi\) arise from arithmetic slips such as using \(r\ell\) or mixing units.

\[ \boxed{16\pi\ \text{cm}^3 \ \text{(Option (c)}} \]