Question

The radius of a sphere (in cm) whose volume is $36\pi \text{ cm}^3$, is :

• A. 3
• B. $3\sqrt{3}$
• C. $3^{\frac{2}{3}}$
• D. $3^{\frac{1}{3}}$

Hint:

When dealing with $\pi$ in solid geometry formulas, always cancel $\pi$ first before doing any cross-multiplication.
Recognizing perfect cubes like $27 = 3^3$, $64 = 4^3$, and $125 = 5^3$ helps to speed up calculations.

Answer 1

The Correct Option is A

Step 1: Cancel pi right away and isolate the cube.
From $V=\frac43\pi r^3=36\pi$, cancel $\pi$ from both sides: $\frac43 r^3=36$, so $r^3=27$.
Step 2: Spot the perfect cube.
27 is a familiar perfect cube: $3\times3\times3=27$, so $r=3$.
Step 3: Quickly check it fits.
With $r=3$, $\frac43\pi(3)^3=\frac43\pi(27)=36\pi$, which matches the given volume.
\[ \boxed{3 \text{ cm}} \]