Question

Find the area of an equilateral triangle whose height is \(12\) cm.

• A. \(24\sqrt{3}\) cm\(^2\)
• B. \(48\) cm\(^2\)
• C. \(48\sqrt{3}\) cm\(^2\)
• D. \(36\sqrt{3}\) cm\(^2\)

Hint:

First find the side using \(h=\frac{\sqrt{3}}{2}a\), then use \(A=\frac{\sqrt{3}}{4}a^2\).

Answer 1

The Correct Option is C

Step 1: Build a direct height to area shortcut.
Rather than finding the side first and then the area in two separate calculations, we can combine both formulas into one shortcut that works for any equilateral triangle once the height is known. Start from $h=\frac{\sqrt{3}}{2}a$, which gives $a=\frac{2h}{\sqrt{3}}$.

Step 2: Substitute this side into the area formula.
The area formula is $A=\frac{\sqrt{3}}{4}a^2$. Replacing $a$ with $\frac{2h}{\sqrt{3}}$:
\[ A = \frac{\sqrt{3}}{4}\left(\frac{2h}{\sqrt{3}}\right)^2 = \frac{\sqrt{3}}{4} \times \frac{4h^2}{3} \]
\[ A = \frac{\sqrt{3}h^2}{3} = \frac{h^2}{\sqrt{3}} \]
This gives a reusable rule: area equals height squared divided by $\sqrt{3}$.

Step 3: Plug in the given height.
Here $h=12$ cm, so $h^2=144$.
\[ A = \frac{144}{\sqrt{3}} = \frac{144\sqrt{3}}{3} = 48\sqrt{3}\ \text{cm}^2 \]

Step 4: Sanity check the size of the answer.
Using $\sqrt{3}\approx1.732$, this area is about $83.1$ cm$^2$, which is reasonable for a triangle with a $12$ cm height. Comparing with the four choices, only option (c) has both the right coefficient and the $\sqrt{3}$ factor together; the rest either lose the $\sqrt{3}$ or scale it wrong.

Final Answer:
The area comes out to $48\sqrt{3}$ cm$^2$, matching option (c). \[ \boxed{48\sqrt{3}\ \text{cm}^2} \]