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: Recall how the height of an equilateral triangle relates to its side.
In an equilateral triangle, the altitude from any vertex bisects the opposite side and meets it at a right angle. This splits the triangle into two right triangles, each with base \(\frac{a}{2}\) and hypotenuse \(a\), where \(a\) is the side of the triangle. Using the Pythagorean theorem on one of these right triangles gives the standard result \(h = \frac{\sqrt{3}}{2}a\).

Step 2: Solve for the side length.
We are told the height is \(12\) cm, so
\[ 12 = \frac{\sqrt{3}}{2}a \]
Multiply both sides by \(2\):
\[ 24 = \sqrt{3}\,a \]
Divide both sides by \(\sqrt{3}\), and rationalise:
\[ a = \frac{24}{\sqrt{3}} = \frac{24\sqrt{3}}{3} = 8\sqrt{3}\ \text{cm} \]

Step 3: Use the area formula for an equilateral triangle.
The area in terms of the side is
\[ A = \frac{\sqrt{3}}{4}a^2 \]
This also comes from the same right triangle logic, since area equals half of base times height and both can be written using \(a\). Now put in \(a = 8\sqrt{3}\) cm:
\[ a^2 = (8\sqrt{3})^2 = 64 \times 3 = 192 \]
\[ A = \frac{\sqrt{3}}{4} \times 192 = 48\sqrt{3}\ \text{cm}^2 \]

Step 4: Check why the other options are wrong.
Option (a), \(24\sqrt{3}\), is half of the correct value, the kind of slip that happens if you square \(4\sqrt{3}\) instead of \(8\sqrt{3}\). Option (b), \(48\), drops the \(\sqrt{3}\) factor entirely and treats the triangle like it has a right angle at the base. Option (d), \(36\sqrt{3}\), comes from using a wrong side value such as \(6\sqrt{3}\) instead of \(8\sqrt{3}\). None of these match the working above.

Final Answer:
The area of the triangle is \(48\sqrt{3}\) cm\(^2\), which is option (c). \[ \boxed{48\sqrt{3}\ \text{cm}^2} \]