The area (in $ ext{cm}^2 $) of an equilateral triangle whose altitude from vertex is $ 12 ext{ cm}$, is:

Question

The area (in $	ext{cm}^2 $) of an equilateral triangle whose altitude from vertex is $ 12	ext{ cm}$, is:

Examoogle · Accepted Answer

The correct answer is Option D: $48\sqrt{3}$. For an equilateral triangle with altitude $h$, the area is simply $\frac{h^2}{\sqrt{3}}$. Since $144/\sqrt{3} = 48\sqrt{3}$, it is computed in seconds.

Step-by-Step Solution & Concepts:
Shortcut Trick: For an equilateral triangle with altitude $h$, the area is simply $\frac{h^2}{\sqrt{3}}$. Since $144/\sqrt{3} = 48\sqrt{3}$, it is computed in seconds.

Formula Used: $$Area = \frac{h^2}{\sqrt{3}}$$

Given: The altitude (h) of an equilateral triangle is 12 cm.

Steps: Identify the relationship between area and altitude: In an equilateral triangle with side $a$, the altitude $h = \frac{\sqrt{3}}{2} a$. Rearranging for $a$, we get $a = \frac{2h}{\sqrt{3}}$. Substituting this into the standard area formula $A = \frac{\sqrt{3}}{4} a^2$ yields the direct formula.. Substitute the given value: Given $h = 12$ cm, substitute this value into the formula.. Rationalize the denominator: Multiply the numerator and denominator by $\sqrt{3}$ to simplify.

Answer: D is correct because the calculated area of the equilateral triangle is $48\sqrt{3}$ cm$^2$.

Exam Tip: This relationship is specific to equilateral triangles; for general triangles, remember $Area = \frac{1}{2} 	imes base 	imes height$.

Examoogle · Answer

Option A: $27\sqrt{3}$

Examoogle · Answer

Option B: $\frac{9\sqrt{3}}{2}$

Examoogle · Answer

Option C: $\frac{9\sqrt{3}}{4}$

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