Bunuel wrote:

What is the area of an equilateral triangle that has an altitude of length 24?

(A) \(24 \sqrt{3}\)

(B) \(48 \sqrt{3}\)

(C) \(96 \sqrt{3}\)

(D) \(192 \sqrt{3}\)

(E) \(384 \sqrt{3}\)

When we drop an altitude in an equilateral triangle we create two 30-60-90 right triangles in which the altitude is opposite the 60 degree angle; thus, the altitude = side√3/2, thus:

side√3/2 = 24

side = 48/√3

We may recall that the area formula for an equilateral triangle is (side^2 * √3)/4, thus:

area = [(48/√3)^2 * √3]/4

area = 768√3/4 = 192√3

Alternate Solution:

When we drop a perpendicular (altitude) in an equilateral triangle, we create two 30-60-90 triangles, each with a ratio of sides of x : 2x : x√3. The altitude is x√3, and so we have:

x√3 = 24

x = 24/√3

x = (24√3)/3

x = 8√3

We see that 8√3 is the base of one of the 30-60-90 triangles, so the base of the entire equilateral triangle is 16√3.

Using the formula for the area of a triangle: A = 1/2 b x h, we have:

A = (1/2) x 16√3 x 24 = 192√3

Answer: D

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