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Bunuel
Find the altitude of an equilateral triangle whose side is 20.

(A) 10
(B) \(10 \sqrt{2}\)
(C) \(10 \sqrt{3}\)
(D) \(20 \sqrt{2}\)
(E) \(20 \sqrt{3}\)
Solution:

If we drop the altitude from a vertex of the equilateral triangle to the base that is opposite the vertex, we split the base into two equal parts. Furthermore, we’ve split the equilateral triangle into two congruent 30-60-90 right triangles. Since the base is a side of the triangle, its length is 20 and thus half its length is 10. Since half its length is the shorter leg of one of the 30-60-90 right triangles and the altitude is the longer leg, the altitude is 10√3 (recall that the ratio of the shorter leg to the longer leg of a 30-60-90 right triangle is x : x√3).

Answer: C
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