1956 AMC 12 Problem 30

Attempt Problem 30 of the 1956 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1956 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

30.

If the altitude of an equilateral triangle is 6,\sqrt6, then the area is:
222\sqrt2
232\sqrt3
333\sqrt3
626\sqrt2
1212
Answer: B
Concepts:equilateral trianglealtitudearea
Small hint:
For side length s,s, an equilateral triangle has altitude s32\frac{s\sqrt3}{2}
Big hint:
Once ss is known, use area sh2\frac{sh}{2}
Solution:
Let the side length be s.s. From s32=6, \frac{s\sqrt3}{2}=\sqrt6, we get s=22.s=2\sqrt2. Hence the area is 12s6=12(22)(6)=23. \frac12s\sqrt6 =\frac12(2\sqrt2)(\sqrt6)=2\sqrt3.
Thus, the correct answer is B.
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