1964 AMC 12 Problem 10

Attempt Problem 10 of the 1964 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 1964 AMC 12 solutions, or check the answer key.

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

10.

Given a square with side of length s.s. On a diagonal as base a triangle with three unequal sides is constructed so that its area equals that of the square. The length of the altitude drawn to the base is:
s2s\sqrt2
s2\dfrac{s}{\sqrt2}
2s2s
2s2\sqrt{s}
2s\dfrac2{\sqrt{s}}
Answer: A
Concepts:square (geometry)diagonaltriangle arearadical
Small hint:
The square’s diagonal has length s2s\sqrt2
Big hint:
Set 12(s2)h=s2\frac12(s\sqrt2)h=s^2
Solution:
The triangle’s base is the square’s diagonal, s2.s\sqrt2. If its altitude is h,h, equality of areas gives 12(s2)h=s2,\frac12(s\sqrt2)h=s^2, so h=s2.h=s\sqrt2.
Therefore, the correct answer is A.
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