1981 AMC 12 Problem 2

Attempt Problem 2 of the 1981 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 1981 AMC 12 solutions, or check the answer key.

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

2.

Point EE is on side ABAB of square ABCD.ABCD. If EBEB has length one and ECEC has length two, then the area of the square is
3\sqrt3
5\sqrt5
33
232\sqrt3
55
Answer: C
Concepts:square (geometry)Pythagorean Theoremarea
Small hint:
Triangle EBCEBC is a right triangle
Big hint:
The unknown leg BCBC is also the side length of the square
Solution:
Let the square’s side length be s.s. In right triangle EBC,EBC, s2+12=22. s^2+1^2=2^2. Hence s2=3,s^2=3, which is exactly the area of the square.
Therefore, the correct answer is C.
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