2013 AMC 12A Problem 1

Attempt Problem 1 of the 2013 AMC 12A 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 2013 AMC 12A solutions, or check the answer key.

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

1.

Square ABCDABCD has side length 10.10. Point EE is on BC‾,\overline{BC}, and the area of △ABE\triangle ABE is 40.40. What is BE?BE?
44
55
66
77
88
Answer: E
Concepts:triangle arearight triangle
Small hint:
The legs of right triangle ABEABE are ABAB and BEBE
Big hint:
Set 12⋅AB⋅BE=40\tfrac12\cdot AB\cdot BE = 40 with AB=10AB = 10
Solution:
The legs of right triangle ABEABE are AB=10AB = 10 and BE.BE. From 12⋅10⋅BE=40,\tfrac12\cdot 10\cdot BE = 40, we get BE=8.BE = 8.
Thus, the correct answer is E.
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