2013 AMC 10A Problem 3

Attempt Problem 3 of the 2013 AMC 10A 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 10A solutions, or check the answer key.

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

3.

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 area
Small hint:
Use ABAB as the height of △ABE\triangle ABE
Big hint:
Set 12⋅AB⋅BE\frac12\cdot AB\cdot BE equal to the given area
Solution:
We have by the formula for the area of a triangle that 40=12⋅10⋅BE. 40 = \dfrac{1}{2} \cdot 10 \cdot BE. This gives us 40=5BE 40 = 5BE BE=8. BE = 8. Thus, E is the correct answer.
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