2016 AMC 8 Problem 2

Attempt Problem 2 of the 2016 AMC 8 below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2016 AMC 8 solutions, or check the answer key.

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

2.

In rectangle ABCD,ABCD, AB=6AB=6 and AD=8.AD=8. Point MM is the midpoint of AD‾.\overline{AD}. What is the area of △AMC?\triangle AMC?
12 12
15 15
18 18
20 20
24 24
Answer: A
Concepts:triangle areamidpoint
Small hint:
Since MM is the midpoint of AD‾\overline{AD}, find AMAM.
Big hint:
Use AMAM as the base and the rectangle’s vertical side length as the height.
Video solution:
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Written solution:
Since MM is the midpoint of AD‾,\overline{AD}, the base AMAM has length 4.4. The altitude from CC to AM‾\overline{AM} has length AB=6.AB=6. The area is therefore 12⋅4⋅6=12.\dfrac{1}{2} \cdot 4 \cdot 6 = 12.
Thus, A is the correct answer.
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