1989 AMC 8 Problem 15

Attempt Problem 15 of the 1989 AMC 8 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 1989 AMC 8 solutions, or check the answer key.

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

15.

In parallelogram ABCDABCD shown, what is the area of the shaded region BEDC?BEDC?
2424
4848
6060
6464
8080
Answer: D
Concepts:parallelogramarea decomposition
Small hint:
The shaded region is the whole parallelogram with the triangle ABEABE removed
Big hint:
The parallelogram area is 10×8;10 \times 8; triangle ABEABE has base AE=10−6AE = 10 - 6 and height 88
Solution:
The parallelogram has base 1010 and height 8,8, so its area is 10×8=80.10 \times 8 = 80.
The unshaded triangle ABEABE has base AE=AD−ED=10−6=4AE = AD - ED = 10 - 6 = 4 and height 8,8, so its area is 12×4×8=16.\frac{1}{2} \times 4 \times 8 = 16. The shaded area is 80−16=64.80 - 16 = 64.
Thus, the correct answer is D .
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