2003 AMC 8 Problem 21

Attempt Problem 21 of the 2003 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 2003 AMC 8 solutions, or check the answer key.

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

21.

The area of trapezoid ABCDABCD is 164 cm2.164\text{ cm}^2. The altitude is 88 cm, ABAB is 1010 cm, and CDCD is 1717 cm. What is BC,BC, in centimeters?
99
1010
1212
1515
2020
Answer: B
Concepts:trapezoidPythagorean Theoremarea decomposition
Small hint:
Drop perpendiculars from BB and CC to the base.
Big hint:
Use the 6−8−106-8-10 and 8−15−178-15-17 right triangles.
Solution:
Drop perpendiculars from BB and CC to AD‾,\overline{AD}, meeting it at EE and F.F.
In the left and right right triangles, AE=102−82=6AE=\sqrt{10^2-8^2}=6 and FD=172−82=15.FD=\sqrt{17^2-8^2}=15.
The two side triangles have areas 12⋅6⋅8=24\frac12\cdot6\cdot8=24 and 12⋅15⋅8=60.\frac12\cdot15\cdot8=60. The middle rectangle has area 8⋅BC.8\cdot BC.
Thus 164=24+60+8BC,164=24+60+8BC, so 8BC=808BC=80 and BC=10.BC=10.
Thus, B is the correct answer.
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