2000 AMC 8 Problem 25

Attempt Problem 25 of the 2000 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 2000 AMC 8 solutions, or check the answer key.

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

25.

The area of rectangle ABCDABCD is 7272 square units. If point AA and the midpoints of BC‾\overline{BC} and CD‾\overline{CD} are joined to form a triangle, the area of that triangle is
2121
2727
3030
3636
4040
Answer: B
Concepts:area decompositiontriangle area
Small hint:
Let the rectangle sides be 2a2a and 2b2b
Big hint:
Subtract the three outside right triangles from the rectangle
Solution:
Let the rectangle have side lengths 2a2a and 2b,2b, so its area is 4ab=724ab=72 and ab=18.ab=18.
The three right triangles outside the desired triangle have areas 12(2a)(b)=ab,\frac12(2a)(b)=ab, 12(2b)(a)=ab,\frac12(2b)(a)=ab, and 12(a)(b)=12ab.\frac12(a)(b)=\frac12ab.
Their total area is 52ab=45.\frac52ab=45. Therefore the desired triangle has area 72−45=27.72-45=27.
Thus, B is the correct answer.
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