2000 AMC 8 Problem 25

Below is the professionally curated solution for Problem 25 of the 2000 AMC 8, 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.

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Concepts:area decomposition triangle area

Difficulty rating: 1520

25.

The area of rectangle $ABCD$ is $72$ units squared. If point $A$ and the midpoints of $\overline{BC}$ and $\overline{CD}$ are joined to form a triangle, the area of that triangle is

$21$

$27$

$30$

$36$

$40$

Solution:

Let the rectangle have side lengths $2a$ and $2b$ , so its area is $4ab=72$ and $ab=18$ .

The three right triangles outside the desired triangle have areas $\frac12(2a)(b)=ab$ , $\frac12(2b)(a)=ab$ , and $\frac12(a)(b)=\frac12ab$ .

Their total area is $\frac52ab=45$ . Therefore the desired triangle has area $72-45=27$ .

Thus, B is the correct answer.

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