1992 AMC 8 Problem 10

Below is the professionally curated solution for Problem 10 of the 1992 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1992 AMC 8 solutions, or check the answer key.

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Difficulty rating: 930

10.

An isosceles right triangle with legs of length $8$ is partitioned into $16$ congruent triangles as shown. The shaded area is

$10$

$20$

$32$

$40$

$64$

The large triangle has area $\dfrac12 \times 8 \times 8 = 32,$ so each of the $16$ congruent small triangles has area $\dfrac{32}{16} = 2.$

Ten of the small triangles are shaded, so the shaded area is $10 \times 2 = 20.$

Thus, the correct answer is B .