1992 AMC 8 Problem 10

Attempt Problem 10 of the 1992 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 1992 AMC 8 solutions, or check the answer key.

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

10.

An isosceles right triangle with legs of length 88 is partitioned into 1616 congruent triangles as shown. The shaded area is
1010
2020
3232
4040
6464
Answer: B
Concepts:triangle areaarea decomposition
Small hint:
The whole triangle has area 12×8×8=32,\dfrac12 \times 8 \times 8 = 32, split into 1616 equal pieces
Big hint:
Each small triangle has area 22; count how many are shaded
Solution:
The large triangle has area 12×8×8=32,\dfrac12 \times 8 \times 8 = 32, so each of the 1616 congruent small triangles has area 3216=2.\dfrac{32}{16} = 2.
Ten of the small triangles are shaded, so the shaded area is 10×2=20.10 \times 2 = 20.
Thus, the correct answer is B .
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