2002 AMC 8 Problem 16

Attempt Problem 16 of the 2002 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 2002 AMC 8 solutions, or check the answer key.

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

16.

Right isosceles triangles are constructed on the sides of a 3−4−53-4-5 right triangle, as shown. A capital letter represents the area of each triangle. Which one of the following is true?
X+Z=W+YX + Z = W + Y
W+X=ZW + X = Z
3X+4Y=5Z3X + 4Y = 5Z
X+W=12(Y+Z)X + W = \dfrac{1}{2}(Y + Z)
X+Y=ZX + Y = Z
Answer: E
Concepts:Pythagorean Theoremarea
Small hint:
Each right isosceles triangle has area half the square of the side it is built on.
Big hint:
Compute the areas on the sides 3,3, 4,4, and 5.5.
Solution:
For a right isosceles triangle built on a side of length s,s, the two legs have length s,s, so its area is s22.\frac{s^2}{2}.
W=3⋅42=6,X=322=4.5,Y=422=8,Z=522=12.5. \begin{aligned} &W=\frac{3\cdot4}{2}=6, \\ &\quad X=\frac{3^2}{2}=4.5, \\ &\quad Y=\frac{4^2}{2}=8, \\ &\quad Z=\frac{5^2}{2}=12.5. \end{aligned}
These values satisfy X+Y=Z,X+Y=Z, and the other listed equations do not.
Thus, E is the correct answer.
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