# 1992 AJHSME Problems/Problem 10

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## Problem

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

$[asy] for (int a=0; a <= 3; ++a) { for (int b=0; b <= 3-a; ++b) { fill((a,b)--(a,b+1)--(a+1,b)--cycle,grey); } } for (int c=0; c <= 3; ++c) { draw((c,0)--(c,4-c),linewidth(1)); draw((0,c)--(4-c,c),linewidth(1)); draw((c+1,0)--(0,c+1),linewidth(1)); } label("8",(2,0),S); label("8",(0,2),W); [/asy]$

$\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 64$

## Solution

### Solution 1

Because the smaller triangles are congruent, the shaded area take $\frac{10}{16}$ of the largest triangles area, which is $\frac{8 \times 8}{2}=32$, so the shaded area is $\frac{10}{16} \times 32= \boxed{\text{(B)}\ 20}$.

### Solution 2

Each of the triangle has side length of $\frac{1}{4} \times 8=2$, so the area is $\frac{1}{2} \times 2 \times 2=2$. Because there are $10$ triangles is the shaded area, its area is $2 \times 10 =\boxed{\text{(B)}\ 20}$.

## See Also

 1992 AJHSME (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byProblem 11 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions

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