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 (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 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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