1992 AJHSME Problems/Problem 10
Problem
An isosceles right triangle with legs of length 
is partitioned into 
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]](http://latex.artofproblemsolving.com/5/0/d/50d24f2652cf461b0ca9941ff2e7e3d38dec6712.png)
Solution
Solution 1
Because the smaller triangles are congruent, the shaded area take 
of the largest triangles area, which is 
, so the shaded area is 
.
Solution 2
Each of the triangle has side length of 
, so the area is 
. Because there are 
triangles is the shaded area, its area is 
.
See Also
| 1992 AJHSME (Problems • Answer Key • Resources) | ||
| 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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