2003 AMC 8 Problems/Problem 25
Contents
Problem
In the figure, the area of square 
is 
. The four smaller squares have sides 1 cm long, either parallel to or coinciding with the sides of the large square. In 
, 
, and when is folded over side 
, point 
coincides with 
, the center of square . What is the area of , in square centimeters?
![[asy] defaultpen(fontsize(8)); size(225); pair Z=origin, W=(0,10), X=(10,10), Y=(10,0), O=(5,5), B=(-4,8), C=(-4,2), A=(-13,5); draw((-4,0)--Y--X--(-4,10)--cycle); draw((0,-2)--(0,12)--(-2,12)--(-2,8)--B--A--C--(-2,2)--(-2,-2)--cycle); dot(O); label("$A$", A, NW); label("$O$", O, NE); label("$B$", B, SW); label("$C$", C, NW); label("$W$",W , NE); label("$X$", X, N); label("$Y$", Y, S); label("$Z$", Z, SE); [/asy]](http://latex.artofproblemsolving.com/f/b/5/fb5de36e57bfb81488edf36e2717cf7b93f0444b.png)
Solution
The side lengths of square 
must be 5 cm, since the area is 
. First, you should determine the height of 
. The distance from 
to line 
must be 2.5 cm, since line 
= 5 cm, and the distance from to 
is half of that. The distance from line to line 
must be 2, since the side lengths of the small squares are 1, and there are two squares from line to line . So, the height of must be 4.5, which is 2.5 + 2. The length of can be determined by subtracting 2 from 5, since the length of is 5, and the two squares in the corners give us 2 together. This gives us the base for , which is 3. Then, we multiply 4.5 by 3 and divide by 2, to get an answer of 
.
Video Solution
https://www.youtube.com/watch?v=4RBCH1rUcSw
See Also
| 2003 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Last Problem | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
