2008 AMC 8 Problems/Problem 23
Contents
Problem
In square , and . What is the ratio of the area of to the area of square ?
Solution 1
The area of is the area of square subtracted by the the area of the three triangles around it. Arbitrarily assign the side length of the square to be .
The ratio of the area of to the area of is
Solution 2
Say that has length , and that from there we can infer that . We also know that , and that . The area of triangle is the square's area subtracted from the area of the excess triangles, which is simply these equations: Thus, the area of the triangle is . We can now put the ratio of triangle 's area to the area of the square as a fraction. We have: Thus, our answer is , .
~Mr.BigBrain_AoPS
Video Solution by OmegaLearn
https://youtu.be/abSgjn4Qs34?t=528
~ pi_is_3.14
Video Solution 2
https://www.youtube.com/watch?v=9cTv1aoGTy0
This video was recorded in vertical accidentally.
See Also
2008 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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.