2010 AMC 8 Problems/Problem 17
Contents
[hide]Problem
The diagram shows an octagon consisting of unit squares. The portion below is a unit square and a triangle with base . If bisects the area of the octagon, what is the ratio ?
Solution 1
We see that half the area of the octagon is . We see that the triangle area is . That means that . Meaning,
Solution 2
Like stated in solution 1, we know that half the area of the octagon is .
After moving the square from the bottom right to the top left, the area of the resulting trapezoid is .
. Solving for , we get .
Subtracting from , we get .
Therefore, the answer comes out to
~kempwood
Video Solution by OmegaLearn
https://youtu.be/j3QSD5eDpzU?t=937
Video by MathTalks
https://www.youtube.com/watch?v=KSYVsSJDX-0&feature=youtu.be
Video Solution by WhyMath
~savannahsolver
See Also
2010 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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.