Difference between revisions of "2010 AMC 8 Problems/Problem 17"
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Like stated in solution 1, we know that half the area of the octagon is <math>5</math>. | Like stated in solution 1, we know that half the area of the octagon is <math>5</math>. | ||
− | + | After moving the square from the bottom right to the top left, the area of the resulting trapezoid is <math>5+1=6</math>. | |
<math>5(XQ+2)/2=6</math>. Solving for <math>XQ</math>, we get <math>XQ=2/5</math>. | <math>5(XQ+2)/2=6</math>. Solving for <math>XQ</math>, we get <math>XQ=2/5</math>. | ||
Line 56: | Line 56: | ||
Therefore, the answer comes out to <math>\boxed{\textbf{(D) }\frac{2}{3}}</math> | Therefore, the answer comes out to <math>\boxed{\textbf{(D) }\frac{2}{3}}</math> | ||
− | ~ | + | ~kempwood |
− | ==Solution | + | ==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== | ||
+ | https://youtu.be/N7Yu9-bLqls | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2010|num-b=16|num-a=18}} | {{AMC8 box|year=2010|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:00, 23 October 2024
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.