Difference between revisions of "2010 AMC 8 Problems/Problem 17"
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Meaning, <math>\frac{\frac{2}{5}}{\frac{3}{5}} = \boxed{\textbf{(D) }\frac{2}{3}}</math> | Meaning, <math>\frac{\frac{2}{5}}{\frac{3}{5}} = \boxed{\textbf{(D) }\frac{2}{3}}</math> | ||
− | ==Solution 2 | + | ==Solution 2== |
Like stated in solution 1, we know that half the area of the octagon is <math>5</math>. That means that the area of the 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>. Subtracting <math>2/5</math> from <math>1</math>, we get <math>QY=3/5</math>. Therefore, the answer comes out to <math>\boxed{\textbf{(D) }\frac{2}{3}}</math> | Like stated in solution 1, we know that half the area of the octagon is <math>5</math>. That means that the area of the 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>. Subtracting <math>2/5</math> from <math>1</math>, we get <math>QY=3/5</math>. Therefore, the answer comes out to <math>\boxed{\textbf{(D) }\frac{2}{3}}</math> | ||
Revision as of 17:54, 29 April 2020
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 . That means that the area of the trapezoid is . . Solving for , we get . Subtracting from , we get . Therefore, the answer comes out to
~Hithere22702
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 |
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