Difference between revisions of "2010 AMC 8 Problems/Problem 17"
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==Solution 2== | ==Solution 2== | ||
− | + | we know that half the area of the octagon is 5. that means that the area of the trapezoid is 5+1=6 5(XQ+2)/2=6 Solving for XQ we get XQ=2/5 Subtracting 2/5 from 1 we get QY=3/5 Therefore, the answer comes out to D:2/3 | |
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− | Subtracting | ||
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− | Therefore, the answer comes out to | ||
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==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}} |
Revision as of 13:03, 15 August 2021
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
we know that half the area of the octagon is 5. that means that the area of the trapezoid is 5+1=6 5(XQ+2)/2=6 Solving for XQ we get XQ=2/5 Subtracting 2/5 from 1 we get QY=3/5 Therefore, the answer comes out to D:2/3
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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