Difference between revisions of "2010 AMC 8 Problems/Problem 17"
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<math> \textbf{(A)}\ \frac{2}{5}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{3}{5}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{3}{4} </math> | <math> \textbf{(A)}\ \frac{2}{5}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{3}{5}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{3}{4} </math> | ||
− | == | + | ==Solution 1== |
We see that half the area of the octagon is <math>5</math>. We see that the triangle area is <math>5-1 = 4</math>. That means that <math>\frac{5h}{2} = 4 \rightarrow h=\frac{8}{5}</math>. | We see that half the area of the octagon is <math>5</math>. We see that the triangle area is <math>5-1 = 4</math>. That means that <math>\frac{5h}{2} = 4 \rightarrow h=\frac{8}{5}</math>. | ||
<cmath>\text{QY}=\frac{8}{5} - 1 = \frac{3}{5}</cmath> | <cmath>\text{QY}=\frac{8}{5} - 1 = \frac{3}{5}</cmath> | ||
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== |
+ | 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> | ||
+ | |||
+ | ~Hithere22702 | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | ==Solution 3== | ||
+ | We can move the cube on the bottom right to the top left, thus creating a rectangle. We thus know that now this has been separated to 2 triangle, with equal area, although, obviously, one is not a perfect triangle. From this we can subtract 1 from one of the 5's (the area of one of the triangle-shapes-polygons) and add it to the other, since we have just moved the block, original in the triangle-shaped-polygon on the bottom, and thus, we get 4/6, or <math>\boxed{\textbf{(D) }\frac{2}{3}}</math> | ||
+ | -RealityWrites | ||
+ | |||
+ | ==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 11:30, 7 September 2021
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
Solution 3
We can move the cube on the bottom right to the top left, thus creating a rectangle. We thus know that now this has been separated to 2 triangle, with equal area, although, obviously, one is not a perfect triangle. From this we can subtract 1 from one of the 5's (the area of one of the triangle-shapes-polygons) and add it to the other, since we have just moved the block, original in the triangle-shaped-polygon on the bottom, and thus, we get 4/6, or -RealityWrites
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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