Difference between revisions of "2010 AMC 8 Problems/Problem 17"
(Undo revision 110442 by Jerryrchang729 (talk)) (Tag: Undo) |
(Undo griefing) |
||
Line 40: | Line 40: | ||
<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 == |
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> |
Revision as of 15:46, 22 October 2019
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
We see that half the area of the octagon is . We see that the triangle area is . That means that . Meaning,
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.