Difference between revisions of "2012 AMC 10A Problems/Problem 15"
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<math> \textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac15 \qquad\textbf{(C)}\ \frac29 \qquad\textbf{(D)}\ \frac13 \qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}</math> | <math> \textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac15 \qquad\textbf{(C)}\ \frac29 \qquad\textbf{(D)}\ \frac13 \qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}</math> | ||
+ | [[Category: Introductory Geometry Problems]] | ||
== Solution 1 == | == Solution 1 == |
Revision as of 11:44, 13 August 2014
Problem
Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of ?
Solution 1
intersects at a right angle, so . The hypotenuse of right triangle BED is .
Since AC=2BC, . is a right triangle so the area is just
Solution 2
Let be the origin. Then,
$\widebar{EB}$ (Error compiling LaTeX. ! Undefined control sequence.) can be represented by the line Also, can be represented by the line
Subtracting the second equation from the first gives us . Thus, . Plugging this into the first equation gives us .
Since , is ,
and .
Thus, . The answer is .
Solution 3
Triangle is similar to triangle ; line
Triangle is similar to triangle and the ratio of line to line .
Based on similarity the length of the height of is thus .
Thus, . The answer is
See Also
2012 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.