2014 AMC 10B Problems/Problem 15
Problem
In rectangle , and points and lie on so that and trisect as shown. What is the ratio of the area of to the area of rectangle ?
$\textbf{(A)}\ \ \frac{\sqrt{3}}{6}\qquad\textbf{(B)}\ \frac{\sqrt{6}}{8}\qquad\textbf{(C)}\ \frac{3\sqrt{3}}{16}\qquad\textbf{(D)}}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}$ (Error compiling LaTeX. Unknown error_msg)
Solution
Let the length of be , so that the length of is and .
Because is a rectangle, , and so . Thus is a right triangle; this implies that , so . Now drop the altitude from of , forming two triangles. Because the length of is , from the properties of a triangle the length of is and the length of is thus . Thus the altitude of is , and its base is , so its area is .
To finish, $\frac{\text{[}\triangle DEF\text{]}}{\text{[}ABCD\text{]}}=\frac{\frac{x^2\sqrt{3}}{3}}{2x^2}=\boxed{\textbf{(D) }}\sqrt{3}{6}}$ (Error compiling LaTeX. Unknown error_msg)
See Also
2014 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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