Difference between revisions of "2014 AMC 10B Problems/Problem 15"
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==Problem== | ==Problem== | ||
| + | In rectangle <math>ABCD</math>, <math>DC = 2CB</math> and points <math>E</math> and <math>F</math> lie on <math>\overline{AB}</math> so that <math>\overline{ED}</math> and <math>\overline{FD}</math> trisect <math>\angle ADC</math> as shown. What is the ratio of the area of <math>\triangle DEF</math> to the area of rectangle <math>ABCD</math>? | ||
| + | |||
| + | [asy] | ||
| + | draw((0, 0)--(0, 1)--(2, 1)--(2, 0)--cycle); | ||
| + | draw((0, 0)--(sqrt(3)/3, 1)); | ||
| + | draw((0, 0)--(sqrt(3), 1)); | ||
| + | label("A", (0, 1), N); | ||
| + | label("B", (2, 1), N); | ||
| + | label("C", (2, 0), S); | ||
| + | label("D", (0, 0), S); | ||
| + | label("E", (sqrt(3)/3, 1), N); | ||
| + | label("F", (sqrt(3), 1), N); | ||
| + | [/asy] | ||
| + | |||
| + | <math> \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}</math> | ||
==Solution== | ==Solution== | ||
Revision as of 17:02, 20 February 2014
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 ?
[asy] draw((0, 0)--(0, 1)--(2, 1)--(2, 0)--cycle); draw((0, 0)--(sqrt(3)/3, 1)); draw((0, 0)--(sqrt(3), 1)); label("A", (0, 1), N); label("B", (2, 1), N); label("C", (2, 0), S); label("D", (0, 0), S); label("E", (sqrt(3)/3, 1), N); label("F", (sqrt(3), 1), N); [/asy]
$\textbf{(A)}\ \ \frac{\sqrt{3}}{6}\qquad\textbf{(B)}\ \frac{\sqrt{6}}{8}\qquad\textbf{(C)}\ \frac{3\sqrt{3}}{16}\qquad\textbf{(D)}}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}$ (Error compiling LaTeX. Unknown error_msg)
Solution
See Also
| 2014 AMC 10B (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. 
