2014 AMC 10B Problems/Problem 15: Difference between revisions
Created page with "==Problem== ==Solution== ==See Also== {{AMC10 box|year=2014|ab=B|num-b=14|num-a=16}} {{MAA Notice}}" |
TheCrafter (talk | contribs) Added problem [credits to AlcumusGuy] |
||
| Line 1: | Line 1: | ||
==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 | ||
These problems are copyrighted © by the Mathematical Association of America.