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 ?
![[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]](http://latex.artofproblemsolving.com/0/b/8/0b8719923caa95ad435933a935115d4f177f5fd2.png)
$\textbf{(A)}\ \ \frac{\sqrt{3}}{6}\qquad\textbf{(B)}\ \frac{\sqrt{6}}{8}\qquad\textbf{(C)}\ \frac{3\sqrt{3}}{16}\qquad\textbf{(D)}}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}$ (Error compiling LaTeX. Unknown error_msg)
Solution
Let the length of 
be 
, so that the length of 
is 
and ![$\text{[}ABCD\text{]}=2x^2$](http://latex.artofproblemsolving.com/d/1/e/d1ee314d4b265b3795d08a3212a0e1697d6657cb.png)
.
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, $\frac{\text{[}\triangle DEF\text{]}}{\text{[}ABCD\text{]}}=\frac{\frac{x^2\sqrt{3}}{3}}{2x^2}=\boxed{\textbf{(D) }}\frac{\sqrt{3}}{6}}$ (Error compiling LaTeX. Unknown error_msg)
See Also
| 2014 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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 | ||
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