Difference between revisions of "2014 AMC 10B Problems/Problem 15"
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<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> | <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> | ||
+ | [[Category: Introductory Geometry Problems]] | ||
==Solution== | ==Solution== |
Revision as of 10:53, 13 August 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
?
$\textbf{(A)}\ \ \frac{\sqrt{3}}{6}\qquad\textbf{(B)}\ \frac{\sqrt{6}}{8}\qquad\textbf{(C)}\ \frac{3\sqrt{3}}{16}\qquad\textbf{(D)}}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}$ (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,
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.