Difference between revisions of "2014 AMC 10A Problems/Problem 16"
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<math> \textbf{(A)}\ \dfrac1{12}\qquad\textbf{(B)}\ \dfrac{\sqrt3}{18}\qquad\textbf{(C)}\ \dfrac{\sqrt2}{12}\qquad\textbf{(D)}\ \dfrac{\sqrt3}{12}\qquad\textbf{(E)}\ \dfrac16 </math> | <math> \textbf{(A)}\ \dfrac1{12}\qquad\textbf{(B)}\ \dfrac{\sqrt3}{18}\qquad\textbf{(C)}\ \dfrac{\sqrt2}{12}\qquad\textbf{(D)}\ \dfrac{\sqrt3}{12}\qquad\textbf{(E)}\ \dfrac16 </math> | ||
+ | [[Category: Introductory Geometry Problems]] | ||
==Solution 1== | ==Solution 1== |
Revision as of 11:52, 13 August 2014
Problem
In rectangle ,
,
, and points
,
, and
are midpoints of
,
, and
, respectively. Point
is the midpoint of
. What is the area of the shaded region?
Solution 1
Denote . Then
. Let the intersection of
and
be
, and the intersection of
and
be
. Then we want to find the coordinates of
so we can find
. From our points, the slope of
is
, and its
-intercept is just
. Thus the equation for
is
. We can also quickly find that the equation of
is
. Setting the equations equal, we have
. Because of symmetry, we can see that the distance from
to
is also
, so
. Now the area of the kite is simply the product of the two diagonals over
. Since the length
, our answer is
.
Solution 2
Let the area of the shaded region be . Let the other two vertices of the kite be
and
with
closer to
than
. Note that
. The area of
is
and the area of
is
. We will solve for the areas of
and
in terms of x by noting that the area of each triangle is the length of the perpendicular from
to
and
to
respectively. Because the area of
=
based on the area of a kite formula,
for diagonals of length
and
,
. So each perpendicular is length
. So taking our numbers and plugging them into
gives us
Solving this equation for
gives us
Solution 3
From the diagram in Solution 1, let be the height of
and
be the height of
. It is clear that their sum is
as they are parallel to
. Let
be the ratio of the sides of the similar triangles
and
, which are similar because
is parallel to
and the triangles share angle
. Then
, as 2 is the height of
. Since
and
are similar for the same reasons as
and
, the height of
will be equal to the base, like in
, making
. However,
is also the base of
, so
where
so
. Subbing into
gives a system of linear equations,
and
. Solving yields
and
, and since the area of the kite is simply the product of the two diagonals over
and
, our answer is
. Solution by hjklhjklhjkl2008.
See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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.