2014 AMC 10A Problems/Problem 16
Contents
[hide]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
Note that the region is a kite; hence its diagonals are perpendicular and it has area for diagonals of length
and
. Since
as both
and
are midpoints of parallel sides of rectangle
and
, we let
. Now all we need to do is to find
.
Let the other two vertices of the kite be and
with
closer to
than
. This gives us
. Now let
. We thus find that the equation of
is
and that of
is
. Solving this system gives us
, so the
-coordinate of
is
; in other words,
is
from
. By symmetry,
is also the same distance from
, so as
we have
. Hence the area of the kite 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
See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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