2012 AMC 10B Problems/Problem 19
Problem
In rectangle ,
,
, and
is the midpoint of
. Segment
is extended 2 units beyond
to point
, and
is the intersection of
and
. What is the area of quadrilateral
?
Solution
Note that the area of equals the area of
.
Since
. Now,
, so
and
so
Therefore,
hence our answer is
Solution 2
Notice that is a trapezoid with height
, so we need to find
.
, so
. Since
,
. The area of
is
Solution 3 (Coordinate Bash)
Let .
We know these points from the problem statement:
We can use the Shoelace Formula to find the area of quadrilateral . We know the coordinates of all of the points in
except
. Since
is the intersection of
and
, we can use a system of equations to solve for the coordinates of
. The line for
is simply
. The line for
passes through the origin so it has a y-intercept of
, and a slope of
. Therefore, the line for
is
. Substituting
for
, we find that
. Therefore,
. Applying Shoelace on these points gives us that
.
See Also
2012 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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All AMC 10 Problems and Solutions |
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