2018 AMC 8 Problems/Problem 22
Problem 22
Point is the midpoint of side
in square
and
meets diagonal
at
The area of quadrilateral
is
What is the area of
=Solution 1
Let the area of be
. Thus, the area of triangle
is
and the area of the square is
.
By AA similarity, with a 1:2 ratio, so the area of triangle
is
. Now consider trapezoid
. Its area is
, which is three-fourths the area of the square. We set up an equation in
:
Solving, we get
. The area of square
is
.
Solution 2
We can use analytic geometry for this problem.
Let us start by giving the coordinate
,
the coordinate
, and so forth.
and
can be represented by the equations
and
, respectively. Solving for their intersection gives point
coordinates
.
Now, ’s area is simply
or
. This means that pentagon
’s area is
of the entire square, and it follows that quadrilateral
’s area is
of the square.
The area of the square is then .
See Also
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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All AJHSME/AMC 8 Problems and Solutions |
Set s to be the bottom left triangle.
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