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 AAA 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 .
Solution 3
There is a trivial solution.
See Also
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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Set s to be the bottom left triangle. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.