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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