1997 AHSME Problems/Problem 9
Contents
Problem 9
In the figure, is a square, is the midpoint of , and is on . If is perpendicular to , then the area of quadrilateral is
Solution 1
Since and , we have .
From those two equations, we find that and
Now that we have and , we can find the area of the bottom triangle :
The area of left triangle is
The area of the square is .
Thus, the area of the remaining quadrilateral is , and the answer is
Solution 2
Place the square on a coordinate grid so that and . Line is . Line goes through and has slope , so it must be
The intersection of the two lines is , and thus has coordinates . The altitude from to thus has length , so the area of the triangle is .
The other triangle has area , and the whole square has area . As above, we find the area of the quadrilateral by subtracting the two triangles, and we get , which is .
See also
1997 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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