2013 AMC 8 Problems/Problem 24
Contents
[hide]Problem
Squares , , and are equal in area. Points and are the midpoints of sides and , respectively. What is the ratio of the area of the shaded pentagon to the sum of the areas of the three squares?
Solution 1
First let (where is the side length of the squares) for simplicity. We can extend until it hits the extension of . Call this point . The area of triangle then is The area of rectangle is . Thus, our desired area is . Now, the ratio of the shaded area to the combined area of the three squares is .
Solution 2
Let the side length of each square be .
Let the intersection of and be .
Since , . Since and are vertical angles, they are congruent. We also have by definition.
So we have by congruence. Therefore, .
Since and are midpoints of sides, . This combined with yields .
The area of trapezoid is .
The area of triangle is .
So the area of the pentagon is .
The area of the squares is .
Therefore, .
See Also
2013 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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