2013 AMC 8 Problems/Problem 24
Contents
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?
Video Solution for Problems 21-25
Video Solution
https://youtu.be/sbBjMvq5GG4 ~savannahsolver
Easiest Solution
We can see that the Pentagon is made of two congruent shapes. We can fit one triangle into the gap in the upper square. Therefore, the answer is just
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
WARNING: Some errors
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, .
Solution 3
Let the intersection of and be .
Now we have and .
Because both triangles has a side on congruent squares therefore .
Because and are vertical angles .
Also both and are right angles so .
Therefore by AAS(Angle, Angle, Side) .
Then translating/rotating the shaded into the position of
So the shaded area now completely covers the square
Set the area of a square as
Therefore, .
Solution 4
Given the information in the problem, we see that the ABCIJ is congruent to ADHGJ. Since it is half of 2/3, it takes up 1/3 of the area.
Video Solution by OmegaLearn
https://youtu.be/j3QSD5eDpzU?t=346
See Also
2013 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
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