Difference between revisions of "2013 AMC 8 Problems/Problem 24"
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filldraw(A--B--C--I--J--cycle,grey); | filldraw(A--B--C--I--J--cycle,grey); | ||
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− | + | dot("$A$", A, NW); | |
− | + | dot("$B$", B, NE); | |
− | + | dot("$C$", C, NE); | |
− | + | dot("$D$", D, NW); | |
− | + | dot("$E$", E, NW); | |
− | + | dot("$F$", F, SW); | |
− | + | dot("$G$", G, S); | |
− | + | dot("$H$", H, N); | |
− | + | dot("$I$", I, NE); | |
− | + | dot("$X$", X,SE); | |
− | + | dot("$J$", J, SE);</asy> | |
Revision as of 22:53, 27 November 2013
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 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.