2014 AMC 12B Problems/Problem 21
Contents
Problem
In the figure, is a square of side length . The rectangles and are congruent. What is ?
Solution 1
Draw the altitude from to and call the foot . Then . Consider . It is the hypotenuse of both right triangles and , and we know , so we must have by Hypotenuse-Leg congruence. From this congruence we have .
Notice that all four triangles in this picture are similar. Also, we have . So set and . Now . This means , so is the midpoint of . So , along with all other similar triangles in the picture, is a 30-60-90 triangle, and we have and subsequently . This means , which gives , so the answer is .
Solution 2
Let . Let . Because and , are all similar. Using proportions and the pythagorean theorem, we find Because we know that , we can set up a systems of equations and solving for , we get Now solving for , we get Plugging into the second equations with , we get
Solution 3
Let , , and . Then and because and , . Furthermore, the area of the four triangles and the two rectangles sums to 1:
By the Pythagorean theorem:
Then by the rational root theorem, this has roots , , and . The first and last roots are extraneous because they imply and , respectively, thus .
Solution 4
Let = and = . It is shown that all four triangles in the picture are similar. From the square side lengths:
Solving for we get:
Solution 5
Note that is a diagonal of , so it must be equal in length to . Therefore, quadrilateral has , and , so it must be either an isosceles trapezoid or a parallelogram. But due to the slope of and , we see that it must be a parallelogram. Therefore, . But by the symmetry in rectangle , we see that . Therefore, . We also know that , hence .
As and , and as is right, we know that must be a 30-60-90 triangle. Therefore, and . But by similarity, is also a 30-60-90 triangle, hence . But , hence . As , this implies that . Thus the answer is .
Solution 6
Let ,
, , ,
, , ,
, , ,
, , ,
, ,
Substitute into we get ,
, as ,
See Also
2014 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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All AMC 12 Problems and Solutions |
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