2015 AMC 8 Problems/Problem 21
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[hide]Problem
In the given figure, hexagon is equiangular, and are squares with areas and respectively, is equilateral and . What is the area of ?
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Solution 1
Clearly, since is a side of a square with area , . Now, since , we have .
Now, is a side of a square with area , so . Since is equilateral, .
Lastly, is a right triangle. We see that , so is a right triangle with legs and . Now, its area is .
Solution 2
Since , and , . Meanwhile, , and since is equilateral, . If is equiangular, , where is the number of sides of the shape. Adding all the angles around gives , so . Because is right, the area of . Therefore, the answer is . ~strongstephen
Note: Another more complicated way to find out if angle KBC is right, is that we can extend line AB to a point L on triangle KBC. Now, since we know that angle IJB is right, and we know that line IJ now acts as a transversal to lines IJ and AL. That means angle IJB(90 degrees) must be equal to angle JBL by the alternate-exterior angle rule. Since we know that triangle JKB is equilateral, angle JBK will be 60 degrees, hence angle KBL will be 90 - 60 = 30 degrees. Now, we move on to figure ABCDEF. Through the solution above, we know that angle ABC is 120 degrees, so we can use the same line AL which will result in the formation of a linear pair of angles: angles ABC and CBL. Since the angles in a linear pair always add up to 180, and angle ABC is 120, angle CBL = 180 - 120 = 60 degrees. Notice that angle CBL and angle KBL together make up angle KBC. Hence, by the value of 30 degrees derived for angle KBL, angle KBC = angle KBL + angle CBL = 30 degrees + 60 degrees = 90 degrees.
Video Solution
~savannahsolver
See Also
2015 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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