2015 AMC 8 Problems/Problem 21
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
Clearly, since is a side of a square with area , $\overlin{FE} = \sqrt{32} = 4 \sqrt{2}$ (Error compiling LaTeX. Unknown error_msg). 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 90 + 120 + \angle CBK + 60 = 360 \rightarrow \angle CBK = 90\Delta KBC3 \sqrt{2}4 \sqrt{2}\dfrac{3 \sqrt{2} \cdot 4 \sqrt{2}}{2} = \dfrac{24}{2} = \dfrac{12}\boxed{\text{C}}$.