# Difference between revisions of "2015 AMC 8 Problems/Problem 21"

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== Solution == | == Solution == | ||

− | Clearly, since <math>\overline{FE}</math> is a side of a square with area <math>32</math>, <math>\ | + | Clearly, since <math>\overline{FE}</math> is a side of a square with area <math>32</math>, <math>\overline{FE} = \sqrt{32} = 4 \sqrt{2}</math>. Now, since <math>\overline{FE} = \overline{BC}</math>, we have <math>\overline{BC} = 4 \sqrt{2}</math>. |

Now, <math>\overline{AB}</math> is a side of a square with area <math>18</math>, so <math>\overline{AB} = \sqrt{18} = 3 \sqrt{2}</math>. Since <math>\Delta JBK</math> is equilateral, <math>\overline{BK} = 3 \sqrt{2}</math>. | Now, <math>\overline{AB}</math> is a side of a square with area <math>18</math>, so <math>\overline{AB} = \sqrt{18} = 3 \sqrt{2}</math>. Since <math>\Delta JBK</math> is equilateral, <math>\overline{BK} = 3 \sqrt{2}</math>. | ||

− | Lastly, <math>\Delta KBC</math> is a right triangle. We see that <math>\angle JBA + \angle ABC + \angle CBK + \angle KBJ = 360 \rightarrow | + | Lastly, <math>\Delta KBC</math> is a right triangle. We see that <math>\angle JBA + \angle ABC + \angle CBK + \angle KBJ = 360 \rightarrow 90 + 120 + \angle CBK + 60 = 360 \rightarrow \angle CBK = 90</math>, so <math>\Delta KBC</math> is a right triangle with legs <math>3 \sqrt{2}</math> and <math>4 \sqrt{2}</math>. Now, its area is <math>\dfrac{3 \sqrt{2} \cdot 4 \sqrt{2}}{2} = \dfrac{24}{2} = 12</math>, and our answer is <math>\boxed{\text{C}}</math>. |

## Revision as of 15:28, 25 November 2015

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 , . 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 , and our answer is .