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

(Created page with "In the given figure hexagon <math>ABCDEF</math> is equiangular, <math>ABJI</math> and <math>FEHG</math> are squares with areas <math>18</math> and <math>32</math> respectively...") |
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label("$18$",(0,6*sqrt(2)+2),N); | label("$18$",(0,6*sqrt(2)+2),N); | ||

</asy> | </asy> | ||

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+ | == Solution == | ||

+ | |||

+ | Clearly, since <math>\overline{FE}</math> is a side of a square with area <math>32</math>, <math>\overlin{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>. | ||

+ | |||

+ | Lastly, <math>\Delta KBC</math> is a right triangle. We see that <math>\angle JBA + \angle ABC + \angle CBK + \angle KBJ = 360 \rightarrow </math>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} = \dfrac{12}<math>, and our answer is </math>\boxed{\text{C}}$. |

## Revision as of 15:27, 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 , $\overlin{FE} = \sqrt{32} = 4 \sqrt{2}$ (Error compiling LaTeX. ! Undefined control sequence.). 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}}$.