Difference between revisions of "2015 AMC 8 Problems/Problem 21"
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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>. | 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{ | + | Now, <math>\overline{JB}</math> is a side of a square with area <math>18</math>, so <math>\overline{JB} = \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 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) 12}}</math>. | 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) 12}}</math>. |
Revision as of 11:59, 30 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 .
See Also
2015 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
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