Difference between revisions of "2015 AMC 8 Problems/Problem 21"
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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>. | 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>. | ||
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+ | ==See Also== | ||
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+ | {{AMC8 box|year=2015|num-b=24|after=Last Problem}} | ||
+ | {{MAA Notice}} |
Revision as of 17:11, 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 .
See Also
2015 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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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