Difference between revisions of "2015 AMC 8 Problems/Problem 21"
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+ | ==Problem== | ||
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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, <math>\triangle JBK</math> is equilateral and <math>FE=BC</math>. What is the area of <math>\triangle KBC</math>? | 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, <math>\triangle JBK</math> is equilateral and <math>FE=BC</math>. What is the area of <math>\triangle KBC</math>? | ||
− | |||
<asy> | <asy> | ||
draw((-4,6*sqrt(2))--(4,6*sqrt(2))); | draw((-4,6*sqrt(2))--(4,6*sqrt(2))); | ||
Line 22: | Line 23: | ||
label("$18$",(0,6*sqrt(2)+2),N); | label("$18$",(0,6*sqrt(2)+2),N); | ||
</asy> | </asy> | ||
+ | |||
+ | <math>\textbf{(A) }6\sqrt{2}\quad\textbf{(B) }9\quad\textbf{(C) }12\quad\textbf{(D) }9\sqrt{2}\quad\textbf{(E) }32</math>. | ||
== Solution == | == Solution == | ||
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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>. |
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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^{\circ} \rightarrow 90^{\circ} + 120^{\circ} + \angle CBK + 60^{\circ} = 360^{\circ} \rightarrow \angle CBK = 90^{\circ}</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} = \boxed{\textbf{(C)}~12}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/mUBhWTlvuLQ | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==See Also== | ||
− | + | {{AMC8 box|year=2015|num-b=20|num-a=22}} | |
+ | {{MAA Notice}} |
Revision as of 17:35, 16 January 2021
Contents
Problem
In the given figure hexagon is equiangular, and are squares with areas and respectively, is equilateral and . What is the area of ?
.
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 .
Video Solution
~savannahsolver
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.