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 10:59, 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 
?
.
![[asy] draw((-4,6*sqrt(2))--(4,6*sqrt(2))); draw((-4,-6*sqrt(2))--(4,-6*sqrt(2))); draw((-8,0)--(-4,6*sqrt(2))); draw((-8,0)--(-4,-6*sqrt(2))); draw((4,6*sqrt(2))--(8,0)); draw((8,0)--(4,-6*sqrt(2))); draw((-4,6*sqrt(2))--(4,6*sqrt(2))--(4,8+6*sqrt(2))--(-4,8+6*sqrt(2))--cycle); draw((-8,0)--(-4,-6*sqrt(2))--(-4-6*sqrt(2),-4-6*sqrt(2))--(-8-6*sqrt(2),-4)--cycle); label("$I$",(-4,8+6*sqrt(2)),dir(100)); label("$J$",(4,8+6*sqrt(2)),dir(80)); label("$A$",(-4,6*sqrt(2)),dir(280)); label("$B$",(4,6*sqrt(2)),dir(250)); label("$C$",(8,0),W); label("$D$",(4,-6*sqrt(2)),NW); label("$E$",(-4,-6*sqrt(2)),NE); label("$F$",(-8,0),E); draw((4,8+6*sqrt(2))--(4,6*sqrt(2))--(4+4*sqrt(3),4+6*sqrt(2))--cycle); label("$K$",(4+4*sqrt(3),4+6*sqrt(2)),E); draw((4+4*sqrt(3),4+6*sqrt(2))--(8,0),dashed); label("$H$",(-4-6*sqrt(2),-4-6*sqrt(2)),S); label("$G$",(-8-6*sqrt(2),-4),W); label("$32$",(-10,-8),N); label("$18$",(0,6*sqrt(2)+2),N); [/asy]](http://latex.artofproblemsolving.com/f/1/1/f11d24db37ed085cbc452224454338f8eabdfb37.png)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
