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

In the given figure hexagon $ABCDEF$ is equiangular, $ABJI$ and $FEHG$ are squares with areas $18$ and $32$ respectively, $\triangle JBK$ is equilateral and $FE=BC$. What is the area of $\triangle KBC$?

$\textbf{(A) }6\sqrt{2}\quad\textbf{(B) }9\quad\textbf{(C) }12\quad\textbf{(D) }9\sqrt{2}\quad\textbf{(E) }32$. $[asy] draw((-4,6*sqrt(2))--(4,6*sqrt(2))); draw((-4,-6*sqrt(2))--(4,-6*sqrt(2))); draw((-8,0)--(-4,6*sqrt(2))); draw((-8,0)--(-4,-6*sqrt(2))); draw((4,6*sqrt(2))--(8,0)); draw((8,0)--(4,-6*sqrt(2))); draw((-4,6*sqrt(2))--(4,6*sqrt(2))--(4,8+6*sqrt(2))--(-4,8+6*sqrt(2))--cycle); draw((-8,0)--(-4,-6*sqrt(2))--(-4-6*sqrt(2),-4-6*sqrt(2))--(-8-6*sqrt(2),-4)--cycle); label("I",(-4,8+6*sqrt(2)),dir(100)); label("J",(4,8+6*sqrt(2)),dir(80)); label("A",(-4,6*sqrt(2)),dir(280)); label("B",(4,6*sqrt(2)),dir(250)); label("C",(8,0),W); label("D",(4,-6*sqrt(2)),NW); label("E",(-4,-6*sqrt(2)),NE); label("F",(-8,0),E); draw((4,8+6*sqrt(2))--(4,6*sqrt(2))--(4+4*sqrt(3),4+6*sqrt(2))--cycle); label("K",(4+4*sqrt(3),4+6*sqrt(2)),E); draw((4+4*sqrt(3),4+6*sqrt(2))--(8,0),dashed); label("H",(-4-6*sqrt(2),-4-6*sqrt(2)),S); label("G",(-8-6*sqrt(2),-4),W); label("32",(-10,-8),N); label("18",(0,6*sqrt(2)+2),N); [/asy]$

## Solution

Clearly, since $\overline{FE}$ is a side of a square with area $32$, $\overline{FE} = \sqrt{32} = 4 \sqrt{2}$. Now, since $\overline{FE} = \overline{BC}$, we have $\overline{BC} = 4 \sqrt{2}$.

Now, $\overline{JB}$ is a side of a square with area $18$, so $\overline{JB} = \sqrt{18} = 3 \sqrt{2}$. Since $\Delta JBK$ is equilateral, $\overline{BK} = 3 \sqrt{2}$.

Lastly, $\Delta KBC$ is a right triangle. We see that $\angle JBA + \angle ABC + \angle CBK + \angle KBJ = 360^{\circ} \rightarrow 90^{\circ} + 120^{\circ} + \angle CBK + 60^{\circ} = 360^{\circ} \rightarrow \angle CBK = 90^{\circ}$, so $\Delta KBC$ is a right triangle with legs $3 \sqrt{2}$ and $4 \sqrt{2}$. Now, its area is $\dfrac{3 \sqrt{2} \cdot 4 \sqrt{2}}{2} = \dfrac{24}{2} = \boxed{\textbf{(C)}~12}$.

~savannahsolver