Question

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 $\triangle KBC?$

$6\sqrt{2}$

$9$

$12$

$9\sqrt{2}$

$32$

Video solution:

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Written solution:

The square with area $18$ has side length $JB=\sqrt{18}=3\sqrt2$ . Since $\triangle JBK$ is equilateral, $BK=3\sqrt2$ .

The square with area $32$ has side length $FE=\sqrt{32}=4\sqrt2$ . Since $FE=BC$ , we have $BC=4\sqrt2$ .

In the equiangular hexagon configuration, $BK\perp BC$ . Therefore $[\triangle KBC]=\dfrac12(3\sqrt2)(4\sqrt2)=12.$

Thus, C is the correct answer.

2015 AMC 8 Problem 21

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Written solution:

Problem 21 in Other Years