Let ABC be an equilateral triangle,and let KL | vedclass

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(B) Given that $ABC$ is an equilateral triangle,all its angles are $60^{\circ}$.Since $KLMN$ is a rectangle,$NK \perp BC$ and $NM \parallel BC$.In $

Vedclass · Accepted Answer

$108$

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(B) Given that $ABC$ is an equilateral triangle,all its angles are $60^{\circ}$.Since $KLMN$ is a rectangle,$NK \perp BC$ and $NM \parallel BC$.In $	riangle BKN$,$\angle B = 60^{\circ}$ and $\angle BKN = 90^{\circ}$,so $\angle KNB = 30^{\circ}$.We have $\frac{AN}{NB} = 2$,which implies $AN = 2NB$. Thus,$AB = AN + NB = 3NB$.In $	riangle BKN$,$NK = BN \sin 60^{\circ} = BN \frac{\sqrt{3}}{2}$ and $BK = BN \cos 60^{\circ} = \frac{BN}{2}$.The area of $	riangle BKN = \frac{1}{2} 	imes BK 	imes NK = \frac{1}{2} 	imes \frac{BN}{2} 	imes \frac{BN \sqrt{3}}{2} = \frac{BN^2 \sqrt{3}}{8}$.Given the area is $6$,we have $\frac{BN^2 \sqrt{3}}{8} = 6$,so $BN^2 = \frac{48}{\sqrt{3}} = 16\sqrt{3}$.The side length of the equilateral triangle $ABC$ is $s = AB = 3BN$.Therefore,$s^2 = 9BN^2 = 9 	imes 16\sqrt{3} = 144\sqrt{3}$.The area of $	riangle ABC = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} 	imes 144\sqrt{3} = \frac{144 	imes 3}{4} = 36 	imes 3 = 108$.

Let $ABC$ be an equilateral triangle,and let $KLMN$ be a rectangle with $K, L$ on $BC$,$M$ on $AC$,and $N$ on $AB$. Suppose $AN / NB = 2$ and the area of $\triangle BKN$ is $6$. The area of the $\triangle ABC$ is:

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