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  4. Let ABC be an equilateral triangle and suppose KLMN be a rectangle with K, L on BC, M on AC and N on AB. If AN = 2NB and area of triangle BKN is 6, then area of triangle ABC is equal to

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Q. Let ABC be an equilateral triangle and suppose KLMN be a rectangle with K, L on BC, M on AC and N on AB. If AN = 2NB and area of triangle BKN is 6, then area of triangle ABC is equal to

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Solution:

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From $\triangle B K N, \frac{a}{x}=\sin 60^{\circ} \Rightarrow a=\frac{\sqrt{3} x}{2}$ ....(1)
Also, area $(\triangle B K N)=6$
$\Rightarrow \frac{1}{2}(a x) \sin 30^{\circ}=6 \Rightarrow a=\frac{24}{x}$ ...(2)
$\therefore$ From (1) and (2), we get $x^2=\frac{48}{\sqrt{3}}$ .....(3)
Now, area $(\triangle ABC )=\frac{\sqrt{3}}{4}(3 x )^2=\frac{\sqrt{3}}{4} \times 9\left(\frac{48}{\sqrt{3}}\right)=108$