Question

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

• A. 48
• B. 54
• C. 96
• D. 108

Answer 1

Answer: (D)

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