Question

Let $ABC$ be an equilateral triangle, 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 $\Delta BKN$ is $6$ the area of the $\Delta ABC$ is

• A. 54
• B. 108
• C. 48
• D. not determinable with the above data

Answer 1

Answer: (B)

Given,

$ABC$ is an equilateral triangle
$\therefore AB=BC=AC$
$KLMN$ be a rectangle
$\therefore KL=MN$
and $NK=LM$
$\frac{AN}{NB}=2$
$\therefore AN=2NB=AM=MN$
$AB=AN+NB=3NB$
Area of $\Delta BKN=6$
In $\Delta BKN$,
$sin60^{\circ }=\frac{NK}{BN}$
$\Rightarrow NK=\frac{\sqrt{3}}{2}BN$
Area of $\Delta BKN=\frac{1}{2}\cdot BN\cdot NKsin30^{\circ }$
$\Rightarrow 6=\frac{1}{2}.BN.\frac{\sqrt{3}}{2}.BN.\frac{1}{2}$
$\Rightarrow B{N}^2=\frac{48}{\sqrt{3}}$
$\therefore$ Area of $\Delta ABC=\frac{\sqrt{3}}{4}\times A{B}^2$
$=\frac{\sqrt{3}}{4}\times 9B{N}^2=\frac{\sqrt{3}}{4}\times 9\times \frac{48}{\sqrt{3}}=108$