Question

$△ABC$ is an equilateral triangle such that the vertices $B$ and $C$ lie on two parallel lines at a distance $6$. If $A$ lies between the parallel lines at a distance $4$ from one of them, then the length of a side of the equilateral triangle is:

• A. $8$
• B. $\sqrt{\frac{88}{3}}$
• C. $\frac{4\sqrt{7}}{\sqrt{3}}$
• D. $88$

Answer 1

The correct option is C $\frac{4\sqrt{7}}{\sqrt{3}}$

Given,

$BO=6$ and $OA=4$

Let length of side of $\Delta ABC=a$

$acos\theta =6$ ----(1)

$\angle ABO=90-(60+\theta )$

$=30-\theta$

So, $asin(30-\theta )=4$ ----(2)

From 1 & 2,

$a(\frac{cos\theta }{2}-\frac{sin\theta \sqrt{3}}{2})=4$

$3-4=\frac{a\sqrt{3}}{2}\sqrt{1-\frac{36}{a^2}}$

$\Rightarrow 1=\frac{3a^2}{4}(1-\frac{36}{a^2})$

$\Rightarrow 1=\frac{3a^2}{4}-27$

$\Rightarrow a^2=\frac{28\times 4}{3}$

$\Rightarrow a=4\sqrt{\frac{7}{3}}$