Question

The distance between the two parallel lines is $1$ unit. A point '$A$' is chosen to lie between the lines at a distance '$d$' from one of them. Triangle $ABC$ is equilateral with $B$ on one line and $C$ on the other parallel line. Find the length of the side of the equilateral triangle is

• A. $\frac{2}{3}\sqrt{d^2+d+1}$
• B. $2\sqrt{\frac{d^2-d+1}{3}}$
• C. $2\sqrt{d^2-d+1}$
• D. $\sqrt{d^2-d+1}$

Answer 1

The correct option is B $2\sqrt{\frac{d^2-d+1}{3}}$

Let the sides of the equilateral triangle be $a$.

From the picture we have for the triangles $\Delta$ ACD and $\Delta$ DAB,

$\angle$ACD $=\angle$ DBA $={60}^{\circ }$, $AC=AB=a$, $AD$ is the common side.

So, $\Delta$ACD $\cong \Delta$ DAB. ($S-A-S$ rule)

Then $\angle$CAD $=\angle$ DAB $\Rightarrow \angle$FAB $=\angle$CAE.

So, $\tan \angle$FAB $=\tan \angle$CAE

$\Rightarrow \frac{\sqrt{a^2-d^2}}{d}=\frac{\sqrt{a^2-(1-d{)}^2}}{(1-d)}$

or, $d^2\{a^2-(1-d{)}^2\}=(1-d{)}^2(a^2-d^2)$

or, $a^2(1-d{)}^2-a^2-d^2=0$

or, $a^2(-2d+1)=0$

$\Rightarrow d=\frac{1}{2}$.

If $d=\frac{1}{2}$ then $CE=BF\Rightarrow a=1$ [Since in this condition $BC$ is nothing but the shortest distance between the given parallel lines.]

For, $d=\frac{1}{2}$, $d^2-d+1=\frac{3}{4}$ and $2\sqrt{\frac{d^2-d+1}{3}}$ $=1$.

So, option (B) is correct.