Question

The distance between two parallel lines is unity. A point $P$ lies between the lines at a distance ${}^{\prime }{a}^{\prime }$ from one of them. The length of a side of an equilateral triangle $PQR$, vertex $Q$ of which lies on one of the parallel lines and vertex $R$ lies on the other is

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

Answer 1

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

(b) Let $PQ=QR=RP=r$ and $\angle PQX=\theta$ then $\angle RQX=\frac{\pi }{3}+\theta$

Given$PL=a,RN=1$

Now $a=PL=r\sin \theta$ and $1=RN=r\sin (\frac{\pi }{3}+\theta )$

$\Rightarrow 1=r\left(\sin \frac{\pi }{3}\cos \theta +\cos \frac{\pi }{3}\sin \theta \right)$ $\Rightarrow r(\frac{\sqrt{3}}{2}\cos \theta +\frac{1}{2}\sin \theta )=1$

$\Rightarrow r(\frac{\sqrt{3}}{2}\sqrt{1-\frac{a^2}{r^2}}+\frac{1}{2}.\frac{a}{r})=1(\because \sin \theta =\frac{a}{r})$

$\frac{\sqrt{3}}{2}\sqrt{r^2-a^2}=1-\frac{a}{2}$ or $r^2-a^2=\frac{{(2-a)}^2}{3}=\frac{4+a^2-4a}{3}$

$\Rightarrow r^2=\frac{4+a^2-4a}{3}+a^2=\frac{4+4a^2-4a}{3}$

$\therefore r=\frac{2}{\sqrt{3}}\sqrt{a^2-a+1}.$