Question

The distance between two parallel lines is unity. A point $P$ lies between the lines at a distance of $m$ from one of them. An equilateral trianlge $PQR$ is drawn where $R$ lies on one of the parallel line and $Q$ lies on the other line. Then

• A. Minimum area of $\Delta PQR=\frac{\sqrt{3}}{4}$
• B. Maximum area of $\Delta PQR=\frac{1}{\sqrt{3}}$
• C. Maximum area of $\Delta PQR$ approches $\frac{1}{\sqrt{3}}$
• D. For minimum area of $\Delta PQR,$ $m=\frac{1}{2}$

Answer 1

Answer: (A), (C), (D)

The correct options are
A Minimum area of $\Delta PQR=\frac{\sqrt{3}}{4}$
C Maximum area of $\Delta PQR$ approches $\frac{1}{\sqrt{3}}$
D For minimum area of $\Delta PQR,$ $m=\frac{1}{2}$

Let one side of the triangle be $x$
then $\sin \theta =\frac{1-m}{x}\sin (60°-\theta )=\frac{m}{x}$
Eliminating $\theta$, we get $x=\frac{2}{\sqrt{3}}\sqrt{m^2-m+1}$
Now, $0<m<1$
$x$ approches the maximum value of $\frac{2}{\sqrt{3}}$ when $m=0\text{or}1$
So, the maximum value of the area of trianlge $PQR$ approches $\frac{1}{\sqrt{3}}$
minimum lenght of the side is $1$ at $m=\frac{1}{2}$
So, the minimum area will be $\frac{\sqrt{3}}{4}$