(b) \(\frac1{\sqrt3}\)
∵ LM || QR
∠PLM = ∠PQR
∠PML = ∠PRQ Corresponding ∠s
⇒ ΔPQR ~ ΔPLM
∴ \(\frac{\text{Area of ΔPLM}}{\text{Area of ΔPQR}}=\frac{PL^2}{PQ^2}\) ......(i)
Area of ΔPQR = Area of ΔPLM + Area of trap. LMRQ
Given, Area of trapezium LMRQ = 2 Area of ΔPLM
∴ Area of ΔPQR = Area of ΔPLM + 2 Area of ΔPLM = 3 Area of ΔPLM
∴ From (i), we have
\(\frac{\text{Area of ΔPLM}}{3\times\text{Area of ΔPLM}}=\frac{PL^2}{PQ^2}\)
⇒ \(=\frac{PL^2}{PQ^2}\) = \(\frac13\) ⇒ \(\frac{PL}{PQ} = \frac1{\sqrt3}.\)