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In the given figure, LM is parallel to QR. If LM divides the ΔPQR such that the area of trapezium LMRQ is two times the area of ΔPLM, then what is \(\frac{PL}{QL}\) equal to?

(a) \(\frac1{\sqrt2}\) 

(b) \(\frac1{\sqrt3}\)

(c) \(\frac12\) 

(d) \(\frac13\)

1 Answer

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Best answer

(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

\(\frac{\text{Area of ΔPLM}}{3\times\text{Area of ΔPLM}}=\frac{PL^2}{PQ^2}\)

⇒ \(=\frac{PL^2}{PQ^2}\) = \(\frac13\) ⇒ \(\frac{PL}{PQ} = \frac1{\sqrt3}.\)

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