Question

In $\Delta PQR$, MN||QR and MN divides the triangle into two parts of equal areas. $\frac{QM}{PQ}=??$

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

Answer 1

The correct option is C

$\frac{\sqrt{2}-1}{\sqrt{2}}$

Since, MN||QR,

$\angle PMN=\angle PQR$ (Corresponding angles)

$\angle PNM=\angle PRQ$ (Corresponding angles)

Hence, by AA similarity,

$\Delta PMN\sim \Delta PQR.$

We know that

$\frac{Area\left(\Delta PMN\right)}{Area\left(\Delta PQR\right)}=\frac{P{M}^2}{P{Q}^2}=\frac{M{N}^2}{Q{R}^2}=\frac{P{N}^2}{P{R}^2}$

Since, Area $\left(\Delta PMN\right)$ =Area (quad. MNRQ) [Given]

We can say.

$\frac{Area\left(\Delta PMN\right)}{Area\left(\Delta PQR\right)}=\frac{1}{2}=\frac{P{M}^2}{P{Q}^2}$

or $\frac{PM}{PQ}=\frac{1}{\sqrt{2}}\Rightarrow 1-\frac{PM}{PQ}=1-\frac{1}{\sqrt{2}}$

$\Rightarrow \frac{PQ-PM}{PQ}=\frac{\sqrt{2}-1}{\sqrt{2}}$

$\Rightarrow \frac{QM}{PQ}=\frac{\sqrt{2}-1}{\sqrt{2}}$