Question

In $△PQR$, $LM\parallel QR$ and $PM:MR=3:4$. Calculate$\frac{Area\left(△LMN\right)}{Area\left(△MNR\right)}$

• A. $3:7$
• B. $3:8$
• C. $3:5$
• D. $3:2$

Answer 1

The correct option is A $3:7$

Given, $LM\parallel QR$, $PM:MR=3:4$
Construct: A perpendicular from M on LR meeting LR at J
$\frac{PM}{MR}=\frac{3}{4}$
$\frac{PM+MR}{MR}=\frac{3+4}{4}$
$\frac{PR}{MR}=\frac{7}{4}$
Now, In $△PLM$ and $△PQR$
$\angle LPM=\angle QPR$ (Common angle)
$\angle PLM=\angle PQR$ (Corresponding angles)
$\angle PML=\angle PRQ$ (Corresponding angles)
Thus, $△PLM\sim △PQR$ (AAA rule)
Hence, $\frac{PL}{PQ}=\frac{LM}{QR}=\frac{PM}{PR}$ (Corresponding sides)
$\frac{PL}{PQ}=\frac{LM}{QR}=\frac{3}{7}$
Now, In $△LMN$ and $△QNR$,
$\angle LNM=\angle QNR$ (Vertically Opposite angles)
$\angle NLM=\angle NRQ$ (Alternate angles)
$\angle NML=\angle NQR$ (Alternate angles)
$△LMN\sim △RQN$ (AAA rule)
thus, $\frac{LN}{NR}=\frac{LM}{QR}$ (Corresponding sides)
$\frac{LN}{NR}=\frac{3}{7}$
Area of triangle = $\frac{1}{2}base\times height$
Now, $\frac{A\left(△LMN\right)}{A\left(△MNR\right)}=\frac{\frac{1}{2}MJ\times LN}{\frac{1}{2}MJ\times NR}$
$\frac{A\left(△LMN\right)}{A\left(△MNR\right)}=\frac{LN}{NR}$
$\frac{A\left(△LMN\right)}{A\left(△MNR\right)}=\frac{3}{7}$