Question

If in a triangles $XYZ,P,Q$ are points on $XY,XZ$ respectively such that $XP=2PY,XQ=2QZ,$ then the ratio, area of $\Delta XPQ:$ area of $\Delta XYZ,$ is:

• A. $4:9$
• B. $2:3$
• C. $3:2$
• D. $9:4$

Answer 1

The correct option is A $4:9$

Given: $\frac{XP}{XY}=\frac{2}{3}=\frac{XQ}{QZ}$

By converse of $BPT$, $PQ\parallel$$YZ$

so $XL\perp RQ⟹XM\perp YZ$

Referring figure,
$\frac{\text{Area of}\Delta XPQ}{\text{Area of}\Delta XYZ}=\frac{\frac{1}{2}\times PQ\times XL}{\frac{1}{2}\times YZ\times XM}$

$=\frac{PQ}{YZ}\times \frac{PQ}{YZ}$

$={\left(\frac{PQ}{YZ}\right)}^2={\left(\frac{2}{3}\right)}^2=\frac{4}{9}$