Question

PQR is a triangle. S is a point on the side QR of $\Delta$PQR such that $\angle$PSR$=\angle$QPR/ Given $QP=8$cm, PR$=6$cm and SR$=3$cm.
(I) Prove $\Delta$PQR$\sim \Delta$ SPR
(II) Find the length of QR and PS
(III) $\frac{areaof\Delta PQR}{areaof\Delta SPR}$.

Answer 1

Given that:

$QP=8cm,PR=6cm$ and $SR=3cm$

$\left(I\right)$ In $△PQR$ and $△SPR$

$\angle PRQ=\angle SRP$ (Common angle)

$\angle QPR=\angle PSR$ (Given that)

$\angle PQR=\angle PSR$ (Properties of triangle )

$\therefore △PQR\sim △SPR$ (By AAA)

$\left(II\right)$ $\frac{PQ}{SP}=\frac{QR}{PR}=\frac{PR}{SR}$ (Properties of similar triangles)

$\Rightarrow \frac{8cm}{SP}=\frac{6cm}{3cm}$

$\Rightarrow SP=4cm$ and

$\Rightarrow \frac{QR}{6cm}=\frac{6cm}{3cm}$

$\Rightarrow QR=12cm$

$\left(III\right)\frac{ar\left(△PQR\right)}{ar\left(△SPR\right)}=\frac{P{Q}^2}{S{P}^2}=\frac{8^2}{4^2}=4$