question_answer

In a \[\Delta PQR,\] S and T are points on the sides PQ and PR respectively, such that \[ST\left\| QR \right.\]. If \[\text{PT}=2\text{cm}\] and \[\text{TR}=\text{4cm},\]then the ratio of the areas of \[\Delta PST\] and \[\Delta PQR\] is:

A) \[1:9\]

B) \[1:3\]

C) \[1:2\]

D) \[1:4\]

Correct Answer: A

Solution :

[a] In \[\Delta PST\] and \[\Delta PQR,\]

\[\angle PST=\angle Q\] (Corresponding angle)

\[\angle P=\angle P\]               (Common)

\[\Delta PST\tilde{\ }\Delta PQR\]           (AA similarity)

So,       \[\frac{ar(\Delta PST)}{ar(\Delta PQR)}=\frac{P{{T}^{2}}}{P{{R}^{2}}}=\frac{P{{T}^{2}}}{{{(PT+TR)}^{2}}}\]

(Given, \[PT=2\,cm,\,\,TR=4cm\])

\[=\frac{{{(2)}^{2}}}{{{(2+4)}^{2}}}=\frac{4}{36}=\frac{1}{9}\]