Question

In the given figure, S and T are points on the sides PQ and PR respectively of ∆PQR such that PT = 2 cm, TR = 4 cm and ST is parallel to QR. Find the ratio of the areas of ∆PST and ∆PQR.

Answer 1

Given: In ΔPQR, S and T are the points on the sides PQ and PR respectively such that PT = 2cm, TR = 4cm and ST is parallel to QR.

To find: Ratio of areas of ΔPST and ΔPQR

\[In ∆ PST and ∆ PQR, \]
\[\angle PST = \angle Q \left( \text{Corresponding angles} \right)\]
\[\angle P = \angle P \left( \text{Common} \right)\]
\[ \therefore ∆ PST ~ ∆ PQR \left( AA \hspace{0.167em} \text{Similarity} \right)\]

Now, we know that the areas of two similar triangles are in the ratio of the squares of the corresponding sides. Therefore,

`(Area(Δ PST))/(Area(Δ PQR))= (PT^2)/(PR^2)`

`(Area(Δ PST))/(Area(Δ PQR))= (PT^2)/(PT+TR)^2`

`(Area(Δ PST))/(Area(Δ PQR))= (2^2)/(2+4)^2`

`(Area(Δ PST))/(Area(Δ PQR))= 4/36=1/9`