Question

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

Answer 1

Answer:

The ratio of the ar(∆PST) : ar(∆PQR) is  1 : 9.

Step-by-step explanation:

Given:

ST || QR

PT = 2 cm

TR = 4cm

In ΔPST and ΔPQR,

∠SPT = ∠QPR (Common)

∠PST = ∠PQR (Corresponding angles)

ΔPST ∼ ΔPQR (By AA similarity criterion)

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

ar(∆PST) /ar(∆PQR) = (PT)²/(PR)²

ar(∆PST) /ar(∆PQR) = 2²/(PT+TR)²

ar(∆PST) /ar(∆PQR) = 4 /(2 + 4)²

ar(∆PST) /ar(∆PQR) = 4 /6²  

ar(∆PST) /ar(∆PQR)= 4/36

ar(∆PST) /ar(∆PQR) = 1/9

ar(∆PST) : ar(∆PQR) = 1 : 9

Hence, the ratio of the ar(∆PST) : ar(∆PQR) is  1 : 9.

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