Question

In given fig. ST parallel RQ,PS=3cm and SR=4cm. Find the ratio of the area of trainglePST to the area of trianglePRQ

Answer 1

Given:

ST || RQ

PS= 3 cm

SR = 4cm

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(∆PRQ)= (PS)²/(PR)²

ar(∆PST) /ar(∆PRQ)= 3²/(PS+SR)²

ar(∆PST) /ar(∆PRQ)= 9/(3+4)²= 9/7²=9/49

Hence, the required ratio ar(∆PST) :ar(∆PRQ)= 9:49

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Answer 2

Answer:

Given:

ST || RQ

PS= 3 cm

SR = 4cm

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(∆PRQ)= (PS)²/(PR)²

ar(∆PST) /ar(∆PRQ)= 3²/(PS+SR)²

ar(∆PST) /ar(∆PRQ)= 9/(3+4)²= 9/7²=9/49

Hence, the required ratio ar(∆PST) :ar(∆PRQ)= 9:49

==================================================================

Hope this will help you...

MARK IT AS AN BRAINLIEST

THANKS..:D