In given figure, ST II RQ, PS = 3 cm and SR = 4 cm. Find the ratio of the area of ΔPST to the area of ΔPRQ.
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the ratio of the area of triangle PST to traingle PRQ is 9:16 by the theorem of areas of similar triangles
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Answer:
9:49
Step-by-step explanation:
Given
ST || RQ
PS= 3 cm
SR = 4cm
Proof :--
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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