In the given question we are to find the ratio o
Given triangle ABC SIMILAR TO triangle PQR if AB/PQ =1/3 then find area ABC/PQR
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Let AD be the perpendicular dropped from A to the side BC and PS be the perpendicular dropped from P to the side QR respectively.
Since, ΔABC∼ΔPQR, we have ΔABD∼ΔPQS.
Therefore,
AB/PQ​=AD/PS​=BC/QR​
⇒arΔABC​/arΔPQR=AD×BC​/PS×QR
=AB/PQ​×AB/PQ
=AB^2​/PQ^2
=1/9​