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In a ΔPQR, angle Q = 90° and S is the mid-point of QR. Prove thatPR2= PS2+ 3RS2 |
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Answer» Answer: As per Pythagorean theorem PT2=PQ2+QT2....(in triangle QPT) ⇒QT2=PT2−PQ2 And PR2=PQ2+QR2 ( in the triangle APR and given T is the mide point QR then QR=2QT) PUT VALUE ⇒PR2=PQ2+(2QT)2 ⇒PR2=PQ2+4QT2 ⇒PR2=PQ2+4(PT2−PQ2) =PQ2+4PT2−4PQ2 ⇒PR2=4PT2−3PQ2. |
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