Mathematics
In the adjoining figure, △PQR is right angled at Q and points S and T trisect side QR. Prove that
8PT2 = 3PR2 + 5PS2.
Pythagoras Theorem
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Answer
Let RT = TS = SQ = x
In right angle triangle PQR,
By pythagoras theorem we get,
⇒ PR2 = QR2 + PQ2
⇒ PR2 = (3x)2 + PQ2
⇒ PR2 = 9x2 + PQ2
Multiplying above equation by 3 we get,
⇒ 3PR2 = 27x2 + 3PQ2 …….(i)
In right angle triangle PTQ,
By pythagoras theorem we get,
⇒ PT2 = QT2 + PQ2
⇒ PT2 = (2x)2 + PQ2
⇒ PT2 = 4x2 + PQ2
Multiplying above equation by 8 we get,
⇒ 8PT2 = 32x2 + 8PQ2 …….(ii)
In right angle triangle PSQ,
By pythagoras theorem we get,
⇒ PS2 = SQ2 + PQ2
⇒ PS2 = (x)2 + PQ2
⇒ PS2 = x2 + PQ2
Multiplying above equation by 5 we get,
⇒ 5PS2 = 5x2 + 5PQ2 …….(iii)
Adding (i) and (iii) we get,
⇒ 3PR2 + 5PS2 = 27x2 + 5x2 + 3PQ2 + 5PQ2
⇒ 3PR2 + 5PS2 = 32x2 + 8PQ2
From (ii) we get,
⇒ 3PR2 + 5PS2 = 8PT2.
Hence, proved that 8PT2 = 3PR2 + 5PS2.
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