Answer:
Since, S and T trisect the side QR QS = TS = RT Let’s assume QS = TS = RT = x QR = 3x and QT = 2x
In Right angled triangle PQR, PR2 = PQ2 + QR2 PR2 = PQ2 +(3x)2 PR2 = PQ2 + 9x2
In Right angled triangle PQS, PS2 = PQ2 +QS2 PS2 = PQ2 +x2
In Right angled triangle PQT, PT2 = PQ2 + OT2 PT2 = PQ2 + (2x)2 PT2 =
PQ2 + 4x2
To prove that, 8PT2 =3 PR2 + 5PS2
LHS = 8PT2 = 8(PQ2 + 4x2)
= 8PQ2 + 32x2 RHS
= 3 PR2 + 5PS2
= 3(PQ2 + 9x2) – 5 (PQ2 + x2)
= 3PQ2 + 27x2 + 5PQ2 + 5x2
= 8PQ2 + 32x2
LHS = RHS
Hence, proved
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Answers & Comments
Answer:
Since, S and T trisect the side QR QS = TS = RT Let’s assume QS = TS = RT = x QR = 3x and QT = 2x
In Right angled triangle PQR, PR2 = PQ2 + QR2 PR2 = PQ2 +(3x)2 PR2 = PQ2 + 9x2
In Right angled triangle PQS, PS2 = PQ2 +QS2 PS2 = PQ2 +x2
In Right angled triangle PQT, PT2 = PQ2 + OT2 PT2 = PQ2 + (2x)2 PT2 =
PQ2 + 4x2
To prove that, 8PT2 =3 PR2 + 5PS2
LHS = 8PT2 = 8(PQ2 + 4x2)
= 8PQ2 + 32x2 RHS
= 3 PR2 + 5PS2
= 3(PQ2 + 9x2) – 5 (PQ2 + x2)
= 3PQ2 + 27x2 + 5PQ2 + 5x2
= 8PQ2 + 32x2
LHS = RHS
Hence, proved
Answer:
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