Q.3 (B)(3) Seg PM is a median of Δ PQR , PM = 9 and PQ² + PR² = 290, then find QR.

Answer: QR = 16 By Apollonius theorem, since PM is a median (QM = MR = 1/2 QR), PQ² + PR² = 2PM² + 2QM² . 290 = 2(9)² + 2QM² = 162 + 2QM² ∴ 2QM² = 128 , so QM² = 64 and QM = 8. ∴ QR = 2 × QM = 2 × 8 = 16.

Seg PM is a median of \(\Delta PQR\), PM = 9 and PQ2 + PR2 = 290, then find QR.

Q.3 (B)(3) Seg PM is a median of \(\Delta PQR\), PM = 9 and PQ² + PR² = 290, then find QR.

Doubtly Asked 25 Oct 2026 ssc 10th Save WhatsApp Report

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Answer: QR = 16

By Apollonius theorem, since PM is a median (QM = MR = \(\tfrac12\)QR),

\(PQ^2 + PR^2 = 2PM^2 + 2QM^2\).

\(290 = 2(9)^2 + 2QM^2 = 162 + 2QM^2\)

∴ \(2QM^2 = 128\), so \(QM^2 = 64\) and QM = 8.

∴ QR = 2 × QM = 2 × 8 = 16.

Doubtly AI · 25 Oct 2026 AI-verified · Oct 2026 Report

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Seg PM is a median of \(\Delta PQR\), PM = 9 and PQ2 + PR2 = 290, then find QR.

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