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In the adjoining figure, ∠QPR = 90° , seg PM ⊥ seg QR and Q – M – R, PM = 10, QM = 8, find QR. |
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Answer» In ∆PQR, ∠QPR = 90° and [Given] seg PM ⊥ seg QR ∴ PM2 = OM × MR [Theorem of geometric mean] ∴ 102 = 8 × MR ∴ MR = 100/8 = 12.5 Now, QR = QM + MR [Q – M – R] = 8 + 12.5 ∴ QR = 20.5 units |
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