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Answer» gure:Since OT is PERPENDICULAR bisector of PQ.So, PM = QM = (1/2) PQ = 8 cm.(i) In ΔOMP:(OP)² = (PM)² + (OM)²(10)² = (8)² + OM²100 - 64 = OM²36 = OM²OM = 6 cm.∴ Let TP = X.(ii) In ΔPMT:(TP)² = (PM)² + (MT)²⇒ x² = 8² + MT²⇒ x² = 64 + MT² (iii) In ΔOPT:(OT)² = (OP)² + (TP)²⇒ (OT)² = 10² + x²⇒ (OM + MT)² = 100 + x²⇒ (6 + MT)² = 100 + x²⇒ 36 + MT² + 2(6)(MT) = 100 + x²⇒ 36 + MT² + 12MT = 100 + (64 + MT²)⇒ 36 + MT² + 12MT - 100 - 64 - MT² = 0⇒ -128 + 12MT = 0⇒ MT = 128/12Substitute in (ii), we get⇒ x² = 64 + (128/12)²⇒ x² = 64 + (16384/144)⇒ x² = (9216 + 16384)/144⇒ x² = (25600/144)⇒ x = 160/12⇒ x = 13.33 cm.Therefore, TP = x = 13.33 cm.Hope it helps! |
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