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24. PQ is a chord of length 8 cm in a circle of radius 5 cm. Tangents at P and Q intersect atT. Find the length of PT. |
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Answer» Given radius, OP = OQ = 5 cmLength of chord, PQ = 4 cmOT ⊥ PQ,∴ PM = MQ =4 cm [Perpendicular draw from the centre of the circle to a chord bisect the chord]In right ΔOPM, OP²= PM²+ OM²⇒ 52= 42+ OM²⇒ OM²= 25 – 16 = 9 Hence OM = 3cmIn right ΔPTM,PT²= TM²+ PM²→(1)∠OPT = 90º [Radius is perpendicular to tangent at point of contact] In right ΔOPT,OT2²= PT²+ OP²→(2) From equations (1) and (2), we get OT²= (TM²+ PM²) + OP2²⇒ (TM + OM)²= (TM²+ PM²) + OP²⇒ TM²+ OM²+ 2 × TM × OM = TM²+ PM²+ OP²⇒ OM²+ 2 × TM × OM = PM2+ OP²⇒ 32+ 2 × TM × 3 = 42+ 52⇒ 9 + 6TM = 16 + 25⇒ 6TM = 32⇒ TM =32/6 = 16/3 Equation (1) becomes, PT²= TM²+ PM² = (16/3)2+ 42 =(256/9) + 16 = (256 + 144)/9 = (400/9) = (20/3)2⇒PT = 20/3∴the length of tangent PT is(20/3) cm |
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