Math, asked by sahelirudraroy, 1 year ago

poeple help me out here-the figure is given below,
and please give the entire method for solving it.​

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Answers

Answered by paarth232
2

Hey mate here is your answer,

Plzz mark as BRAINLIEST if it was helpful!! <3

Given radius, OP = OQ = 5 cm

Length of chord, PQ = 4 cm

OT ⊥ 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 = 3cm

In 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


mukundshrivas004: PQ=8cm not 4cm . And also buddy , you haven't explained where M is ....So it is kind of confusing...
sahelirudraroy: @paarth232 can you please help out here,which pt. do you assume to be M?
paarth232: I have assumed point R as M...sry for the mistake buddy!
paarth232: I hv drawn the wrong figure but the method out here will be same
paarth232: ALSO PR=4cm as its half of PQ
paarth232: I can't edit it otherwise I would have corrected it :(
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