Question

please solve this question
class 10th circles​

Answer 1

Answer:

Your answer is 20 / 3

Step-by-step explanation:

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

Answer 2

Answer:

Hey,

The answer of ur question is:

Step-by-step explanation:

Hope you will get help from this...

Have a nice day..✌☀☀