Question

PQ is a chord of length 16 cm of a circle of radius 10 cm the tangent at P and Q intersect at point P find the length of TP

Answer 1

Bhai point T mark kaha hai

Answer 2

Answer:

Let LP be x

AB ➳ AL = LB = 8 cm

OA = OB = 10 cm

Now, In ∆ OAL, where ∠ L = 90°

By Phythagoras theorem :

➳(Hypotenuse)² = (Perpendicular)² + (Base)²

➳ (OA)² = (OL)² + (AL)²

➳ 10² = (OL)² + 8²

➳ 100 = (OL)² + 64

➳ OL² = 100 - 64

➳ OL² = 36

➳ OL = √36

➳ OL = 6 cm

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In ∆ LAP, where ∠ L = 90°

By Phythagoras theorem :

⇢(AP)² = (AL)² + (LP)²

⇢(AP)² = 8² + (LP)²

⇢AP² = 8² + x².........[Equation (i)]

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In ∆ OAP, where ∠ A = 90°

By Phythagoras theorem :

➙ (OP)² = (OA)² + (AP)²

➙ (OL + LP)² = 10² + (AP)²

➙ AP² = -100 + (6 + x)²

➙ AP² = -100 + 36 + x² + 12x........[Equation (ii)]

Now,from equation (i) and (ii) we get,

➙ 64 + x² = -100 + 36 + x² + 12x

➙ x² - x² = -100 + 36 - 64 + 12x

➙ 12x = 128

➙ x = 128/12

➙ x = 32/3 cm

So,

➥ AP² = 64 + x²

➥ AP² = 64 + (32/3)²

➥ AP² = 576 + 1024/9

➥ AP = √1600/√9

➥ AP = 40/3 cm

Therefore, AP is 40/3 cm