Question

in the given figure PQ is a chord of length 16 cm of a circle of radius 10 centimetre the tangents at P and Q intersect at a point t find the length of TP ​

Answer 1

Answer:

Given : A circle with center O and radius 3 cm and PQ is a chord of length 4.8 cm. The tangents at P and Q intersect at point T

To Find : Length of PT

Construction : Join OQ

Now in △OPT and △OQT

OP = OQ

[radii of same circle]

PT = PQ

[tangents drawn from an external point to a circle are equal]

OT = OT

[Common]

△OPT ≅ △OQT

[By Side - Side - Side Criterion]

∠POT = ∠OQT

[Corresponding parts of congruent triangles are congruent]

or ∠POR = ∠OQR

Now in △OPR and △OQR

OP = OQ

[radii of same circle]

OR = OR [Common]

∠POR = ∠OQR [Proved Above]

△OPR ≅ △OQT

[By Side - Angle - Side Criterion]

∠ORP = ∠ORQ

[Corresponding parts of congruent triangles are congruent]

Now,

∠ORP + ∠ORQ = 180°

[Linear Pair]

∠ORP + ∠ORP = 180°

∠ORP = 90°

⇒ OR ⏊ PQ

⇒ RT ⏊ PQ

As OR ⏊ PQ and Perpendicular from center to a chord bisects the chord we have

PR = QR = PQ/2 = 16/2 = 8 cm

∴ In right - angled △OPR,

By Pythagoras Theorem

[i.e. (hypotenuse)2 = (perpendicular)2 + (base)2 ]

(OP)2 = (OR)2 + (PR)2

(10)2 = (OR)2 + (8)2

100 = (OR)2 + 64

(OR)2= 36

OR = 6 cm

Now,

In right angled △TPR, By Pythagoras Theorem

(PT)2 = (PR)2 + (TR)2 [1]

Also, OP ⏊ OT

[Tangent at any point on the circle is perpendicular to the radius through point of contact]

In right angled △OPT, By Pythagoras Theorem

(PT)2 + (OP)2 = (OT)2

(PR)2 + (TR)2 + (OP)2= (TR + OR)2 [From 1]

(8)2 + (TR)2 + (10)2 = (TR + 6)2

64 + (TR)2 + 100 = (TR)2 + 2(6)TR + (6)2

164 = 12TR + 36

12TR = 128

TR = 10.7 cm [Appx]

Using this in [1]

PT2 = (8)2 + (10.7)2

PT2 = 64 + 114.49

PT2 = 178.49

PT = 13.67 cm [Appx]