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20

PQ is a chord of length 16 cm of a circle of radius 10 cm. The tangents at P and Q intersect at a point T as shown in the figure.

Find the length of TP.

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]

PT^{2} = (8)^{2} + (10.7)^{2}

PT^{2} = 64 + 114.49

PT^{2} = 178.49

PT = 13.67 cm [Appx]