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 TP

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

∴ In right - angled △OPR,

By Pythagoras Theorem

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

(OP)^{2} = (OR)^{2} + (PR)^{2}

(3)^{2} = (OR)^{2} + (2.4)^{2}

9 = (OR)^{2} + 5.76

(OR)^{2}= 3.24

OR = 1.8 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]

(2.4)^{2} + (TR)^{2} + (3)^{2} = (TR + 1.8)^{2}

4.76 + (TR)^{2} + 9 = (TR)^{2} + 2(1.8)TR + (1.8)^{2}

13.76 = 3.6TR + 3.24

3.6TR = 10.52

TR = 2.9 cm [Appx]

Using this in [1]

PT^{2} = (2.4)^{2} + (2.9)^{2}

PT^{2} = 4.76 + 8.41

PT^{2} = 13.17

PT = 3.63 cm [Appx]

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