Q. 133.8( 24 Votes )

PQ is a chord of

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]


PT2 = (2.4)2 + (2.9)2


PT2 = 4.76 + 8.41


PT2 = 13.17


PT = 3.63 cm [Appx]

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