Q. 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.


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]


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