Q. 134.4( 52 Votes )

PQ is a chord of

Answer :


Given: PQ is a chord and PQ = 8cm


Radius of circle, r = 5cm


To find: Length of TP


Construction: Join OT


Let OT intersect PQ at R


We know that,


Lengths of tangents from external point are equal


So, TP = TQ


In Δ TPQ, TP = TQ


This means TPQ is an isosceles triangle as two sides are equal.


Here, OT is bisector ofPTQ,


So, OTPQ


[Angle bisector and altitude of an isosceles triangle are same]


So, PR = RQ


[Perpendicular from center to a chord bisects the chord]



In ΔORP, Using Pythagoras theorem


(Hypotenuse)2 = (Perpendicular)2 + (Base)2


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


(5)2 = (4)2 + OR2


25 = 16 + OR2


OR2 = 25 – 16


OR2 = 9


OR = √9


OR = 3


OR = 3cm …(i)



In ΔPRT, Using Pythagoras theorem


(Hypotenuse)2 = (Perpendicular)2 + (Base)2


(TP)2 = (PR)2 + (RT)2


(TP)2 = (4)2 + RT2


TP2 = 16 + RT2 …(ii)


In ΔOPT, Using Pythagoras theorem


(Hypotenuse)2 = (Perpendicular)2 + (Base)2


(OT)2 = (TP)2 + (OP)2


(OT)2 = 16 + RT2 + (5)2 [from (ii)]


(OR + RT)2 = 16 + RT2 + 25


(3 + RT)2 = 41 + RT2


9 + RT2 + 2 × 3 × RT = 41 + RT2


9 + 6RT = 41


6RT = 41 – 9


6RT = 32




Now, substituting the value of RT in eq. (ii), we get


TP2 = 16 + RT2









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