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       Rate this question :

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