Question

PQ is a chord of length 8 cm of a
circle of radius 5 cm. The tangents at P and Q
intersect at a point T . Find the
length TP. class X ​

Answer 1

Answer:

Joint OT.

Let it meet PQ at the point R.

Then ΔTPQ is isosceles and TO is the angle bisector of ∠PTO.

[∵TP=TQ= Tangents from T upon the circle]

∴OT⊥PQ

∴OT bisects PQ.

PR=RQ=4 cm

Now,

OR=

OP

2

−PR

2

=

5

2

−4

2

=3 cm

Now,

∠TPR+∠RPO=90

∘

(∵TPO=90

∘

)

=∠TPR+∠PTR(∵TRP=90

∘

)

∴∠RPO=∠PTR

∴ Right triangle TRP is similar to the right triangle

PRO. [By A-A Rule of similar triangles]

∴

PO

TP

=

RO

RP

⇒

5

TP

=

3

4

⇒TP=

3

20

cm.

solution