Question

A circle of radius 5 cm has a chord of length 8 cm.
Tangents drawn at the end points of the chord meet at
a point. Find the perimeter of the triangle formed by
the chord and the tangents.​

Answer 1

Answer:

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THIS WAS HARD TO TYPE ✌️✌️

Step-by-step explanation:

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.