Question

pq is a chord of length 8cm of a circle of radius 5cm . The length at p and q intersect at a point T . find the length TP ?​

Answer 1

Answer:

$\large \blue{Solution}$☟︎︎︎

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= ⟌OP² - PR² = ⟌5² - 4² -3cm.

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{Upon} -RP__RO⇒TP__5 -4By3.

⇒TP= 3upon20 cm.

$\large \blue{Answer}$☞︎︎︎TP= 3upon20 cm.

$\large \orange{Mabelrose♡︎}$