Question

PQ is a chord of length 8cn of a circle of radius 5cm. the tagents P and Q intersect at a point T. find the length TP.​

Answer 1

Answere: ⬇️⬇️

➡️ 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,

Now, ∠TPR+∠RPO=90° (∵TPO=90° )

=∠TPR+∠PTR(∵TRP=90°)

∴∠RPO=∠PTR

∴ Right triangle TRP is similar to the right triangle

∴∠RPO=∠PTR

∴ Right triangle TRP is similar to the right triangle PRO.

[By A-A Rule of similar triangles]

∴ TP/PO = RP/RO⇒ TP/5 = 4/3

⇒TP = 20/3 cm.

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Answer 2

Step-by-step explanation:

In ∆ TRP and ∆ TRQ

TP=TQ ( tangents)

TR=TR ( common)

L TPR= L TQR (PQ is a straight line connecting tangent

Also OT passes through chord PQ which is a point of tangents

hence OT is perpendicular to PQ

and ∆TRP ≈ ∆ TRQ

SO,PR=RQ= 1/2 ×8cm= 4cm

In ∆ TRP

TP²= PR²+TR²

TP²= (4cm)²+TR²

=16cm²+TR²(1)

In ∆TPO

TO²=TP²+PO²

(TR+RO)²=(5cm)²+TP²

TR²+RO²+2TR.RO= 25cm²+TP²

from (1) (TP²= 16cm²+TR²)

TR²+[(5cm)²-(4cm)²]+ 2TR.[√(5cm)²-(4cm)² =25cm²+16cm²+TR² (in ∆PRO;PO²=RO²+RP²)

(25cm²-16cm²)+2TR(√25cm²-16cm²)=41cm²

9cm²+2TR×(√9cm²)= 41cm²

2TR×3cm= 41cm²-9cm²

TR= 32cm²/6cm= 16cm/3

Now, from eq (1)

TP²= 16cm²+ TR²

TP²= 16cm²+(16cm/3)²

TP= √16cm²+256cm²/9

=√(144cm²+256cm²)/9

= √400cm²/9

= 20cm/3

You can also find the value of RO from ∆ RPO in the beginning and avoid all those lengthy steps.