PQ is a chord of length 4.8 cm of a circle of radius 3 cm. The tangents at P and Q intersect at a point T as shown in the figure. Find the length of TP.

Question

PQ is a chord of length 4.8 cm of a circle of radius 3 cm. The tangents at P and Q intersect at a point T as shown in the figure. Find the length of TP.

Filo · Accepted Answer

Radius of the circle is 3cm and PQ=4.8cm.PQ is a chord of the circle with centre O.The tangents at P and Q intersect at point TSo, TP and TQ are tangentsOP and OT are joined.Join OQWe have two right triangles: ΔOPT and ΔOQTHere, OT=OT (Common)PT=QT (tangents of the circle)OP=OQ (radius of the same circle)By Side - Side - Side CriterionΔOPT≅ΔOQTTherefore, ∠POT=∠OQTAgain, from triangles ΔOPR and ΔOQROR=OR (Common)OP=OQ (radius of the same circle)∠POR=∠OQR (from above result)By Side - Angle - Side CriterionΔOPR≅ΔOQTTherefore, ∠ORP=∠ORQNow, ∠ORP+∠ORQ=180∘(Sum of linear angles =180 degrees)∠ORP+∠ORP=180∘∠ORP=90∘This implies, OR⊥PQ and RT⊥PQAlso OR perpendicular from center to a chord bisects the chord,PR=QR=PQ/2=4.8/2=2.4cmApplying Pythagoras Theorem on right triangle ΔOPR,(OP)2=(OR)2+(PR)2(3)2=(OR)2+(2.4)2OR=1.8cmApplying Pythagoras Theorem on right angled ΔTPR,(PT)2=(PR)2+(TR)2 ___(1)Also, OP⊥OTApplying Pythagoras Theorem on right ΔOPT,(PT)2+(OP)2=(OT)2[(PR)2+(TR)2]+(OP)2=(TR+OR)2(Using equation (1) and from figure)(2.4)2+(TR)2+(3)2=(TR+1.8)24.76+(TR)2+9=(TR)2+2(1.8)TR+(1.8)213.76=3.6TR+3.24TR=2.9cm [approx.]From (1) =PT2=(2.4)2+(2.9)2PT2=4.76+8.41or PT=3.63cm [approx.]Answer: Length of PT is 3.63cm.

Question Text	PQ is a chord of length 4.8 cm of a circle of radius 3 cm. The tangents at P and Q intersect at a point T as shown in the figure. Find the length of TP.
Answer Type	Text solution:1
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PQ is a chord of length 4.8 cm of a circle of radius 3 cm. The tangents at P and Q intersect at a point T as shown in the figure. Find the length of TP.

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