Question
In the following figure, PQ is the diameter of the circle with centre O and QR is the tangent to the circle at Q.
PR intersects the circle at S and the tangent to the circle as S intersects QR at T. If the diameter of the circle is 20 cm and TR=6 cm, find the approximate length of the line segment PR.
PR intersects the circle at S and the tangent to the circle as S intersects QR at T. If the diameter of the circle is 20 cm and TR=6 cm, find the approximate length of the line segment PR.
Answer: Option D
:
D
Clearly. ∠PSQ=90∘[Angle in a semi circle]
Also, TQ=TS [Tangents from an external point are equal ] ... (1)
Let ∠TQS=∠TSQ=x
Clearly, ∠TSR=90∘ - x
Also in ΔQSR, ∠QRS=90∘-x
:.TR=TS ... (2)
From (1) and (2), QT=TR
Given that TR=6 cm.
⇒ QR=QR+TR =6+6=12cm
Now, the length of the line segment PR = √PQ2+QR2
=√202+122=√544=√16×34
=4√34 cm=22.5 cm
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:
D
Clearly. ∠PSQ=90∘[Angle in a semi circle]
Also, TQ=TS [Tangents from an external point are equal ] ... (1)
Let ∠TQS=∠TSQ=x
Clearly, ∠TSR=90∘ - x
Also in ΔQSR, ∠QRS=90∘-x
:.TR=TS ... (2)
From (1) and (2), QT=TR
Given that TR=6 cm.
⇒ QR=QR+TR =6+6=12cm
Now, the length of the line segment PR = √PQ2+QR2
=√202+122=√544=√16×34
=4√34 cm=22.5 cm
Was this answer helpful ?
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