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In figure, OQ : PQ = 3 : 4 and perimeter of ΔPDQ = 60 cm. Determine PQ, QR and OP.

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Question

In figure, OQ : PQ = 3 : 4 and perimeter of ΔPDQ = 60 cm. Determine PQ, QR and OP.

Sum
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Solution

Given OQ: PQ = 3 : 4

Let OQ = 3x PQ = 4x

OP = y

∠OQP = 90° [since at point of contact, tangent is perpendicular to radius]

In ΔOQP, by Pythagoras theorem

𝑂𝑃2 = 𝑂𝑄2 + 𝑄𝑃2

⇒ 𝑦2 = (3𝑥)2 + (4𝑥)2

⇒ 𝑦2 = 9𝑥2 + 16𝑥2 = 25𝑥2

⇒ 𝑦 = `sqrt(25x^2)` = 5𝑥

Perimeter = OQ + PQ + OP = 3x + 4x + 5x = 12x

According to problem perimeter = 60

∴ 12x = 60

x =`60/12`

= 5𝑐𝑚

OQ = 3 × 5 = 15𝑐𝑚

PQ = 4 × 5 = 20 𝑐𝑚

OP = 5 × 5 = 25𝑐𝑚

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Chapter 8: Circles - EXERCISE 8.2 [Page 8.32]

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R.D. Sharma Mathematics [English] Class 10
Chapter 8 Circles
EXERCISE 8.2 | Q 20. | Page 8.32
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