In Figure Oq : Pq = 3 : 4 and Perimeter of δPdq = 60cm. Determine Pq, Qr and Op. - Mathematics

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

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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 [Page 37]

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RD Sharma Class 10 Maths
Chapter 8 Circles
Exercise 8 | Q 10 | Page 37

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