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CBSE Class 9 Mathematics Question 6 of 12

Circles — Question 3

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Question 3

In Fig. ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.

In Fig. ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR. NCERT Class 9 Mathematics CBSE Solutions.
Answer

From figure,

Consider PR as a chord of the circle.

Steps of construction :

  1. Mark any point S on the major arc of the circle. (on the side opposite to point Q)

  2. Join PS and SR.

PQRS is a cyclic quadrilateral.

In Fig. ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR. NCERT Class 9 Mathematics CBSE Solutions.

⇒ ∠PQR + ∠PSR = 180° (Sum of opposite angles in a cyclic quadrilateral = 180°)

⇒ 100° + ∠PSR = 180°

⇒ ∠PSR = 180° - 100° = 80°.

We know that,

The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

∴ ∠POR = 2∠PSR

⇒ ∠POR = 2 × 80°

⇒ ∠POR = 160°.

In ∆ OPR,

⇒ OP = OR (Radius of the circle)

We know that,

Angles opposite to equal sides are equal.

∴ ∠OPR = ∠ORP = a (let).

We know that,

Sum of all angles in a triangle is 180°.

⇒ ∠OPR + ∠POR + ∠ORP = 180°

⇒ a + 160° + a = 180° [∵ ∠OPR = ∠ORP]

⇒ 2a = 180° - 160°

⇒ 2a = 20°

⇒ a = 20°2\dfrac{20°}{2} = 10°

⇒ ∠OPR = 10°.

Hence, ∠OPR = 10°.

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Circles — Interactive Study Guide

Circle Theorems Quick Reference

  1. Equal chords ⇔ equal central angles
  2. Perpendicular from centre bisects the chord
  3. Equal chords ⇔ equidistant from centre
  4. Central angle = 2 × inscribed angle (same arc)
  5. Angles in same segment are equal
  6. Angle in semicircle = 90°
  7. Opposite angles of cyclic quad = 180°

Problem-Solving Toolkit

For chord problems: Drop perpendicular from centre → bisects chord → use Pythagoras.
For angle problems: Identify if angle is at centre or circumference → apply the 2x rule.
For cyclic quad: Opposite angles add up to 180°.

Quick Self-Check

  1. Inscribed angle = 35°. Central angle for the same arc? (70°)
  2. Chord = 10 cm, distance from centre = 12 cm. Radius? (√(25+144) = √169 = 13 cm)
  3. ABCD is cyclic, ∠A = 95°. Find ∠C. (85°)

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