Class 9 MathematicsChapter 9: Circles
Exercise 9.3 – Stepwise Solutions
Important Concepts Used
1. Angle at the centre is twice the angle at the circumference.
2. Angles in the same segment of a circle are equal.
3. Sum of opposite angles of a cyclic quadrilateral is 180°.
4. Angle in a semicircle is 90°.
5. The angle subtended by an arc at the centre is twice the angle subtended at any point on the circle.
Q1. Find ∠ADC.
Given:
∠AOB = 60°, ∠BOC = 30°
So,
∠AOC = 60° + 30° = 90°
Angle at the centre = 90°
Angle at circumference (same arc AC):
∠ADC = ½ × 90° = 45°
Q2. A chord of a circle is equal to the radius. Find the angle subtended by the chord.
Let the chord be AB and centre be O.
Given:
OA = OB = AB
So, triangle AOB is equilateral.
Therefore,
∠AOB = 60°
Angle on minor arc = ½ × 60° = 30°
Angle on major arc = ½ × (360° − 60°) = ½ × 300° = 150°
Q3. Find ∠QPR.
Given:
∠PQR = 100°
Angle at centre:
∠POR = 2 × 100° = 200°
Remaining angle (minor arc PR):
= 360° − 200° = 160°
Now, ∠QPR = ½ × 160° = 80°
Q4. Find ∠BDC.
Given:
∠ABC = 69°, ∠ACB = 31°
In triangle ABC:
∠BAC = 180° − (69° + 31°) = 80°
Angles in the same segment are equal, so:
∠BDC = ∠BAC = 80°
Q5. Find ∠BAC.
Given:
∠BEC = 130°, ∠ECD = 20°
Using intersecting chords property:
∠BEC = ½ (arc BC + arc AD)
So,
130° = ½ (arc BC + arc AD)
arc BC + arc AD = 260°
Also,
∠ECD = ½ arc ED
20° = ½ arc ED
arc ED = 40°
Remaining arc for BAC:
= 360° − (260° + 40°) = 60°
Therefore,
∠BAC = ½ × 60° = 30°
Q6. Find ∠BCD and ∠ECD.
Given:
∠DBC = 70°, ∠BAC = 30°
Angles in the same segment:
∠DAC = ∠DBC = 70°
In triangle ABC:
∠ABC = 180° − (30° + 70°) = 80°
Opposite angles of cyclic quadrilateral:
∠BCD = 180° − 80° = 100°
Given AB = BC → triangle ABC is isosceles
So,
∠BAC = ∠ACB = 30°
Thus,
∠ECD = 30°
Q7. Prove that the quadrilateral is a rectangle.
If diagonals are diameters, then each angle subtended in a semicircle is 90°.
So, all angles of the quadrilateral are 90°.
Hence, the quadrilateral is a rectangle.
Q8. Prove that the trapezium is cyclic.
If non-parallel sides are equal, trapezium becomes isosceles.
In an isosceles trapezium, base angles are equal.
So, opposite angles sum to 180°.
Hence, the quadrilateral is cyclic.