Chapter 11: Circles (Exercise 11.2)

Chapter 11: Circles (Exercise 11.2)

Question 1:
In the given figure, O is the center of the circle. If ∠AOB = 100°, find the measure of ∠ACB.

Solution:
* According to the Circle Theorem, 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.

∠AOB = 2 × ∠ACB
100° = 2 × ∠ACB
∠ACB = 100° / 2
∠ACB = 50°

Question 2:
In a circle, ABCD is a cyclic quadrilateral. If ∠A = 80°, find the measure of the opposite angle ∠C.

Solution:
* The sum of opposite angles of a cyclic quadrilateral is always supplementary (180°).

∠A + ∠C = 180°
80° + ∠C = 180°
∠C = 180° - 80°
∠C = 100°

Question 3:
ABCD is a cyclic quadrilateral in which AB ∥ CD. If ∠B = 70°, find the measures of all the remaining angles.

Solution:
1. Finding ∠D:
Opposite angles of a cyclic quadrilateral sum to 180°:
∠B + ∠D = 180°
70° + ∠D = 180° ⇒ ∠D = 110°

2. Finding ∠C:
Since AB ∥ CD, consecutive interior angles on the same side of transversal BC sum to 180°:
∠B + ∠C = 180°
70° + ∠C = 180° ⇒ ∠C = 110°

3. Finding ∠A:
Opposite angles ∠A and ∠C sum to 180°:
∠A + ∠C = 180°
∠A + 110° = 180° ⇒ ∠A = 70°

* Final Angles: ∠A = 70°, ∠B = 70°, ∠C = 110°, ∠D = 110°.

Question 4:
Prove that any cyclic parallelogram is a rectangle.

Solution:
* Let ABCD be a cyclic parallelogram.

1. Opposite angles of a parallelogram are equal:
∠A = ∠C --- (Equation 1)

2. Opposite angles of a cyclic quadrilateral sum to 180°:
∠A + ∠C = 180° --- (Equation 2)

* Substituting Equation 1 into Equation 2:
∠A + ∠A = 180°
2 ∠A = 180°
∠A = 90°

* Since a parallelogram with one right angle is a rectangle, ABCD is a rectangle.
(Hence proved.)