CBSE Class 9 Mathematics Circles MCQs Set 07

Answer: (a) 5 cm

Explanation:
1. Half the chord = 12 cm.
2. d² = 13² − 12² = 169 − 144 = 25.
3. d = 5 cm.

Teacher's Note:
1. Pythagorean triple 5, 12, 13.

Answer: (d) \( \frac{1}{\sqrt{2}} \) AB

Explanation:
1. Let the radius be r. Then AB = 2r.
2. In right triangle AOC, AC = \( \sqrt{r^2 + r^2} = r\sqrt{2} \).
3. AC / AB = \( \frac{r\sqrt{2}}{2r} = \frac{1}{\sqrt{2}} \).

Teacher's Note:
1. Triangle AOC is right-angled and isosceles.

Answer: (b) 30°

Explanation:
1. OY = OZ = YZ, so triangle OYZ is equilateral and ∠YOZ = 60°.
2. ∠YXZ = ½ × 60° = 30°.

Teacher's Note:
1. Equilateral triangle formed by two radii and the chord.

Answer: (b) One

Explanation:
1. At a point on a circle only one line touches the circle without cutting it.
2. That line is the tangent, perpendicular to the radius at the point.

Teacher's Note:
1. From an outside point there are two tangents; from a point on the circle, one.

Answer: (b) Chord

Explanation:
1. A chord is a line segment joining two points of the circle.
2. A diameter joins two points and passes through the centre, so it is the longest chord.

Teacher's Note:
1. A diameter is the longest chord.

Answer: (c) 16 cm

Explanation:
1. r² = 6² + 8² = 100, so r = 10 cm.
2. Half of the new chord = \( \sqrt{100 − 36} = 8 \) cm.
3. The chord is 16 cm.

Teacher's Note:
1. First find the radius from the given chord.

Answer: (d) 135°

Explanation:
1. Adjacent angles on a leg of a trapezium are supplementary.
2. x + 3x = 180°, so x = 45°.
3. The greater angle is 135°.

Teacher's Note:
1. A cyclic trapezium is isosceles.
2. Co-interior angles add to 180°.

Answer: (c) 25 cm

Explanation:
1. Half the chord = 24 cm.
2. r² = 24² + 7² = 576 + 49 = 625.
3. r = 25 cm.

Teacher's Note:
1. Pythagorean triple 7, 24, 25.

Answer: (a) 41°

Explanation:
1. The angle in a semicircle is 90°, so ∠PQR = ∠PSR = 90°.
2. ∠QRP = 90° − 67° = 23° and ∠PRS = 90° − 72° = 18°.
3. ∠QRS = 23° + 18° = 41°.

Teacher's Note:
1. Use the right angles at Q and S.

Answer: (d) All of (A), (B) and (C)

Explanation:
1. Opposite angles of a cyclic quadrilateral are supplementary.
2. The exterior angle equals the interior opposite angle.
3. In a parallelogram opposite angles are equal and, being supplementary, each is 90°, so it is a rectangle.

Teacher's Note:
1. All three are standard properties.

Answer: (b) 15 cm

Explanation:
1. A 90° arc is one quarter of the circle.
2. Length = ¼ × 60 = 15 cm.

Teacher's Note:
1. Arc length is proportional to the angle at the centre.

Answer: (a) 30°

Explanation:
1. ∠AOD = 180° − 120° = 60° (AOB is a straight line).
2. The angle at the circumference is half the central angle: ∠ACD = 30°.

Teacher's Note:
1. AOB is a straight line through the centre.

Answer: (b) 65°

Explanation:
1. Equal chords are equidistant from the centre, so OD = OE.
2. In the isosceles triangle ODE, ∠ODE = ∠OED = (180° − 130°)/2 = 25°.
3. ∠PDE = ∠PDO − ∠ODE = 90° − 25° = 65°.

Teacher's Note:
1. Equal chords are equidistant from the centre.
2. Subtract the base angle from the right angle.

Answer: (c) 17

Explanation:
1. The perpendicular from the centre bisects the chord, so half the chord is 8 cm.
2. r² = 8² + 15² = 64 + 225 = 289.
3. r = 17 cm.

Teacher's Note:
1. The perpendicular from the centre to a chord bisects it.
2. Form a right triangle with the radius as hypotenuse.

Answer: (a) 25°

Explanation:
1. The angle in a semicircle is 90°, so ∠BAC = 90°.
2. ∠ABC = 180° − 90° − 65° = 25°.

Teacher's Note:
1. BC being a diameter gives a right angle at A.

Answer: (a) 98°

Explanation:
1. PQ ∥ SR, so the co-interior angles ∠P and ∠S add up to 180°.
2. ∠S = 180° − 82° = 98°.

Teacher's Note:
1. Co-interior angles between parallel lines are supplementary.

Answer: (a) 12 cm

Explanation:
1. The products of the segments of intersecting chords are equal: AP × PB = CP × PD = 48.
2. Let AP = x, then PB = 16 − x and x(16 − x) = 48.
3. x² − 16x + 48 = 0 gives x = 12 or 4; since AP > PB, AP = 12 cm.

Teacher's Note:
1. Use AP × PB = CP × PD.

Answer: (b) 120°

Explanation:
1. ∠QPR = 60°.
2. ∠QOR = 2 × 60° = 120°.

Teacher's Note:
1. Central angle is twice the inscribed angle.

Answer: (b) 120°

Explanation:
1. Each arc cut off by a side subtends 120° at the centre.
2. ∠XOY = 2 × ∠XZY = 2 × 60° = 120°.

Teacher's Note:
1. The three central angles are equal and add to 360°.

Answer: (c) 48°

Explanation:
1. ∠SPQ = 180° − 138° = 42° (opposite angles of a cyclic quadrilateral).
2. ∠PSQ = 90° (angle in a semicircle).
3. ∠PQS = 180° − 90° − 42° = 48°.

Teacher's Note:
1. The angle on the diameter is a right angle.

Answer: (b) 63°

Explanation:
1. The angle at the circumference is half the angle at the centre.
2. ∠ACB = ½ × 126° = 63°.

Teacher's Note:
1. The angle at the centre is double the angle on the remaining circle.