CBSE Class 9 Mathematics Circles MCQs Set 08

Answer: (a) two straight lines

Explanation:
1. Each centre stays at a constant height equal to its radius above the road.
2. So each centre moves along a straight line parallel to the road: two straight lines.

Teacher's Note:
1. A point at a fixed distance from a line moves on a parallel line.

Answer: (b) 16 cm

Explanation:
1. The perpendicular from the centre bisects the chord. Half chord = √(10² − 6²) = 8 cm.
2. Chord = 2 × 8 = 16 cm.

Teacher's Note:
1. Use Pythagoras in the right triangle formed by radius, distance and half-chord.

Answer: (c) 10 cm

Explanation:
1. Half-chord = 4 cm, so radius = √(4² + 3²) = 5 cm.
2. Diameter = 10 cm.

Teacher's Note:
1. Remember to double the radius.

Answer: (d) equal in length

Explanation:
1. Equal chords are equidistant from the centre, and conversely chords at equal distances from the centre are equal.

Teacher's Note:
1. Length of a chord depends only on its distance from the centre.

Answer: (a) 12 cm

Explanation:
1. Half-chord = 5 cm. Distance = √(13² − 5²) = 12 cm.

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

Answer: (b) 1 cm

Explanation:
1. Distance of AB from O = √(25 − 9) = 4 cm and of CD = √(25 − 16) = 3 cm.
2. On the same side of O, the distance between them is 4 − 3 = 1 cm.

Teacher's Note:
1. On opposite sides the distance would be 4 + 3 = 7 cm.

Answer: (c) 13 cm

Explanation:
1. Let r be the radius: √(r² − 25) + √(r² − 144) = 17.
2. r = 13 gives 12 + 5 = 17, which fits.

Teacher's Note:
1. Test the Pythagorean triples 5-12-13.

Answer: (d) 9.6 cm

Explanation:
1. AO ⟂ BC at M and AD, the diameter through A, is 10 cm. In right △ABD, AB² = AM × AD, so AM = 36 ÷ 10 = 3.6 cm.
2. BM = √(36 − 3.6²) = 4.8 cm, so BC = 9.6 cm.

Teacher's Note:
1. The diameter through A is the perpendicular bisector of BC.

Answer: (a) \( 8+2\sqrt7 \) cm ≈ 13.29 cm

Explanation:
1. Distances of the centres from the common chord are √(100 − 36) = 8 cm and √(64 − 36) = 2√7 cm.
2. On opposite sides they add: 8 + 2√7 ≈ 13.29 cm.

Teacher's Note:
1. The line of centres is the perpendicular bisector of the common chord.

Answer: (b) \( 4\sqrt6 \) cm

Explanation:
1. AB = 5 − 3 = 2 cm, so the midpoint M of AB is 1 cm from A.
2. PQ is a chord of the bigger circle at distance 1 cm from its centre: PQ = 2√(25 − 1) = 4√6 cm.

Teacher's Note:
1. Internal contact: distance of centres = difference of radii.

Answer: (c) 14 cm if on opposite sides

Explanation:
1. Distance of PQ from O = √(100 − 64) = 6 cm and of RS = √(100 − 36) = 8 cm.
2. Same side: 8 − 6 = 2 cm. Opposite sides: 8 + 6 = 14 cm.

Teacher's Note:
1. Treat both cases whenever the chords are not stated to be on the same side.

Answer: (d) 2 cm and 10 cm

Explanation:
1. The perpendicular from the centre bisects both chords, so AB = CD = (12 − 8) ÷ 2 = 2 cm.
2. AC = AB + BC = 2 + 8 = 10 cm.

Teacher's Note:
1. The common midpoint of AD and BC is the foot of the perpendicular from O.

Answer: (a) 120°

Explanation:
1. ∠ACB = 180° − 80° − 40° = 60°.
2. ∠ADB = ∠ACB = 60° (same segment), so ∠DAB + ∠ABD = 180° − 60° = 120°.

Teacher's Note:
1. Angles in the same segment are equal.

Answer: (b) 80°

Explanation:
1. CP = CQ = CR, so ∠CQP = 25° and ∠CRP = 15°.
2. ∠QPR = 25° + 15° = 40°, and the angle at the centre is double: ∠QCR = 80°.

Teacher's Note:
1. Angle at the centre is twice the angle at the circumference.

Answer: (c) 60°

Explanation:
1. Opposite angles are supplementary: ∠B + ∠D = 180°.
2. With ∠B − ∠D = 60°, ∠B = 120° and ∠D = 60°.

Teacher's Note:
1. Add and subtract the two equations.

Answer: (d) AC = 2OD

Explanation:
1. D is the midpoint of AB and O is the midpoint of BC, so OD is parallel to AC and OD = ½AC.
2. Therefore AC = 2OD.

Teacher's Note:
1. Midpoint theorem in △ABC.

Answer: (a) \( 4\sqrt3 \) cm

Explanation:
1. The distance between the centres is 4 cm, so the common chord is at distance 2 cm from each centre.
2. Half-chord = √(16 − 4) = 2√3 cm, so the chord is 4√3 cm.

Teacher's Note:
1. The centres and a common point form an equilateral triangle.

Answer: (b) 30°

Explanation:
1. ∠ABC = 180° − 85° = 95°, so ∠PBC = 85°.
2. In △PBC, ∠BCP = 180° − 85° − 40° = 55°, so ∠BCD = 125° and ∠DCQ = 55°. In △DCQ, ∠CDQ = 95°.
3. ∠CQD = 180° − 95° − 55° = 30°.

Teacher's Note:
1. Use the exterior angle property and the angle sum of triangles.

Answer: (c) 20°

Explanation:
1. ∠ADB = ∠ACB = 40° (same segment AB).
2. In △PBD, ∠PBD = 180° − 120° − 40° = 20°.

Teacher's Note:
1. Angles in the same segment are equal.

Answer: (d) rectangle

Explanation:
1. Opposite angles are equal in a parallelogram and supplementary in a cyclic quadrilateral, so each is 90°.

Teacher's Note:
1. A cyclic rhombus is a square, a special rectangle.

Answer: (a) 125°

Explanation:
1. ∠A + ∠C = 180° and ∠A − ∠C = 70°, so ∠A = 125°.

Teacher's Note:
1. Angle ∠C = 55°.

Answer: (b) 30°

Explanation:
1. The chord and the two radii form an equilateral triangle, so the angle at the centre is 60°.
2. The angle on the major arc is half of it: 30°.

Teacher's Note:
1. Angle at the circumference is half the angle at the centre.

Answer: (c) its non-parallel sides are equal

Explanation:
1. Angles on the same base are supplementary to the opposite ones, so the base angles are equal.
2. This makes the trapezium isosceles, hence the non-parallel sides are equal.

Teacher's Note:
1. Cyclic trapezium means isosceles trapezium.