CBSE Class 9 Mathematics Circles MCQs Set 09

Answer: (a) two straight lines

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

Teacher's Note:
1. A rolling wheel's centre keeps a constant height.
2. Two wheels give two parallel lines.

Answer: (c) a circle

Explanation:
1. The centre is always at distance r from the fixed point.
2. All points at a fixed distance r from a point lie on a circle.

Teacher's Note:
1. Locus of points at a fixed distance from a point is a circle.
2. Its radius is r.

Answer: (c) 16 cm

Explanation:
1. Half the chord = \(\sqrt{10^2 - 6^2} = \sqrt{64} = 8\) cm.
2. Chord = 2 × 8 = 16 cm.

Teacher's Note:
1. The perpendicular from the centre bisects the chord.
2. Use Pythagoras for half the chord.

Answer: (a) 10 cm

Explanation:
1. Half the chord = 4 cm, so radius = \(\sqrt{4^2 + 3^2} = 5\) cm.
2. Diameter = 2 × 5 = 10 cm.

Teacher's Note:
1. Radius is the hypotenuse of the right triangle.
2. Diameter is twice the radius.

Answer: (d) equal

Explanation:
1. Chords equidistant from the centre of a circle are equal in length.

Teacher's Note:
1. Equal distance from the centre means equal chords, and the converse also holds.
2. This is a standard theorem.

Answer: (b) 12 cm

Explanation:
1. Half the chord = 5 cm.
2. Distance = \(\sqrt{13^2 - 5^2} = \sqrt{144} = 12\) cm.

Teacher's Note:
1. Distance, half-chord and radius form a right triangle.
2. 5, 12, 13 is a Pythagorean triple.

Answer: (b) 60°

Explanation:
1. Opposite angles of a cyclic quadrilateral add to 180°, so B + D = 180°.
2. With B − D = 60°: B = 120° and D = 60°.
3. The smaller angle is 60°.

Teacher's Note:
1. B + D = 180° in a cyclic quadrilateral.
2. Solve the two equations together.

Answer: (a) AC = 2OD

Explanation:
1. OD bisects AB, so D is the mid-point of AB. O is the mid-point of BC.
2. In triangle ABC the line DO joins mid-points of AB and BC, so DO = AC ÷ 2.
3. Hence AC = 2OD.

Teacher's Note:
1. Use the mid-point theorem.
2. The perpendicular from the centre bisects a chord.

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

Explanation:
1. The centres are 4 cm apart, and the common chord is the perpendicular bisector of the line joining the centres.
2. Half-chord = \(\sqrt{4^2 - 2^2} = \sqrt{12} = 2\sqrt3\) cm.
3. Common chord = \(4\sqrt3\) cm.

Teacher's Note:
1. Distance from a centre to the chord is 2 cm.
2. Use Pythagoras with radius 4 cm.

Answer: (b) 30°

Explanation:
1. \(\angle ABC = 180^\circ - 85^\circ = 95^\circ\), so \(\angle PBC = 85^\circ\). In triangle PBC, \(\angle BCP = 180^\circ - 85^\circ - 40^\circ = 55^\circ\).
2. \(\angle BCD = 180^\circ - 55^\circ = 125^\circ\), so \(\angle BAD = 180^\circ - 125^\circ = 55^\circ\).
3. In triangle ABQ: \(\angle Q = 180^\circ - 55^\circ - 95^\circ = 30^\circ\).

Teacher's Note:
1. Use opposite angles of the cyclic quadrilateral twice.
2. Finish with the angle sum of triangle ABQ.

Answer: (a) rectangle

Explanation:
1. Opposite angles of a parallelogram are equal, and for a cyclic quadrilateral they add to 180°.
2. So each is 90°, which makes the parallelogram a rectangle.

Teacher's Note:
1. Equal and supplementary means 90° each.
2. A square is a special rectangle.

Answer: (a) 125°

Explanation:
1. A + C = 180° and A − C = 70°.
2. Adding: 2A = 250°, so A = 125°.

Teacher's Note:
1. Opposite angles add to 180°.
2. Solve the two equations.

Answer: (a) 30°

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

Teacher's Note:
1. Angle at the centre is double the angle on the remaining arc.
2. Equal chord and radius gives an equilateral triangle.

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

Explanation:
1. For a cyclic trapezium, the angles on the same base are equal, which makes it isosceles.
2. So its non-parallel sides are equal (and its diagonals are equal too).

Teacher's Note:
1. A cyclic trapezium is always isosceles.
2. Its diagonals are equal.