CBSE Class 10 Mathematics Circles MCQs Set 18

Answer: (b) 30 cm

Explanation:
1. The perpendicular from the centre bisects the chord, so AC = \( \sqrt{17^2 - 8^2} \) = \( \sqrt{225} \) = 15 cm.
2. AB = 2 × 15 = 30 cm.

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

Answer: (a) 3 cm

Explanation:
1. Half the chord is 4 cm and the radius is 5 cm.
2. Distance = \( \sqrt{5^2 - 4^2} \) = 3 cm.

Teacher's Note:
1. Half-chord, distance and radius form a right triangle.
2. The triple 3, 4, 5 appears often in such questions.

Answer: (c) 90°

Explanation:
1. O and O' are each equidistant from A and B, so both lie on the perpendicular bisector of AB.
2. Hence line OO' is the perpendicular bisector of AB and angle OMA = 90°.

Teacher's Note:
1. The line joining the centres of two intersecting circles is the perpendicular bisector of the common chord.
2. M is also the midpoint of AB.

Answer: (c) 1 : 1

Explanation:
1. OO' is the perpendicular bisector of the common chord AB.
2. So M is the midpoint of AB and AM : BM = 1 : 1.

Teacher's Note:
1. The perpendicular bisector of a chord passes through the centre of each circle.
2. Common chord and line of centres are perpendicular.

Answer: (a) 1 : 1

Explanation:
1. Equal chords AC and BC are equidistant from the centre, and C is equidistant from A and B.
2. Both O and C lie on the perpendicular bisector of AB, so line OC bisects AB at M and AM : BM = 1 : 1.

Teacher's Note:
1. Points equidistant from A and B lie on the perpendicular bisector of AB.
2. O is the centre, so OA = OB as well.

Answer: (c) 12 cm

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

Teacher's Note:
1. The perpendicular from the centre bisects the chord.
2. The triple 5, 12, 13 is used here.

Answer: (a) 230°

Explanation:
1. The minor and major arcs together make the full circle of 360°.
2. Major arc = 360° − 130° = 230°.

Teacher's Note:
1. The two arcs cut by a chord add up to 360°.
2. The central angle equals the measure of its arc.

Answer: (c) 70°

Explanation:
1. OA = OB, so angle OAB = angle OBA = 55° and angle AOB = 180° − 110° = 70°.
2. Equal chords subtend equal angles at the centre, so angle COD = 70°.

Teacher's Note:
1. Equal chords subtend equal angles at the centre.
2. Triangle OAB is isosceles with OA = OB.

Answer: (d) cyclic quadrilateral

Explanation:
1. Let the bisectors form a quadrilateral PQRS. Because the opposite angles of the cyclic quadrilateral add up to 180°, the opposite angles of PQRS add up to 180° too.
2. A quadrilateral whose opposite angles are supplementary is cyclic.

Teacher's Note:
1. Opposite angles of a cyclic quadrilateral are supplementary.
2. The converse also holds and gives the shape here.

Answer: (a) AP × PC = BP × PD

Explanation:
1. Equal angles on the same side of BC mean A, B, C, D are concyclic.
2. AC and BD are chords meeting at P, so AP × PC = BP × PD.

Teacher's Note:
1. If a segment subtends equal angles at two points on the same side, the four points are concyclic.
2. Chords meeting inside a circle obey the intersecting chords theorem.

Answer: (b) 40°

Explanation:
1. In triangle BDC, angle BDC = 180° − 30° − 110° = 40°.
2. Angles in the same segment are equal, so angle BAC = angle BDC = 40°.

Teacher's Note:
1. Angles in the same segment of a circle are equal.
2. First find the third angle of the triangle.

Answer: (d) 30°

Explanation:
1. OA = OB so angle OAB = 60° and triangle OAB is equilateral; angle AOB = 60°.
2. The angle at the circle is half the central angle: angle ACB = 30°.

Teacher's Note:
1. The angle subtended at the circumference is half the angle at the centre.
2. Check the triangle OAB first.

Answer: (a) 120°

Explanation:
1. ABEC is a cyclic quadrilateral with angle BAC = 60°.
2. Angle BEC = 180° − 60° = 120°.

Teacher's Note:
1. Opposite angles of a cyclic quadrilateral add up to 180°.
2. E and A lie on opposite sides of BC.

Answer: (c) 45°

Explanation:
1. Angle AOD = 90°, so the arc AD measures 90°.
2. Angle ACD at the circumference is half of that: 45°.

Teacher's Note:
1. The angle at the circumference is half the central angle on the same arc.
2. Perpendicular radius gives a 90° arc.

Answer: (a) 90°

Explanation:
1. A cyclic quadrilateral with bisecting diagonals is a parallelogram, hence a rectangle.
2. So angle BCD = 90°.

Teacher's Note:
1. A cyclic parallelogram is a rectangle.
2. Check the given angles are consistent with a rectangle: angle ACB = 60° and angle BAC = 30° give angle ABC = 90°.

Answer: (c) 50°

Explanation:
1. Opposite angles add up to 180°, so angle ABC = 40°.
2. Angle ACB = 90° (angle in a semicircle), so angle BAC = 180° − 90° − 40° = 50°.

Teacher's Note:
1. The angle in a semicircle is a right angle.
2. Use the cyclic quadrilateral property first.

Answer: (b) 20°

Explanation:
1. Angle CAB at the circumference is half of the central angle COB on the same arc BC.
2. Angle CAB = 20°.

Teacher's Note:
1. The angle at the circumference is half the angle at the centre.
2. The diameter is not needed once the arc BC is known.

Answer: (b) 80°

Explanation:
1. Angle BOC = 360° − 80° − 120° = 160° (the angle subtended by the arc BC not containing A).
2. Angle BAC = 160° ÷ 2 = 80°.

Teacher's Note:
1. The three central angles around a point add up to 360°.
2. Then apply the half-angle theorem.

Answer: (c) 5 : 13

Explanation:
1. Angle ADC = 100° ÷ 2 = 50°.
2. ABCD is cyclic, so angle ABC = 180° − 50° = 130°. The ratio is 50 : 130 = 5 : 13.

Teacher's Note:
1. Angles on the major and minor arcs are supplementary.
2. Reduce the ratio to its simplest form.

Answer: (b) 30°

Explanation:
1. Angle ABC = 180° − 120° = 60° (opposite angles of a cyclic quadrilateral).
2. Angle ACB = 90°, so angle CAB = 180° − 90° − 60° = 30°.

Teacher's Note:
1. Use angle in a semicircle = 90°.
2. Find angle ABC first.