CBSE Class 9 Mathematics Triangles and Circles MCQs Set 11

Answer: (a) 90° − \( \frac12 \)∠A

Explanation:
1. The exterior angles at B and C are (180° − B) and (180° − C), and the bisectors cut them in half: 90° − B/2 and 90° − C/2.
2. In triangle BOC: ∠BOC = 180° − (90° − B/2) − (90° − C/2) = (B + C)/2.
3. Since B + C = 180° − A, ∠BOC = 90° − A/2.

Teacher's Note:
1. The bisectors of two exterior angles meet at the excentre opposite A.
2. This result, 90° − A/2, is worth remembering as a standard property.

Answer: (c) 100°

Explanation:
1. In triangle ABD: ∠BAD = 180° − 70° − 30° = 80°.
2. Opposite angles of a cyclic quadrilateral add up to 180°: ∠BCD = 180° − 80° = 100°.

Teacher's Note:
1. A, B, C, D lie on a circle, and ∠BAD and ∠BCD are opposite angles.
2. First find ∠BAD from the triangle formed by the diagonal BD.

Answer: (c) Circumcentre

Explanation:
1. The perpendicular bisectors of the three sides meet at one point, which is equidistant from the three vertices.
2. That point is the centre of the circle through the vertices, so it is the circumcentre.

Teacher's Note:
1. Incentre comes from angle bisectors, orthocentre from altitudes, and centroid from medians.
2. Remember: circumcentre goes with perpendicular bisectors.

Answer: (a) the midpoint of the hypotenuse

Explanation:
1. The circumcentre of a right triangle is the centre of the circle through its vertices.
2. By the angle in a semicircle property, the hypotenuse is a diameter of that circle, so the centre is the midpoint of the hypotenuse.

Teacher's Note:
1. For an acute triangle the circumcentre is inside, for an obtuse triangle it is outside.
2. For a right triangle it lies exactly on the hypotenuse.

Answer: (a) 115°, 115°, 65°

Explanation:
1. AB || CD, so ∠B + ∠C = 180° (co-interior angles): ∠C = 115°.
2. Opposite angles of a cyclic quadrilateral add up to 180°: ∠D = 180° − ∠B = 115°, and ∠A = 180° − ∠C = 65°.

Teacher's Note:
1. A cyclic quadrilateral with a pair of parallel sides is an isosceles trapezium.
2. Co-interior angles between parallel lines add to 180°, and so do opposite angles of a cyclic quadrilateral.

Answer: (b) \( \frac49 \)

Explanation:
1. Area is proportional to the square of the radius (or diameter).
2. \( \frac{\pi 4^2}{\pi 6^2}=\frac{16}{36}=\frac49 \).

Teacher's Note:
1. Doubling a diameter multiplies the area by 4, so areas go as the square of the diameters.
2. The ratio of diameters is 2 : 3, so the ratio of areas is 4 : 9.

Answer: (a) the circumcircle

Explanation:
1. A point equidistant from the three vertices is the circumcentre.
2. The circle drawn with the circumcentre as centre and passing through the vertices is the circumcircle.

Teacher's Note:
1. The incircle touches the three sides from inside, while an excircle touches one side and the extensions of the other two.
2. Only the circumcircle passes through the vertices.

Answer: (d) CP, BP, AP

Explanation:
1. ∠A = 180° − 45° − 65° = 70°, so ∠BAP = ∠PAC = 35°.
2. In triangle ABP: BP is opposite ∠BAP = 35° and AP is opposite ∠B = 45°, so BP < AP.
3. In triangle APC: CP is opposite ∠PAC = 35° and AP is opposite ∠C = 65°, so CP < AP.
4. In triangle ABC, AC is opposite ∠B (45°) and AB is opposite ∠C (65°), so AC < AB. By the angle bisector theorem BP/CP = AB/AC > 1, so BP > CP.
5. Hence CP < BP < AP.

Teacher's Note:
1. In any triangle the larger side is opposite the larger angle.
2. The angle bisector theorem divides BC in the ratio AB : AC.

Answer: (c) 1 : 1

Explanation:
1. Congruent triangles have the same shape and size, so their areas are equal.
2. Ratio of areas = 1 : 1.

Teacher's Note:
1. Congruent triangles can be placed exactly on top of each other.
2. Equal area follows from equal base and equal height.

Answer: (b) 10°

Explanation:
1. Q lies on the minor arc PR, so the major arc PR (the one not containing Q) subtends the reflex angle 2 × 100° = 200° at the centre.
2. Hence the ordinary angle ∠POR = 360° − 200° = 160°.
3. OP = OR (radii), so ∠OPR = ∠ORP = (180° − 160°)/2 = 10°.

Teacher's Note:
1. The angle at the centre is double the angle at the circumference on the same arc.
2. Triangle OPR is isosceles because OP and OR are radii.

Answer: (a) SQ > SR

Explanation:
1. PQ > PR means the angle opposite PQ is larger: ∠R > ∠Q.
2. Hence ∠SRQ = R/2 > Q/2 = ∠SQR.
3. In triangle SQR, the side opposite the larger angle is longer: SQ (opposite ∠SRQ) is longer than SR.

Teacher's Note:
1. The larger side of a triangle faces the larger angle.
2. Halving both angles keeps their order.

Answer: (a) 110°

Explanation:
1. ∠DEC = 180° − 130° = 50° (linear pair along BD).
2. In triangle DEC: ∠EDC = 180° − 50° − 20° = 110°, so ∠BDC = 110°.
3. Angles in the same segment are equal: ∠BAC = ∠BDC = 110°.

Teacher's Note:
1. Angles on the same side of chord BC, such as ∠BAC and ∠BDC, are equal.
2. Use the triangle DEC to find ∠BDC first.