CBSE Class 9 Mathematics Plane Geometry II MCQs Set 19

Answer: (a) 2

Explanation:
1. The perpendicular from the centre bisects the chord, so D is the mid-point of AB.
2. O is the mid-point of BC, so OD joins the mid-points of two sides of triangle ABC and OD = ½ CA.
3. So CA = 2 × OD and k = 2.

Teacher's Note:
1. The line joining mid-points of two sides is half the third side.
2. A perpendicular from the centre always bisects a chord.

Answer: (c) 5 cm

Explanation:
1. Distance of a chord of length l from the centre = √(r² − l²/4).
2. The longer chord (8 cm) is nearer the centre: √(r² − 9) − √(r² − 16) = 1.
3. Try r = 5: 4 − 3 = 1 ✓. So r = 5 cm.

Teacher's Note:
1. Each half-chord, the radius and the perpendicular from the centre form a right triangle.
2. The longer chord is closer to the centre.

Answer: (d) 1/4

Explanation:
1. Let AC = c, so AB = 2c. Then a² = r² − (2c/2)² = r² − c² and b² = r² − c²/4.
2. a² + 3r² = 4r² − c², and b² = (4r² − c²)/4.
3. So b² = ¼ (a² + 3r²) and k = 1/4.

Teacher's Note:
1. Use the half-chord relation for each chord and compare.
2. Express both distances in terms of r and AC.

Answer: (d) 16 units

Explanation:
1. The common chord is the perpendicular bisector of AB, so it is 15 units from each centre.
2. Half the chord = √(17² − 15²) = √64 = 8.
3. Common chord = 16 units.

Teacher's Note:
1. The line of centres is the perpendicular bisector of the common chord.
2. Use the 8-15-17 right triangle.

Answer: (d) 72 cm

Explanation:
1. If O is the centre, OAPB is a rectangle since the diameters are perpendicular.
2. The diagonals of a rectangle are equal, so AB = OP = radius = 36 cm.
3. Diameter = 72 cm.

Teacher's Note:
1. The diagonal AB of the rectangle equals the other diagonal OP.
2. OP is a radius.

Answer: (a) 13°

Explanation:
1. Since OR ⟂ PQ and OP = OR, ∠OPR = 45°, so ∠QPR = 45°.
2. ∠QPS = ∠QPR − ∠RPS = 45° − 32° = 13°.
3. Angles in the same segment on chord QS are equal: ∠QRS = ∠QPS = 13°.

Teacher's Note:
1. Angles subtended by the same arc at the circumference are equal.
2. Break the angle at P into two known parts.

Answer: (b) 50°

Explanation:
1. ∠RPS = ∠RQS = 50° (angles in the same segment on chord RS).
2. In triangle PSR: ∠PSR = 180° − 80° − 50° = 50°.

Teacher's Note:
1. Angles in the same segment are equal.
2. Then apply the angle sum of a triangle.

Answer: (a) 30°

Explanation:
1. PQRS is cyclic, so ∠PQR = 180° − ∠PSR = 60°.
2. PQ is a diameter, so ∠PRQ = 90°.
3. ∠QPR = 180° − 90° − 60° = 30°.

Teacher's Note:
1. Opposite angles of a cyclic quadrilateral add to 180°.
2. An angle in a semicircle is a right angle.

Answer: (d) 120°

Explanation:
1. ∠BAC = 60°.
2. ABEC is a cyclic quadrilateral, so ∠BEC = 180° − ∠BAC = 120°.

Teacher's Note:
1. The angle on the minor arc is supplementary to the angle on the major arc.
2. Opposite angles of a cyclic quadrilateral add to 180°.

Answer: (d) 25°

Explanation:
1. AC is a diameter, so ∠ABC = 90°, hence ∠ABE = 90° − 65° = 25°.
2. ∠ACE = ∠ABE = 25° (angles in the same segment).
3. ED ∥ AC, so ∠DEC = ∠ECA = 25° (alternate angles).

Teacher's Note:
1. Use the right angle in a semicircle first.
2. Alternate angles on parallel lines are equal.

Answer: (c) 50π

Explanation:
1. By the intersecting chords theorem, AP × PB = CP × PD, so 24 = 2 × PD and PD = 12.
2. Put P at the origin, A(−4, 0), B(6, 0), C(0, −2), D(0, 12). The centre lies on the perpendicular bisectors: x = 1 and y = 5.
3. r² = (1 + 4)² + 5² = 50.
4. Area = πr² = 50π.

Teacher's Note:
1. The centre lies on the perpendicular bisector of every chord.
2. Coordinates make the radius easy to find.

Answer: (b) \( 12\sqrt{5} \)

Explanation:
1. AM = MB = 12. By the intersecting chords theorem, AM × MB = CM × MD, so 144 = CM × 4CM and CM = 6, DM = 24.
2. BD² = BM² + DM² = 144 + 576 = 720.
3. BD = √720 = 12√5.

Teacher's Note:
1. The product of the segments of intersecting chords is equal.
2. Triangle BMD is right-angled at M.

Answer: (b) 35°

Explanation:
1. In quadrilateral ABOC, ∠B = ∠C = 90°, so ∠BOC = 180° − 70° = 110°.
2. Triangle OBC is isosceles (OB = OC), so ∠OBC = (180° − 110°) ÷ 2 = 35°.

Teacher's Note:
1. A tangent is perpendicular to the radius at the point of contact.
2. The angle at the centre and the angle between tangents are supplementary.

Answer: (d) 60°

Explanation:
1. ∠OQA = ∠ORA = 90°.
2. In quadrilateral AQOR: ∠QAR = 360° − 90° − 90° − 120° = 60°.

Teacher's Note:
1. The four angles of a quadrilateral add to 360°.
2. Radius ⟂ tangent.

Answer: (c) 26

Explanation:
1. For a direct common tangent, d² = t² + (R − r)² = 24² + 10² = 676.
2. d = 26 cm.

Teacher's Note:
1. Draw a line from the smaller centre parallel to the tangent to make a right triangle.
2. The legs are t and R − r.

Answer: (b) 95

Explanation:
1. The tangent length from A to the incircle of ABD is (AB + AD − BD)/2, and for ADC it is (AC + AD − DC)/2.
2. The circles touch AD at the same point, so these are equal: AB − BD = AC − DC, giving BD − DC = AB − AC = −70.
3. BD + DC = 260, so BD = (260 − 70)/2 = 95.

Teacher's Note:
1. Two circles touching the same line at the same point have equal tangent lengths from A.
2. Solve the pair of linear equations.

Answer: (b) 2 cm

Explanation:
1. AC = √(36 + 64) = 10 cm.
2. Inradius of a right triangle = (a + b − c)/2 = (6 + 8 − 10)/2 = 2 cm.

Teacher's Note:
1. Also r = area/semi-perimeter = 24/12 = 2.
2. The shortcut (a + b − c)/2 only works for right triangles.

Answer: (c) 24 cm

Explanation:
1. Check: PQ + SR = 50 = PS + QR, as required for a circle to be inscribed.
2. Drop perpendiculars from P and Q to SR: each foot is (32 − 18)/2 = 7 cm from the nearer end.
3. Height = √(25² − 7²) = 24 cm, and the diameter equals the height between the parallel tangents = 24 cm.

Teacher's Note:
1. A circle touching all four sides forces the sum of opposite sides to be equal.
2. The diameter equals the distance between the parallel sides.

Answer: (c) 10

Explanation:
1. Tangent-secant theorem: XR² = RQ × RP.
2. 144 = (x − 2)(x − 2 + x) = (x − 2)(2x − 2).
3. 2x² − 6x + 4 = 144, so x² − 3x − 70 = 0, i.e. (x − 10)(x + 7) = 0.
4. x = 10 (reject the negative root).

Teacher's Note:
1. The square of the tangent equals the product of the whole secant and its external part.
2. RP = RQ + QP.

Answer: (d) √2100 units

Explanation:
1. The shortest chord through a point is the one perpendicular to the line joining the point to the centre.
2. Half the chord = √(25² − 10²) = √525.
3. Chord = 2√525 = √2100.

Teacher's Note:
1. The chord is shortest when it is farthest from the centre.
2. 2√525 = √(4 × 525).