CBSE Class 10 Mathematics Circles MCQs Set 16

Answer: (a) one

Explanation:
1. A tangent touches the circle at exactly one point, called the point of contact.

Teacher's Note:
1. Recall the definition of a tangent.
2. A line meeting the circle at two points is a secant.

Answer: (b) secant

Explanation:
1. A secant cuts the circle at two distinct points.
2. A chord is the segment between those two points.

Teacher's Note:
1. A tangent meets the circle once.
2. The line (not the segment) is a secant.

Answer: (a) point of contact

Explanation:
1. It is the single point where the tangent touches the circle.
2. The radius through it is perpendicular to the tangent.

Teacher's Note:
1. Recall the tangent terminology.
2. It is also called the point of tangency.

Answer: (b) two

Explanation:
1. For a given direction there are exactly two tangents, at the two ends of the diameter perpendicular to that direction.

Teacher's Note:
1. Think of the tangents at the ends of a diameter.
2. They are parallel to each other.

Answer: (c) a circle with centre O

Explanation:
1. The perpendicular from O to a chord meets it at its midpoint.
2. Equal chords are equidistant from the centre, so every midpoint is at the same distance from O.
3. Hence the midpoints lie on a circle centred at O.

Teacher's Note:
1. Equal chords are equally far from the centre.
2. A fixed distance from O gives a circle.

Answer: (b) 6r

Explanation:
1. A regular hexagon is made of six equilateral triangles with side r.
2. Perimeter = 6r.

Teacher's Note:
1. Each side subtends 60° at the centre.
2. The triangle formed with the centre is equilateral.

Answer: (a) CE = CD

Explanation:
1. By the tangent-chord theorem, angle CED = angle ACD = 50° and angle CDE = angle BCE = 50°.
2. In triangle CDE two angles are equal, so CD = CE.
3. Angle DCE = 80°, so the triangle is not right-angled, and ED is parallel to AB (alternate angles).

Teacher's Note:
1. Use the angle in the alternate segment.
2. Equal angles give equal opposite sides.

Answer: (b) 60°

Explanation:
1. The triangle is inscribed in the circle with l₃ as the diameter, so the angle opposite l₃ is 90°.
2. cos(angle between l₁ and l₃) = l₁/l₃ = 1/2.
3. The angle is 60°.

Teacher's Note:
1. The angle in a semicircle is a right angle.
2. Use the cosine in the right triangle.

Answer: (b) \( 4\sqrt{3} \) cm

Explanation:
1. The distance between the centres is 4 cm and the chord is the perpendicular bisector of the line of centres.
2. Half the chord = √(4² - 2²) = √12 = 2√3.
3. Chord = 4√3 cm.

Teacher's Note:
1. The common chord is perpendicular to the line of centres.
2. Use Pythagoras with half the chord.

Answer: (d) 21 cm

Explanation:
1. Tangents from C are equal: CR = CS = 27 cm, so BR = 38 - 27 = 11 cm, and BQ = BR = 11 cm.
2. Since angle A = 90° and the circle touches AB and AD, AQ = AP = radius = 10 cm.
3. AB = AQ + QB = 10 + 11 = 21 cm.

Teacher's Note:
1. Equal tangents from an external point.
2. The tangent lengths from a right-angle vertex equal the radius.

Answer: (a) 90°

Explanation:
1. OA bisects the angle between the two tangents from A, and OB bisects the angle between those from B.
2. Angle A + angle B = 180° (co-interior angles of parallel lines).
3. Angle OAB + angle OBA = 90°, so angle AOB = 90°.

Teacher's Note:
1. Use the bisecting property of tangents from a point.
2. Co-interior angles add up to 180°.

Answer: (b) 2AD = AB + BC + CA

Explanation:
1. BD = BF and CE = CF where F is the point of contact of BC, and AD = AE.
2. AB + BC + CA = AB + BF + FC + CA = AB + BD + CE + CA = AD + AE = 2AD.

Teacher's Note:
1. Use equal tangent lengths from B and from C.
2. Group the terms as AD and AE.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. A is true: the radius to the point of contact is perpendicular to the tangent.
2. R is true: the perpendicular is the shortest distance.
3. R explains A: every other point of the tangent is outside the circle, so M is the point nearest to O, which means OM is the perpendicular.

Teacher's Note:
1. Every point of the tangent other than M lies outside the circle.
2. So OM is the shortest distance from O to the line.

Answer: (a) 7 cm

Explanation:
1. s = 15 cm. AD = AF = s - BC = 15 - 8 = 7 cm.

Teacher's Note:
1. Tangent lengths from a vertex = s minus the opposite side.
2. Check AD + DB = 12: 7 + 5.

Answer: (b) 5 cm

Explanation:
1. BE = BD = s - CA = 15 - 10 = 5 cm.

Teacher's Note:
1. Equal tangents from B.
2. BD = 12 - 7 = 5 cm.

Answer: (c) 3 cm

Explanation:
1. CF = CE = s - AB = 15 - 12 = 3 cm.
2. Check: BE + EC = 5 + 3 = 8 cm.

Teacher's Note:
1. Equal tangents from C.
2. Use s = 15.

Answer: (d) 150°

Explanation:
1. The chord and the two radii form an equilateral triangle, so the angle at the centre is 60°.
2. The angle at a point on the major arc is 30°, so the angle at a point on the minor arc is 180° - 30° = 150°.

Teacher's Note:
1. Find the angle at the centre first.
2. Opposite angles of a cyclic quadrilateral add up to 180°.

Answer: (a) 1 : 4

Explanation:
1. Angle CBE = angle CBA = angle CDA = angle EDA (angles on the same arc CA), and angle BEC = angle DEA (vertically opposite).
2. So triangles CBE and ADE are similar with ratio BC : AD = 12 : 24.
3. Ratio of areas = (1/2)² = 1 : 4.

Teacher's Note:
1. Prove the triangles similar by AA.
2. Areas are in the ratio of the squares of the sides.

Answer: (c) 93 cm

Explanation:
1. The distance between two centres = 2 × 20 - overlap = 40 - overlap.
2. Perimeter = 3 × 40 - (5 + 10 + 12) = 120 - 27 = 93 cm.

Teacher's Note:
1. Distance between centres = sum of radii minus the overlap.
2. Add the three distances.

Answer: (c) 60°

Explanation:
1. In the right triangle BCD (angle C = 90°) angle BDC = 90° - 30° = 60°.
2. Angles BAC and BDC stand on the same arc BC, so angle BAC = 60°.

Teacher's Note:
1. Angles in the same segment are equal.
2. Use the angle sum in triangle BCD.

Answer: (a) \( \frac{5\sqrt{17}}{4} \) cm

Explanation:
1. The distances from the centre are √(r² - 16) and √(r² - 25), and they differ by 2.
2. √(r² - 16) = 2 + √(r² - 25) gives r² - 16 = 4 + 4√(r² - 25) + r² - 25, so 4√(r² - 25) = 5.
3. r² - 25 = 25/16, so r² = 425/16 and r = 5√17/4.

Teacher's Note:
1. Use the perpendicular from the centre to each chord.
2. Square the equation once.

Answer: (a) \( 8\sqrt{2} \) cm

Explanation:
1. AD = AC = 6 cm, so AB = BC = 3 cm and DB = 9 cm. The radius BF = 3 cm.
2. DF = √(DB² - BF²) = √(81 - 9) = 6√2 cm.
3. Angle DEC = 90° (angle in a semicircle) and BF is perpendicular to DE, so BF is parallel to CE and DF/DE = DB/DC = 9/12.
4. DE = 6√2 × 12/9 = 8√2 cm.

Teacher's Note:
1. The radius to the point of contact is perpendicular to the tangent.
2. Use similar triangles DFB and DEC.

Answer: (c) 3 cm

Explanation:
1. The smaller circle has diameter AC = 6 cm, so its radius BF = 3 cm.

Teacher's Note:
1. BF is a radius of the smaller circle.
2. AC = 6 cm is its diameter.

Answer: (a) \( 9\sqrt{2} \) cm²

Explanation:
1. Triangle BDF is right-angled at F with BF = 3 cm and DF = 6√2 cm.
2. Area = ½ × 3 × 6√2 = 9√2 cm².

Teacher's Note:
1. Angle BFD = 90° (radius and tangent).
2. Use half the product of the legs.