CBSE Class 10 Mathematics Circles MCQs Set 15

Answer: (a) \( 2\sqrt7\ \text{cm} \)

Explanation:
1. The tangent is perpendicular to the radius, so length = sqrt(8^2 - 6^2) = sqrt28 = 2 sqrt7 cm.

Teacher's Note:
1. A tangent to a circle is perpendicular to the radius through the point of contact, so the tangent length from a point at distance d from the centre of a circle of radius r is sqrt(d^2 - r^2). Tangents from an external point are equal in length and subtend equal angles at the centre.

Answer: (d) \( \sqrt{119}\ \text{cm} \)

Explanation:
1. OP is perpendicular to PQ, so PQ = sqrt(OQ^2 - OP^2) = sqrt(144 - 25) = sqrt119 cm.

Teacher's Note:
1. A tangent to a circle is perpendicular to the radius through the point of contact, so the tangent length from a point at distance d from the centre of a circle of radius r is sqrt(d^2 - r^2). Tangents from an external point are equal in length and subtend equal angles at the centre.

Answer: (c) 50 degrees

Explanation:
1. In quadrilateral OAPB the angles at A and B are 90 degrees and angle P = 80, so angle AOB = 100. OP bisects it, so angle POA = 50 degrees.

Teacher's Note:
1. A tangent to a circle is perpendicular to the radius through the point of contact, so the tangent length from a point at distance d from the centre of a circle of radius r is sqrt(d^2 - r^2). Tangents from an external point are equal in length and subtend equal angles at the centre.

Answer: (b) 90 degrees

Explanation:
1. The tangent at C meets AB at a point M with MA = MC = MB (equal tangents). So M is the centre of a circle through A, C, B with AB as diameter, and angle ACB = 90 degrees.

Teacher's Note:
1. A tangent to a circle is perpendicular to the radius through the point of contact, so the tangent length from a point at distance d from the centre of a circle of radius r is sqrt(d^2 - r^2). Tangents from an external point are equal in length and subtend equal angles at the centre.

Answer: (b) 2 cm

Explanation:
1. AC = 10 cm. r = (6 + 8 - 10)/2 = 2 cm. Check: area 24 = r x semi-perimeter 12.

Teacher's Note:
1. For a quadrilateral circumscribing a circle, the sum of one pair of opposite sides equals the sum of the other pair: AB + CD = BC + AD. A right triangle with legs a, b and hypotenuse c has inradius r = (a + b - c)/2.

Answer: (a) 14 cm

Explanation:
1. BQ = BP = 27, so CQ = 38 - 27 = 11 and CR = 11, giving DR = 25 - 11 = 14. Since angle D = 90 degrees and OS, OR are perpendicular to DA and DC, the quadrilateral ORDS has four right angles and OR = OS, so it is a square and r = DR = 14 cm.

Teacher's Note:
1. A tangent to a circle is perpendicular to the radius through the point of contact, so the tangent length from a point at distance d from the centre of a circle of radius r is sqrt(d^2 - r^2). Tangents from an external point are equal in length and subtend equal angles at the centre.

Answer: (d) 8 cm

Explanation:
1. The chord touches the smaller circle at its mid-point M with OM = 3. Half the chord = sqrt(25 - 9) = 4, so the chord is 8 cm.

Teacher's Note:
1. A tangent to a circle is perpendicular to the radius through the point of contact, so the tangent length from a point at distance d from the centre of a circle of radius r is sqrt(d^2 - r^2). Tangents from an external point are equal in length and subtend equal angles at the centre.

Answer: (c) 9.6 cm

Explanation:
1. AO is the perpendicular bisector of BC. Let it meet BC at M. By similar triangles AB^2 = AM x AD = AM x 10, so AM = 3.6 cm. Then BM = sqrt(36 - 12.96) = 4.8 and BC = 9.6 cm.

Teacher's Note:
1. Use the right angle in a semicircle (angle ABD = 90 degrees) together with similar triangles.

Answer: (c) 13 cm and 15 cm

Explanation:
1. Let the tangent length from the opposite vertex be x. The sides are 14, 6 + x and 8 + x with semi-perimeter s = 14 + x. Area = r s = 4(14 + x) = sqrt(s x 6 x 8), so 16(14 + x) = 48x and x = 7. The other sides are 13 and 15.

Teacher's Note:
1. For a quadrilateral circumscribing a circle, the sum of one pair of opposite sides equals the sum of the other pair: AB + CD = BC + AD. A right triangle with legs a, b and hypotenuse c has inradius r = (a + b - c)/2.

Answer: (b) 2 cm

Explanation:
1. PR = 13 cm. x = (12 + 5 - 13)/2 = 2 cm.

Teacher's Note:
1. For a quadrilateral circumscribing a circle, the sum of one pair of opposite sides equals the sum of the other pair: AB + CD = BC + AD. A right triangle with legs a, b and hypotenuse c has inradius r = (a + b - c)/2.

Answer: (a) 3 cm

Explanation:
1. AB + CD = BC + AD, so 6 + 4 = 7 + AD and AD = 3 cm.

Teacher's Note:
1. For a quadrilateral circumscribing a circle, the sum of one pair of opposite sides equals the sum of the other pair: AB + CD = BC + AD. A right triangle with legs a, b and hypotenuse c has inradius r = (a + b - c)/2.

Answer: (d) 12 cm

Explanation:
1. Tangent = sqrt(13^2 - 5^2) = sqrt144 = 12 cm.

Teacher's Note:
1. A tangent to a circle is perpendicular to the radius through the point of contact, so the tangent length from a point at distance d from the centre of a circle of radius r is sqrt(d^2 - r^2). Tangents from an external point are equal in length and subtend equal angles at the centre.

Answer: (d) 30 degrees

Explanation:
1. The tangents are equally inclined to the line joining the point to the centre, so each makes 60/2 = 30 degrees with it.

Teacher's Note:
1. A tangent to a circle is perpendicular to the radius through the point of contact, so the tangent length from a point at distance d from the centre of a circle of radius r is sqrt(d^2 - r^2). Tangents from an external point are equal in length and subtend equal angles at the centre.

Answer: (c) 9 cm

Explanation:
1. AB + CD = AD + BC with AB = CD and AD = BC gives AB = AD. The parallelogram is a rhombus, so AD = 9 cm.

Teacher's Note:
1. For a quadrilateral circumscribing a circle, the sum of one pair of opposite sides equals the sum of the other pair: AB + CD = BC + AD. A right triangle with legs a, b and hypotenuse c has inradius r = (a + b - c)/2.

Answer: (b) 28 cm

Explanation:
1. AQ = AR, BQ = BP and CP = CR. Perimeter = AB + BC + CA = (AB + BQ) + (CR + CA) = AQ + AR = 2AQ = 28 cm.

Teacher's Note:
1. A tangent to a circle is perpendicular to the radius through the point of contact, so the tangent length from a point at distance d from the centre of a circle of radius r is sqrt(d^2 - r^2). Tangents from an external point are equal in length and subtend equal angles at the centre.

Answer: (a) 50 degrees

Explanation:
1. PA = PB, so triangle PAB is isosceles and angle PAB = (180 - 80)/2 = 50 degrees.

Teacher's Note:
1. A tangent to a circle is perpendicular to the radius through the point of contact, so the tangent length from a point at distance d from the centre of a circle of radius r is sqrt(d^2 - r^2). Tangents from an external point are equal in length and subtend equal angles at the centre.

Answer: (a) 12 cm

Explanation:
1. The segment joining the two points of contact passes through the centre and is perpendicular to both tangents, so it is a diameter of length 12 cm.

Teacher's Note:
1. A tangent to a circle is perpendicular to the radius through the point of contact, so the tangent length from a point at distance d from the centre of a circle of radius r is sqrt(d^2 - r^2). Tangents from an external point are equal in length and subtend equal angles at the centre.

Answer: (d) 4 cm

Explanation:
1. AB + CD = BC + AD gives 8 + 5 = 9 + AD, so AD = 4 cm.

Teacher's Note:
1. For a quadrilateral circumscribing a circle, the sum of one pair of opposite sides equals the sum of the other pair: AB + CD = BC + AD. A right triangle with legs a, b and hypotenuse c has inradius r = (a + b - c)/2.

Answer: (c) 60 degrees

Explanation:
1. Angle TPQ = 90 - 30 = 60 degrees, and TP = TQ so angle TQP = 60 too. Hence angle PTQ = 60 degrees = 2 x angle OPQ.

Teacher's Note:
1. A tangent to a circle is perpendicular to the radius through the point of contact, so the tangent length from a point at distance d from the centre of a circle of radius r is sqrt(d^2 - r^2). Tangents from an external point are equal in length and subtend equal angles at the centre.

Answer: (b) 70 degrees

Explanation:
1. The angles at the centre by opposite sides are supplementary: 180 - 110 = 70 degrees.

Teacher's Note:
1. For a quadrilateral circumscribing a circle, the sum of one pair of opposite sides equals the sum of the other pair: AB + CD = BC + AD. A right triangle with legs a, b and hypotenuse c has inradius r = (a + b - c)/2.

Answer: (b) 24 cm

Explanation:
1. Half the chord = sqrt(13^2 - 5^2) = 12, so the chord is 24 cm.

Teacher's Note:
1. A tangent to a circle is perpendicular to the radius through the point of contact, so the tangent length from a point at distance d from the centre of a circle of radius r is sqrt(d^2 - r^2). Tangents from an external point are equal in length and subtend equal angles at the centre.

Answer: (a) 90 degrees

Explanation:
1. OS bisects angle POQ and OT bisects angle QOR. Since POQ + QOR = 180 degrees (PR is a diameter), angle SOT = 90 degrees.

Teacher's Note:
1. A tangent to a circle is perpendicular to the radius through the point of contact, so the tangent length from a point at distance d from the centre of a circle of radius r is sqrt(d^2 - r^2). Tangents from an external point are equal in length and subtend equal angles at the centre.

Answer: (d) passes through the centre of the circle

Explanation:
1. The radii to the points of contact are each perpendicular to the parallel tangents, so they lie along one line: the segment passes through the centre and is a diameter.

Teacher's Note:
1. A tangent to a circle is perpendicular to the radius through the point of contact, so the tangent length from a point at distance d from the centre of a circle of radius r is sqrt(d^2 - r^2). Tangents from an external point are equal in length and subtend equal angles at the centre.

Answer: (c) 3 cm

Explanation:
1. AB = 15 cm. r = (9 + 12 - 15)/2 = 3 cm.

Teacher's Note:
1. For a quadrilateral circumscribing a circle, the sum of one pair of opposite sides equals the sum of the other pair: AB + CD = BC + AD. A right triangle with legs a, b and hypotenuse c has inradius r = (a + b - c)/2.