CBSE Class 10 Mathematics Circles MCQs Set 10

Answer: (b) 9 cm

Explanation:
1. Tangents from an external point are equal: CQ = CR = 3 cm.
2. BQ = BC − CQ = 7 − 3 = 4 cm, so BP = BQ = 4 cm.
3. AP = AS = 5 cm, so AB = AP + PB = 5 + 4 = 9 cm.

Teacher's Note:
1. Write down the tangent length from each vertex, one vertex at a time.
2. Each side is the sum of two tangent lengths.

Answer: (d) 90°

Explanation:
1. Draw the common tangent at P; let it meet AB at O.
2. OA = OP and OP = OB (tangents from O), so ∠OAP = ∠OPA and ∠OBP = ∠OPB.
3. In △APB, ∠OAP + ∠OBP + ∠APB = 180°, that is, ∠OPA + ∠OPB + ∠APB = 180°, so 2∠APB = 180° and ∠APB = 90°.

Teacher's Note:
1. The common tangent at the point of contact meets the other common tangent at its mid-point.
2. Equal tangents make two isosceles triangles.

Answer: (b) 7.6

Explanation:
1. P is an external point for both circles, so PQ = PT and PR = PT.
2. Hence PQ = PR = 3.8 cm.
3. QR = PQ + PR = 3.8 + 3.8 = 7.6 cm.

Teacher's Note:
1. Tangents from one external point to a circle are equal.
2. Use the point P once for each circle.

Answer: (b) 45°

Explanation:
1. For a quadrilateral circumscribing a circle, the angles subtended at the centre by opposite sides are supplementary: ∠AOD + ∠BOC = 180°.
2. ∠BOC = 180° − 135° = 45°.

Teacher's Note:
1. Opposite sides of a circumscribed quadrilateral subtend supplementary angles at the centre.
2. This follows from AB + CD = AD + BC and the equal-tangent theorem.

Answer: (b) 4 cm

Explanation:
1. CA ⟂ PA and CB ⟂ PB (radius ⟂ tangent), and ∠APB = 90°, so ∠ACB = 360° − 270° = 90°.
2. APBC has four right angles and AP = BP, so it is a square.
3. Each tangent equals the radius, 4 cm.

Teacher's Note:
1. Three right angles in a quadrilateral force the fourth to be 90° as well.
2. A rectangle with two equal adjacent sides is a square.

Answer: (a) 11

Explanation:
1. DR = DS = 5 cm, so AR = AD − DR = 23 − 5 = 18 cm.
2. AQ = AR = 18 cm, so QB = AB − AQ = 29 − 18 = 11 cm.
3. OQ ⟂ AB, OP ⟂ BC and ∠B = 90°, so OQBP is a square. Radius = OQ = QB = 11 cm.

Teacher's Note:
1. Find the tangent lengths from D, then A, then B.
2. A square is formed at the right-angled vertex, so radius = tangent length from B.

Answer: (c) 5

Explanation:
1. The radius OP is perpendicular to the tangent QP, so triangle OPQ is right-angled at P with OQ = 13 and PQ = 12.
2. OP² = OQ² − PQ² = 169 − 144 = 25, so the radius = 5 cm.

Teacher's Note:
1. The distance from the centre is the hypotenuse.
2. 13, 12, 5 is a Pythagorean triple.

Answer: (a) 7.5

Explanation:
1. Let BC touch the circle at D. Then BP = BD and CQ = CD, and AP = AQ.
2. AP + AQ = (AB + BP) + (AC + CQ) = AB + AC + (BD + CD) = 5 + 6 + 4 = 15.
3. Since AP = AQ, 2AP = 15 and AP = 7.5 cm.

Teacher's Note:
1. AP is half the perimeter of triangle ABC.
2. Replace BP by BD and CQ by CD.

Answer: (b) 10

Explanation:
1. AR = AP = 4 cm, so CR = AC − AR = 11 − 4 = 7 cm.
2. CQ = CR = 7 cm and BQ = BP = 3 cm.
3. BC = BQ + CQ = 3 + 7 = 10 cm.

Teacher's Note:
1. Each vertex has one fixed tangent length.
2. A side is the sum of the tangent lengths of its two end vertices.

Answer: (b) 9 cm

Explanation:
1. PA = PB = 12 cm (tangents from P).
2. CA = CQ = 3 cm (tangents from C).
3. PC = PA − CA = 12 − 3 = 9 cm.

Teacher's Note:
1. C lies on PA, so PC and CA are parts of PA.
2. The tangent length from C is CQ = CA.

Answer: (c) 90°

Explanation:
1. The tangent at any point of a circle is perpendicular to the radius through the point of contact.
2. The angle is 90°.

Teacher's Note:
1. This is the basic tangent theorem.
2. Many other results in the chapter start from this right angle.

Answer: (b) 45°

Explanation:
1. OP ⟂ PQ, so ∠OPQ = 90°.
2. A triangle can have only one right angle, so the two equal angles are ∠POQ and ∠OQP.
3. Each is (180° − 90°) ÷ 2 = 45°.

Teacher's Note:
1. The equal angles of an isosceles triangle cannot include the right angle.
2. Use the angle sum of a triangle.

Answer: (b) 2 cm

Explanation:
1. AC = √(8² + 6²) = 10 cm.
2. For a right triangle, the inradius r = (AB + BC − AC) ÷ 2 = (8 + 6 − 10) ÷ 2 = 2 cm.
3. Check by areas: ½ × 8 × 6 = 24 = r × (24 ÷ 2), so r = 2 cm.

Teacher's Note:
1. The inradius meets the two legs in a square at B.
2. Tangent lengths from B equal r.

Answer: (a) 30 cm

Explanation:
1. The circle touches AB at R, BC at P and CA at Q, with AQ = 4 cm, BR = 6 cm and CP = 5 cm.
2. AR = AQ = 4, BP = BR = 6, CQ = CP = 5.
3. AB = 4 + 6 = 10, BC = 6 + 5 = 11, CA = 5 + 4 = 9, so the perimeter = 10 + 11 + 9 = 30 cm.

Teacher's Note:
1. The perimeter is twice the sum of the three tangent lengths.
2. Every vertex has two equal tangent segments.

Answer: (b) 10 m

Explanation:
1. MT = MQ and NT = NR (tangents from M and from N).
2. Perimeter of △PMN = PM + PN + MN = PM + PN + (MT + NT) = (PM + MQ) + (PN + NR) = PQ + PR.
3. Perimeter = 5 + 5 = 10 m.

Teacher's Note:
1. Split MN into MT and TN and replace each by an equal tangent.
2. The perimeter equals the sum of the two outer tangent lengths.

Answer: (a) 100°

Explanation:
1. PA = PB (tangents from P), so ∠PBA = ∠PAB = 50°.
2. In △APB, ∠APB = 180° − 50° − 50° = 80°.
3. ∠AOB = 180° − ∠APB = 180° − 80° = 100°.

Teacher's Note:
1. The angle between the tangents and the angle between the radii are supplementary.
2. Find ∠APB first using the isosceles triangle APB.

Answer: (b) 5 cm

Explanation:
1. For a quadrilateral circumscribing a circle, AB + CD = BC + AD (equal tangents).
2. 6 + 8 = 9 + AD, so AD = 14 − 9 = 5 cm.

Teacher's Note:
1. The sums of opposite sides of a circumscribed quadrilateral are equal.
2. Add the tangent lengths from each vertex to prove it.

Answer: (b) 8 cm

Explanation:
1. AR = AQ = 7 cm, so BR = AB − AR = 10 − 7 = 3 cm.
2. BP = BR = 3 cm and CP = CQ = 5 cm.
3. BC = BP + PC = 3 + 5 = 8 cm.

Teacher's Note:
1. Work out the tangent length from each vertex.
2. A side is the sum of the tangent lengths from its two end points.

Answer: (a) 5 cm

Explanation:
1. AP = BP, so ∠PAB = ∠PBA = (180° − 60°) ÷ 2 = 60°.
2. All three angles of △APB are 60°, so it is equilateral and AB = AP = 5 cm.

Teacher's Note:
1. Equal tangents make △APB isosceles.
2. An isosceles triangle with a 60° angle is equilateral.

Answer: (c) 24 cm

Explanation:
1. AP = √(13² − 5²) = 12 cm, and AQ = AP = 12 cm.
2. BR = BP and CR = CQ (tangents from B and C), so the perimeter = AB + BR + RC + CA = AB + BP + CQ + CA = AP + AQ.
3. Perimeter = 12 + 12 = 24 cm.

Teacher's Note:
1. Replace BC = BR + RC by BP + CQ using equal tangents.
2. The perimeter is the sum of the two tangents from A.