Circles CBQs — CBSE Class 10 (2027) with step-wise solutions

Sample Circles CBQs

1. Question 1MCQ1 mark
   A carpenter makes a wooden quadrilateral frame PQRS for a round wall clock. The circular clock face fits inside the frame so that it touches all four sides PQ, QR, RS and SP. If PQ = 18 cm, QR = 21 cm and RS = 25 cm, then the length of side SP is
   • 18 cm
   • 22 cm
   • 24 cm
   • 28 cm
   Show step-wise solution

   1. [1 mark] The frame circumscribes the circle, so by equal tangents from each vertex PQ + RS = QR + SP ⇒ 18 + 25 = 21 + SP ⇒ SP = 43 − 21 = 22 cm.

   Answer: SP = 22 cm

2. Question 2Assertion-Reason1 mark
   A bicycle wheel with centre O and radius 35 cm rests on a straight, level road and touches the road at the point P. Assertion (A): The spoke OP, which joins the centre of the wheel to the point where it touches the road, is perpendicular to the road. Reason (R): The tangent at any point of a circle is perpendicular to the radius through the point of contact.
   • (A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
   • (B) Both Assertion and Reason are true but Reason is NOT the correct explanation of Assertion.
   • (C) Assertion is true but Reason is false.
   • (D) Assertion is false but Reason is true.
   Show step-wise solution

   1. [1 mark] (A): The road touches the rim only at P, so the road is a tangent to the circle at P and OP is the radius through the point of contact; by the theorem in R, OP ⊥ road. Both are true and R explains A.

   Answer: (A)

3. Question 3Short answer2 marks
   A garden plot ABCD is in the shape of a quadrilateral with ∠B = 90°. A circular fountain with centre O is built inside it so that its boundary touches the sides AB, BC, CD and DA at P, Q, R and S respectively. If AD = 17 m, AB = 20 m and DS = 4 m, find the radius of the fountain.
   Show step-wise solution

   1. [1 mark] Tangents from an external point are equal: AS = AD − DS = 17 − 4 = 13 m, so AP = AS = 13 m and PB = AB − AP = 20 − 13 = 7 m.
   2. [1 mark] OP ⊥ AB and OQ ⊥ BC (tangent ⊥ radius) and ∠B = 90°, so OPBQ is a rectangle with OP = OQ = r, i.e. a square; hence r = PB = 7 m.

   Answer: 7 m

4. Question 4Case study4 marks
   In a village in Gujarat, the old circular well has a low parapet wall whose outer edge is a circle with centre O and radius 1.5 m. To keep cattle away, the panchayat fixes a bamboo post at a point P on the ground, 2.5 m from O. Two straight bamboo rails PA and PB are fixed from the post so that each rail just touches the outer edge of the parapet, at A and B respectively. The region enclosed by the rails PA, PB and the radii OA, OB is to be paved with stone. Based on the above information, answer the following questions: (i) What is the measure of ∠OAP? Give a reason. (ii) Find the length of the rail PA. (iii) Find the area of the quadrilateral OAPB that is to be paved.
   Show step-wise solution

   1. [1 mark] (i) ∠OAP = 90°, because PA is a tangent at A and the tangent at any point of a circle is perpendicular to the radius through the point of contact.
   2. [1 mark] (ii) In right △OAP, PA = √(OP² − OA²) = √(2.5² − 1.5²) = √(6.25 − 2.25) = √4 = 2 m.
   3. [1 mark] (iii) PB = PA = 2 m (tangents from an external point) and OB = OA, so △OAP ≅ △OBP (SSS); area of △OAP = ½ × OA × PA = ½ × 1.5 × 2 = 1.5 m².
   4. [1 mark] (iii) Area of OAPB = 2 × 1.5 = 3 m².

   Answer: (i) 90° (ii) PA = 2 m (iii) 3 m²

5. Question 5Long answer5 marks
   A drive belt leaves a circular pulley of radius 10 cm (centre O) at the points A and B and runs in two straight pieces PA and PB to a small hook P, each piece just touching the pulley. The two pieces of belt meet at P at an angle ∠APB = 60°. (a) Prove that ∠AOB + ∠APB = 180° and find ∠AOB. [2 marks] (b) Find the distance OP of the hook from the centre of the pulley. [2 marks] (c) Find the length PA of one straight piece of belt. (Use √3 = 1.732.) [1 mark]
   Show step-wise solution

   1. [1 mark] (a) OA ⊥ PA and OB ⊥ PB (tangent ⊥ radius), so ∠OAP = ∠OBP = 90°. Angle sum of quadrilateral OAPB: ∠AOB + ∠APB = 360° − 90° − 90° = 180°.
   2. [1 mark] ∠AOB = 180° − 60° = 120°.
   3. [1 mark] (b) △OAP ≅ △OBP (RHS: OA = OB, OP common, right angles at A and B), so OP bisects ∠APB and ∠OPA = 30°.
   4. [1 mark] In right △OAP, sin 30° = OA/OP ⇒ 1/2 = 10/OP ⇒ OP = 20 cm.
   5. [1 mark] (c) PA² = OP² − OA² = 400 − 100 = 300 ⇒ PA = 10√3 = 10 × 1.732 = 17.32 cm.

   Answer: ∠AOB = 120°; OP = 20 cm; PA = 10√3 ≈ 17.32 cm

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