Class 10 Maths Chapter 10: Circles worksheet with answer key
A 20-mark worksheet on Circles for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.
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Answer: (b) 120°
OP = OQ = PQ, so ΔOPQ is equilateral and ∠POQ = 60°. ∠PTQ and ∠POQ are supplementary: ∠PTQ = 180° − 60° = 120°.
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Answer: (c) a tangent
A line that meets a circle in exactly one point is a tangent, and K is its point of contact.
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Answer: (b) 1
OP = 6 cm equals the radius, so P lies on the circle. At a point on the circle there is exactly one tangent.
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Answer: (b) 37 m
The radius to the point of contact is perpendicular to the lane, so the distance from K to the centre is the hypotenuse: √(35² + 12²) = √1369 = 37 m. (√(35² − 12²) treats the distance as a leg.)
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Answer: (a) 38°
The tangent is perpendicular to the radius, so ∠OPT = 90°. In ΔOPT, ∠OTP = 180° − 90° − 52° = 38°.
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∠OAP = ∠OBP = 90°. In right ΔOAP, PA² = OP² − OA²; in right ΔOBP, PB² = OP² − OB². OA = OB (radii), so PA² = PB² and PA = PB.
- PA² = OP² − OA² and PB² = OP² − OB² (1)
- OA = OB, so PA = PB (1)
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TP = TQ and ∠PTO = ∠QTO (ΔOPT ≅ ΔOQT by RHS). In ΔTPM and ΔTQM, TP = TQ, ∠PTM = ∠QTM and TM is common, so they are congruent (SAS). Hence PM = QM and ∠TMP = ∠TMQ = 90°, as the two angles are equal and add to 180°.
- TP = TQ, ∠PTM = ∠QTM, ΔTPM ≅ ΔTQM (SAS) (1)
- PM = QM and ∠TMP = ∠TMQ = 90° (1)
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OA ⟂ tangent at A and OB ⟂ tangent at B. The tangents are parallel, so OA and OB are both perpendicular to the same direction and both pass through O; they lie on one line through O. Hence A, O, B are collinear and AB is a diameter.
- OA and OB perpendicular to the two parallel tangents (1)
- so OA, OB lie on one line through O: AB is a diameter (1)
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The nearest point C lies on PO, so OP = r + 32. ∠OAP = 90°, so (r + 32)² = r² + 56². Then 64r + 1024 = 3136, 64r = 2112, r = 33 m.
- OP = r + 32 (1)
- (r + 32)² = r² + 56² with the right angle at A (1)
- r = 33 m (1)
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TP = TQ, so ∠TQP = 62° and ∠PTQ = 180° − 124° = 56°. ∠POQ = 180° − 56° = 124°. ∠OPQ = 90° − 62° = 28°.
- ∠PTQ = 56° (1)
- ∠POQ = 124° (supplementary) (1)
- ∠OPQ = 28° (1)
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OP ⟂ AB, so P is the midpoint of AB: AP = √(41² − 9²) = √1600 = 40 cm and AB = 80 cm. OP is the height of ΔOAB on the base AB, so its area = ½ × 80 × 9 = 360 cm².
- OP ⟂ AB, AP = 40 cm (1)
- P bisects AB, AB = 80 cm (1)
- area = ½ × AB × OP = 360 cm² (1)
← Chapter 9: Some Applications of TrigonometryChapter 11: Areas Related to Circles →