Q1. [1] § 10.2 Tangent to a Circle § 10.4 Summary
In the given figure, tangents PA and PB to the circle centred at O, from point P are perpendicular to each other. If $PA = 5$ cm, then length of AB is equal to
- (A) $5$ cm
- (B) $5\sqrt{2}$ cm
- (C) $2\sqrt{5}$ cm
- (D) $10$ cm
Previously asked in CBSE board exam
2024 30/2/1 Q14
Generated by claude-sonnet-4-6 · 2026-06-15 10:22 · grounding rag
Model Answer
(B) $5\sqrt{2}$ cm
Since PA ⊥ PB and PA = PB = 5 cm (equal tangents), triangle APB is a right isosceles triangle. By Pythagoras theorem: $AB = \sqrt{5^2 + 5^2} = \sqrt{50} = 5\sqrt{2}$ cm.
Explanation
Key steps: (1) PA = PB because tangents from an external point are equal (Theorem 10.2). (2) ∠APB = 90° is given. (3) Apply Pythagoras in △APB directly. Students often forget that equal tangents make △APB isosceles, so both legs are 5 cm.
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