Mathematics — Standard (041) — CBSE Class 10 board question

Given: PQ and PR are tangents to circle with centre O; ∠OPR = 45°.

To prove: ORPQ is a square.

Proof:

Since PQ and PR are tangents, by Theorem 10.1:
$$\angle OQP = \angle ORP = 90°$$

By Theorem 10.2, OP bisects ∠QPR, so:
$$\angle QPR = 2 \times \angle OPR = 2 \times 45° = 90°$$

Sum of angles in quadrilateral ORPQ:
$$\angle QOR = 360° - 90° - 90° - 90° = 90°$$

So all four angles are 90°, making ORPQ a rectangle.

Also, OQ = OR (radii) and PQ = PR (equal tangents). Since ∠OQP = 90°, in △OQP:
$$PQ = OQ \cdot \tan 45° = OQ \times 1 \Rightarrow PQ = OQ$$

Thus all sides are equal. Hence ORPQ is a square. $\blacksquare$

Source: Chapter 10, Sections 10.2–10.3 (Theorem 10.1, Theorem 10.2)

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• Examiners look for: stating ∠OQP = ∠ORP = 90° (radius ⊥ tangent), finding ∠QPR = 90° using OP as angle bisector, and then showing all sides equal (tangent length = radius when angle = 45°).
• The key insight is tan 45° = 1, so PQ = OQ, proving the rectangle is actually a square.
• Mention both theorems (10.1 and 10.2) explicitly for full marks.