In figure 3.92, O is the centre of a circle, chord PQ ≅ chord RS. If ∠ POR = 70^° and (arc RS) = 80^° , find - (1) m(arc PR) (2) m(arc QS) (3) m(arc QSR) Figure: Going round the circle the points are in the order Q, P, R, S; radii OP and OR are drawn.

Answer: (1) 70^° (2) 130^° (3) 210^° (1) m(arc PR) = central angle ∠ POR = 70^° . (2) Chord PQ ≅ chord RS, so m(arc PQ) = m(arc RS) = 80^° . The four arcs QP, PR, RS, SQ make the whole circle: m(arc QS) = 360^° - (80^° + 70^° + 80^°) = 130^° (3) m(arc QSR) = m(arc QS) + m(arc SR) = 130^° + 80^° = 210^°

In figure 3.92, O is the centre of a circle, chord PQ ≅ chord RS. If ∠POR = 70° and (arc RS) = 80°, find - (1) m(arc PR) (2) m(arc QS) (3) m(arc QSR)

In figure 3.92, O is the centre of a circle, chord PQ ≅ chord RS. If \(\angle POR = 70^\circ\) and (arc RS) = \(80^\circ\), find -

(1) m(arc PR) (2) m(arc QS) (3) m(arc QSR)

Figure: Going round the circle the points are in the order Q, P, R, S; radii OP and OR are drawn.

Doubtly Asked 24 Oct 2026 Save WhatsApp Report

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Answer: (1) \(70^\circ\) (2) \(130^\circ\) (3) \(210^\circ\)

(1) m(arc PR) = central angle \(\angle POR = 70^\circ\).

(2) Chord PQ ≅ chord RS, so m(arc PQ) = m(arc RS) = \(80^\circ\). The four arcs QP, PR, RS, SQ make the whole circle:

\[m(\text{arc QS}) = 360^\circ - (80^\circ + 70^\circ + 80^\circ)\] \[= 130^\circ\]

(3)

\[m(\text{arc QSR}) = m(\text{arc QS}) + m(\text{arc SR})\] \[= 130^\circ + 80^\circ = 210^\circ\]

Doubtly AI · 24 Oct 2026 AI-verified · Oct 2026 Report

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In figure 3.92, O is the centre of a circle, chord PQ ≅ chord RS. If ∠POR = 70° and (arc RS) = 80°, find - (1) m(arc PR) (2) m(arc QS) (3) m(arc QSR)

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