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The chords PQ and RS of a circle with centre O subtend equal angles of 72° at O…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

The chords $PQ$ and $RS$ of a circle with centre $O$ subtend equal angles of $72^\circ$ at $O$ . Show that $PQ = RS$ . Given $PQ = 9\ \text{cm}$ , find $RS$ and $\angle OPQ$ .

Need a reminder? The key ideaShow

Equal chords of a circle subtend equal angles at the centre, and chords that subtend equal angles at the centre are equal. The proofs use the radii, which are equal, and the SSS and SAS rules. Triangle $OAB$ is isosceles, so $\angle OAB = \dfrac{180^\circ - \angle AOB}{2}$ .

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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