Question

In the adjoining figure, $O$ is the centre of a circle, chord $PQ\cong$ chord $RS$.
If $\angle POR={70}^{\circ }$ and $\left(arcRS\right)={80}^{\circ }$, find
i. $m$ (arc PR)
ii. $m$ (arc QS)
iii. $m(arcQSR)$.

Answer 1

i. $m\left(\mathrm{arc}PR\right)=m\angle POR$ [Definition of measure of arc]
$\therefore m\left(\mathrm{arc}PR\right)={70}^{\circ }$
ii. chord $PQ$ chord RS [Given]
$\therefore m\left(arcPQ\right)=m\left(arcRS\right)={80}^{\circ }$ [Corresponding arcs of congruents chords of a circle are congruent]
Now, $m\left(arcQS\right)+m\left(\mathrm{arc}PQ\right)+m\left(\mathrm{arc}PR\right)+m\left(\mathrm{arcRS}\right)={360}^{\circ }$
$\therefore m\left(\mathrm{arc}QS\right)+{80}^{\circ }+{70}^{\circ }+{80}^{\circ }={360}^{\circ }$ [Measure of a circle is ${360}^{\circ }$ ]
$\therefore m\left(arcQS\right)+{230}^{\circ }={360}^{\circ }$
$\therefore m\left(\mathrm{arc}QS\right)={130}^{\circ }$
iii. $m\left(arcQSR\right)=m\left(arcQS\right)+m\left(arcSR\right)[$ Arc addition property]
$={130}^{\circ }+{80}^{\circ }$
$\therefore m\left(\mathrm{arc}QSR\right)={210}^{\circ }$