Question Number 210789 by peter frank last updated on 19/Aug/24
Answered by mr W last updated on 19/Aug/24
$${Area}\:{of}\:{sector}\:{O}\overset{\frown} {{PQ}}: \\ $$$${A}_{\mathrm{1}} =\frac{\theta}{\mathrm{2}\pi}×\pi{r}^{\mathrm{2}} =\frac{{r}^{\mathrm{2}} \theta}{\mathrm{2}} \\ $$$${Area}\:{of}\:{triangle}\:{OPQ}: \\ $$$${A}_{\mathrm{2}} =\frac{{r}×{r}\:\mathrm{sin}\:\theta}{\mathrm{2}}=\frac{{r}^{\mathrm{2}} \:\mathrm{sin}\:\theta}{\mathrm{2}} \\ $$$${Area}\:{of}\:{segmeny}\:{P}\overset{\frown} {{QP}}: \\ $$$${A}={A}_{\mathrm{1}} −{A}_{\mathrm{2}} =\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\theta−\mathrm{sin}\:\theta\right)\:\checkmark \\ $$
Commented by peter frank last updated on 20/Aug/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{Mr}\:\mathrm{w} \\ $$