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Question Number 210789 by peter frank last updated on 19/Aug/24

Answered by mr W last updated on 19/Aug/24

$$AreaofsectorO\stackrel{⌢}{PQ}:$$$$A_1=\frac{\theta }{2\pi }\times \pi r^2=\frac{r^2\theta }{2}$$$$AreaoftriangleOPQ:$$$$A_2=\frac{r\times r\sin \theta }{2}=\frac{r^2\sin \theta }{2}$$$$AreaofsegmenyP\stackrel{⌢}{QP}:$$$$A=A_1-A_2=\frac{r^2}{2}(\theta -\sin \theta )✓$$

Commented by peter frank last updated on 20/Aug/24

$$\mathrm{thank}\mathrm{you}\mathrm{Mr}\mathrm{w}$$

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