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Question Number 134878 by bemath last updated on 08/Mar/21

Answered by EDWIN88 last updated on 08/Mar/21

$$\mathrm{B}\left(\theta \right)=\frac{1}{2}.100.\sin \theta =50\sin \theta$$$$\left(\mathrm{i}\right)\mathrm{radius}\mathrm{of}\mathrm{semi}\mathrm{circle}\mathrm{is}\mathrm{r}=10.\sin \left(\frac{1}{2}\theta \right)$$$$\mathrm{so}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{semi}\mathrm{circle}\mathrm{A}\left(\theta \right)=\frac{1}{2}\pi (100.\sin {}^2\left(\frac{1}{2}\theta \right))$$$$\mathrm{A}\left(\theta \right)=50\pi \sin {}^2\left(\frac{1}{2}\theta \right)$$$$\underset{\theta \to 0^{+}}{\lim }\frac{50\pi \sin {}^2\left(\frac{1}{2}\theta \right)}{50\sin \theta }=\underset{\theta \to 0^{+}}{\lim }\frac{\pi \sin \left(\frac{1}{2}\theta \right)}{2\cos \left(\frac{1}{2}\theta \right)}=0$$

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