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Question Number 202760 by Mastermind last updated on 02/Jan/24

Answered by ajfour last updated on 02/Jan/24

$$\frac{\pi }{2}$$

Commented by Mastermind last updated on 02/Jan/24

$$\mathrm{your}\mathrm{workings}\mathrm{please}!$$

Commented by ajfour last updated on 02/Jan/24

$$reqarea=\frac{\pi r^2}{2}$$$$sincer=1,requiredregion=\frac{\pi }{2}$$

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