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Question Number 218491 by Spillover last updated on 10/Apr/25

Commented by Nicholas666 last updated on 10/Apr/25

$$R=\sqrt{\frac{116}{\pi }}$$

Answered by Nicholas666 last updated on 11/Apr/25

$$a\Rightarrow a^2+a^2={\left(2R\right)}^2\Rightarrow a^2=2R^2$$$$S_1=a^2=2R^2$$$$4S=\pi R^2-S_1=\pi R^2-2R^2=R^2(\pi -2)$$$$4S+S_1=116$$$$R^2(\pi -2+2)=116$$$$R^2\pi =116$$$$R^2=116/\pi =\sqrt{\frac{116}{\pi }}=6.076$$$$$$

Answered by mr W last updated on 11/Apr/25

$$S_1={\left(\sqrt{2}R\right)}^2=2R^2$$$$S=a\times a$$$$R^2={(\frac{R}{\sqrt{2}}+a)}^2+{\left(\frac{a}{2}\right)}^2$$$$5a^2+4\sqrt{2}Ra-2R^2=0$$$$\Rightarrow a=\frac{\sqrt{2}R}{5}$$$$\Rightarrow S=a^2=\frac{2R^2}{25}$$$$4S+S_1=116$$$$4\times \frac{2R^2}{25}+2R^2=116$$$$\Rightarrow R=5\sqrt{2}✓$$

Commented by Spillover last updated on 11/Apr/25

$$correct$$

Answered by Spillover last updated on 11/Apr/25

Answered by Spillover last updated on 11/Apr/25

Answered by Spillover last updated on 11/Apr/25

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