Question Number 180423 by mr W last updated on 12/Nov/22
Commented by Rasheed.Sindhi last updated on 12/Nov/22
$$\mathcal{I}\:{think}\:{sir}\:{that}\:{the}\:{area}\:{depends} \\ $$$${upon}\:{m}\overline {\mathrm{O}_{\mathrm{1}} \mathrm{O}_{\mathrm{2}} } \\ $$
Commented by mr W last updated on 12/Nov/22
$${O}_{\mathrm{1}} \:{and}\:{O}_{\mathrm{2}} \:{are}\:{clearly}\:{defined}:\: \\ $$$${the}\:{center}\:{of}\:{one}\:{circle}\:{lies}\:{on}\:{the}\: \\ $$$${other}\:{circle}. \\ $$
Commented by mr W last updated on 12/Nov/22
$${find}\:{the}\:{area}\:{of}\:{the}\:{square}. \\ $$
Commented by Rasheed.Sindhi last updated on 12/Nov/22
$$\mathcal{R}{ight}\:\boldsymbol{{sir}}! \\ $$
Answered by JDamian last updated on 12/Nov/22
$$\left(\frac{\sqrt{\mathrm{7}}−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} ? \\ $$
Commented by mr W last updated on 12/Nov/22
$${yes} \\ $$
Answered by mr W last updated on 12/Nov/22
$$\mathrm{1}^{\mathrm{2}} =\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{a}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} +\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}×\frac{{a}}{\:\sqrt{\mathrm{2}}}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{a}−\mathrm{3}=\mathrm{0} \\ $$$${a}=\frac{−\mathrm{1}+\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$$$\Rightarrow{area}\:{a}^{\mathrm{2}} =\mathrm{2}−\frac{\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$
Commented by mr W last updated on 12/Nov/22
Commented by mr W last updated on 12/Nov/22
$$\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{a}−\mathrm{3}=\mathrm{0} \\ $$$${a}=\frac{−\mathrm{1}+\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$
Commented by ajfour last updated on 12/Nov/22
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Commented by ajfour last updated on 12/Nov/22
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Commented by mr W last updated on 12/Nov/22
$${thanks}\:{sir}! \\ $$$${nice}\:{to}\:{see}\:{that}\:{you}\:{are}\:{back}\:{to}\:{here}! \\ $$