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Question Number 187587 by Mingma last updated on 19/Feb/23

Commented by Mingma last updated on 19/Feb/23

Blue area=?

Answered by mr W last updated on 19/Feb/23

Commented by mr W last updated on 19/Feb/23

$$\frac{s}{s+\sqrt{2}s}=\frac{1}{\sqrt{{\left(\sqrt{2}s\right)}^2-1^2}}$$$$\Rightarrow s^2=2+\sqrt{2}=bluearea$$

Commented by Rupesh123 last updated on 22/Feb/23

Great work!

Answered by HeferH last updated on 19/Feb/23

Commented by HeferH last updated on 19/Feb/23

$$d=\sqrt{a^2[{(1+\sqrt{2})}^2+1]}=a\sqrt{4+2\sqrt{2}}$$$$a\sqrt{2}=\sqrt{4+2\sqrt{2}}$$$$a^2=2+\sqrt{2}$$

Commented by Rupesh123 last updated on 22/Feb/23

Great work!

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