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Question Number 202581 by Frix last updated on 30/Dec/23

Commented by Frix last updated on 30/Dec/23

$$\mathrm{An}\mathrm{easy}\mathrm{one}:\mathrm{find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{rectangle}.$$

Answered by mr W last updated on 30/Dec/23

Commented by mr W last updated on 30/Dec/23

$$R=radiusofcircumcircleof\mathrm{\Delta }ABC$$$$\frac{\sqrt{5}a}{\sin 135°}=2R=10$$$$\Rightarrow a=\frac{\sqrt{10}R}{5}=\sqrt{10}$$$$areaofrectangle=2a^2=20$$

Commented by Frix last updated on 30/Dec/23

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Answered by ajfour last updated on 30/Dec/23

Commented by ajfour last updated on 30/Dec/23

$$r^2={(a+\frac{a}{2})}^2+k^2$$$$r^2={(a+k)}^2+{\left(\frac{a}{2}\right)}^2$$$$\Rightarrow k=\frac{a}{2}$$$$r^2=\frac{5a^2}{2}$$$$Area=2a^2=\frac{4r^2}{5}=\frac{4}{5}\times 25=20$$

Commented by mr W last updated on 30/Dec/23

$$right!igotthesame.$$

Commented by Frix last updated on 30/Dec/23

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