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Question Number 207683 by efronzo1 last updated on 22/May/24

Answered by mr W last updated on 23/May/24

Commented by mr W last updated on 22/May/24

$$MK=\sqrt{9^2+{(6-3)}^2}=3\sqrt{10}$$$$MN=\sqrt{9^2+{(6+3)}^2}=9\sqrt{2}$$$$\sin \alpha =\frac{9}{9\sqrt{2}}=\frac{1}{\sqrt{2}}$$$$R=\frac{MK}{2\sin \alpha }=\frac{3\sqrt{10}}{2\times \frac{1}{\sqrt{2}}}=3\sqrt{5}$$$$areaofsquare={\left(2R\right)}^2={\left(6\sqrt{5}\right)}^2=180$$

Commented by Tawa11 last updated on 21/Jun/24

$$\mathrm{weldone}\mathrm{sir}$$

Answered by efronzo1 last updated on 23/May/24

$$\mathrm{Let}\{\begin{array}{l}\mathrm{CH}=a\\ A\mathrm{E}=\mathrm{b}\end{array}\mathrm{then}\{\begin{array}{l}36=a(9+b)\\ 9=b(9+a)\end{array}$$$$\{\begin{array}{l}a=b+3\\ (b+12)b-9=0\end{array}\Rightarrow b=3\sqrt{5}-6$$$$\mathrm{and}a=3\sqrt{5}-3$$$$\mathrm{S}\left[\mathrm{ABCD}\right]={(9+a+b)}^2={\left(6\sqrt{5}\right)}^2=180{cm}^2$$

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