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Question Number 49748 by Pk1167156@gmail.com last updated on 10/Dec/18

Answered by MJS last updated on 11/Dec/18

$$\mid AB\mid =2a$$$$x^2+y^2=2a^2\Rightarrow y_1=\sqrt{2a^2-x^2}$$$$x^2+{(y-a)}^2=a^2\Rightarrow y_2=a+\sqrt{a^2-x^2}$$$$E=\left(\begin{array}{c}p\\ a+\sqrt{a^2-p^2}\end{array}\right)=\left(\begin{array}{c}p\\ a+5\end{array}\right)\Rightarrow a+\sqrt{a^2-p^2}=a+5\left(1\right)$$$$F=\left(\begin{array}{c}p\\ \sqrt{2a^2-p^2}\end{array}\right)=\left(\begin{array}{c}p\\ a+2\end{array}\right)\Rightarrow \sqrt{2a^2-p^2}=a+2\left(2\right)$$$$\left(1\right)\Rightarrow a^2-p^2=25\Rightarrow p^2=a^2-25$$$$\left(2\right)\sqrt{2a^2-(a^2-25)}=a+2$$$$\sqrt{a^2+25}=a+2$$$$a^2+25=a^2+4a+4\Rightarrow a=\frac{21}{4}$$$$\left(1\right)p^2=\frac{41}{16}\Rightarrow p=\frac{\sqrt{41}}{4}$$$$\mid AB\mid =\frac{21}{2}\Rightarrow \mathrm{area}\left(ABCD\right)=\frac{441}{4}$$

Commented by Pk1167156@gmail.com last updated on 12/Dec/18

thank you Sir

Commented by MJS last updated on 12/Dec/18

$${\mathrm{you}}^{\prime }\mathrm{re}\mathrm{welcome}$$

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