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Question Number 201353 by sonukgindia last updated on 05/Dec/23

Commented by mr W last updated on 05/Dec/23

$$doyouhavetheanswer?$$

Answered by mr W last updated on 05/Dec/23

Commented by mr W last updated on 05/Dec/23

$$\frac{z}{a}=\frac{1}{x}\Rightarrow a=xz$$$$\frac{z}{a}=\frac{2}{y}$$$$\Rightarrow \frac{2}{y}=\frac{1}{x}\Rightarrow y=2x$$$$$$$$4{(a+x)}^2=2[{(a+y)}^2+z^2]-2^2$$$$2x^2-a^2+z^2-2=0$$$$(1-x^2)(z^2-2)=0$$$$\Rightarrow z^2=2\Rightarrow z=\sqrt{2}$$$$$$$$CF=\frac{\sqrt{z\times 1\times (z+1+a+x)\times (z+1-a-x)}}{z+1}$$$$=\frac{\sqrt{z[{(z+1)}^2-{(a+x)}^2]}}{z+1}=\sqrt{z(1-x^2)}$$$$BF=\frac{\sqrt{z\times 2\times (z+2+a+y)(z+2-a-y)}}{z+2}$$$$=\frac{\sqrt{2z[{(z+2)}^2-{(a+2x)}^2]}}{z+2}=\sqrt{2z(1-x^2)}$$$$BF=CF+a$$$$\sqrt{2z(1-x^2)}=\sqrt{z(1-x^2)}+xz$$$$(\sqrt{2}-1)\sqrt{1-x^2}=x\sqrt{z}$$$$\Rightarrow {(\sqrt{2}-1)}^2(1-x^2)=x^{2}z$$$$\Rightarrow {(\sqrt{2}-1)}^2=(3-\sqrt{2})x^2$$$$\Rightarrow x=\frac{\sqrt{2}-1}{\sqrt{3-\sqrt{2}}}$$$$\Rightarrow a=xz=\frac{2-\sqrt{2}}{\sqrt{3-\sqrt{2}}}\approx 0.4652$$$$shadedarea=\frac{\sqrt{3}{a}^2}{4}=\frac{5\sqrt{3}-3\sqrt{6}}{14}✓$$

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