Question Number 201353 by sonukgindia last updated on 05/Dec/23
Commented by mr W last updated on 05/Dec/23
$${do}\:{you}\:{have}\:{the}\:{answer}? \\ $$
Answered by mr W last updated on 05/Dec/23
Commented by mr W last updated on 05/Dec/23
![(z/a)=(1/x) ⇒a=xz (z/a)=(2/y) ⇒(2/y)=(1/x) ⇒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 ⇒z^2 =2 ⇒z=(√2) CF=((√(z×1×(z+1+a+x)×(z+1−a−x)))/(z+1)) =((√(z[(z+1)^2 −(a+x)^2 ]))/(z+1))=(√(z(1−x^2 ))) BF=((√(z×2×(z+2+a+y)(z+2−a−y)))/(z+2)) =((√(2z[(z+2)^2 −(a+2x)^2 ]))/(z+2))=(√(2z(1−x^2 ))) BF=CF+a (√(2z(1−x^2 )))=(√(z(1−x^2 )))+xz ((√2)−1)(√(1−x^2 ))=x(√z) ⇒((√2)−1)^2 (1−x^2 )=x^2 z ⇒((√2)−1)^2 =(3−(√2))x^2 ⇒x=(((√2)−1)/( (√(3−(√2))))) ⇒a=xz=((2−(√2))/( (√(3−(√2)))))≈0.4652 shaded area=(((√3)a^2 )/4)=((5(√3)−3(√6))/(14)) ✓](https://www.tinkutara.com/question/Q201389.png)
$$\frac{{z}}{{a}}=\frac{\mathrm{1}}{{x}}\:\Rightarrow{a}={xz} \\ $$$$\frac{{z}}{{a}}=\frac{\mathrm{2}}{{y}} \\ $$$$\Rightarrow\frac{\mathrm{2}}{{y}}=\frac{\mathrm{1}}{{x}}\:\Rightarrow{y}=\mathrm{2}{x} \\ $$$$ \\ $$$$\mathrm{4}\left({a}+{x}\right)^{\mathrm{2}} =\mathrm{2}\left[\left({a}+{y}\right)^{\mathrm{2}} +{z}^{\mathrm{2}} \right]−\mathrm{2}^{\mathrm{2}} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −{a}^{\mathrm{2}} +{z}^{\mathrm{2}} −\mathrm{2}=\mathrm{0} \\ $$$$\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left({z}^{\mathrm{2}} −\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow{z}^{\mathrm{2}} =\mathrm{2}\:\Rightarrow{z}=\sqrt{\mathrm{2}} \\ $$$$ \\ $$$${CF}=\frac{\sqrt{{z}×\mathrm{1}×\left({z}+\mathrm{1}+{a}+{x}\right)×\left({z}+\mathrm{1}−{a}−{x}\right)}}{{z}+\mathrm{1}} \\ $$$$\:\:=\frac{\sqrt{{z}\left[\left({z}+\mathrm{1}\right)^{\mathrm{2}} −\left({a}+{x}\right)^{\mathrm{2}} \right]}}{{z}+\mathrm{1}}=\sqrt{{z}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)} \\ $$$${BF}=\frac{\sqrt{{z}×\mathrm{2}×\left({z}+\mathrm{2}+{a}+{y}\right)\left({z}+\mathrm{2}−{a}−{y}\right)}}{{z}+\mathrm{2}} \\ $$$$\:\:=\frac{\sqrt{\mathrm{2}{z}\left[\left({z}+\mathrm{2}\right)^{\mathrm{2}} −\left({a}+\mathrm{2}{x}\right)^{\mathrm{2}} \right]}}{{z}+\mathrm{2}}=\sqrt{\mathrm{2}{z}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)} \\ $$$${BF}={CF}+{a} \\ $$$$\sqrt{\mathrm{2}{z}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}=\sqrt{{z}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}+{xz} \\ $$$$\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }={x}\sqrt{{z}} \\ $$$$\Rightarrow\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)={x}^{\mathrm{2}} {z} \\ $$$$\Rightarrow\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} =\left(\mathrm{3}−\sqrt{\mathrm{2}}\right){x}^{\mathrm{2}} \\ $$$$\Rightarrow{x}=\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\:\sqrt{\mathrm{3}−\sqrt{\mathrm{2}}}} \\ $$$$\Rightarrow{a}={xz}=\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}−\sqrt{\mathrm{2}}}}\approx\mathrm{0}.\mathrm{4652} \\ $$$${shaded}\:{area}=\frac{\sqrt{\mathrm{3}}{a}^{\mathrm{2}} }{\mathrm{4}}=\frac{\mathrm{5}\sqrt{\mathrm{3}}−\mathrm{3}\sqrt{\mathrm{6}}}{\mathrm{14}}\:\checkmark \\ $$
Commented by mr W last updated on 05/Dec/23
$${formulas}: \\ $$
Commented by mr W last updated on 05/Dec/23
