This question was uploaded on 17/01/22 on social media accounts.
Solution:
Here, y is assumed and 2y is due to intersecting chord theorem.
2y length is divided into two parts of x, (2y-x).
Red triangle is right angled triangle with sides (2x+y), (2y-x), 2r.
By intersecting chord theorem:
\[\color{blue} {(2x)(y)=(x)(2y)=(3)(1)}\]
\[\Rightarrow\color{blue} {y=\frac3{2x}}\]
By right angled triangle (Red Triangle):
\[\color{red} {(2x+y)^2+(2y-x)^2=4^2}\]
\[\Rightarrow\color{red} {5(x^2+y^2)=16}\]
\[\Rightarrow\color{red} {5\left(x^2+\left(\color{blue}{\frac3{2x}}\right)^2\right)=16}\]
\[\Rightarrow\color{red} {20x^4-64x^2+45=0}\]
\[\Rightarrow\color{red} {x^2=\frac{16\pm\sqrt{31}}{10}}\]
Blue Area (A):
\[A=\color{blue} {2x^2=\frac{16\pm\sqrt{31}}5}\]
Blue Area cannot be much small, so we will neglect negative symbol:
\[A=\color{blue} {\frac{16+\sqrt{31}}5}\]
