ABCD is a square of side $l$. A line parallel to the diagonal BD at a distance 'x' from the vertex A cuts two adjacent sides. Express the area of the segment of the square with A at a vertex, as a function of x. Find this area at $x=1/\sqrt{2}$ and at x = 2, when $l$ = 2.

$s(x)=\left\{\begin{matrix}\frac{1}{2}&at\,x=\frac{1}{\sqrt{2}}\\8(\sqrt{2}-1)&at\,x=2\end{matrix}\right.$

There are two cases

Case-I: when $x = AP ≤ OP$, i.e., $x≤\frac{l}{\sqrt{2}}$

$∴ar(ΔAEF)=\frac{1}{2}x.2x=x^2$  $∵PE=PF=AP=x$

Case-II: when $x = AP > OA$, i.e., $x>\frac{l}{\sqrt{2}}$ but $x≤\sqrt{2}l$

$ar(ABEFDA)=ar(ABCD)-ar(ΔCFE)$

$=l^2-\frac{1}{2}(\sqrt{2}l-x).2(\sqrt{2}l-x)$   $[∵CP=\sqrt{2}l-x]$

$=l^2-(2l^2+x^2-2\sqrt{2}lx)=2\sqrt{2}lx-x^-l^2$

So, the required function s(x) is

$s(x)=\left\{\begin{matrix}x^2,&0≤x≤\frac{l}{\sqrt{2}}\\2\sqrt{2}lx-x^2-l^2,&\frac{l}{\sqrt{2}}<x≤\sqrt{2}l\end{matrix}\right.$

$∴s(x)=\left\{\begin{matrix}\frac{1}{2}&at\,x=\frac{1}{\sqrt{2}}\\8(\sqrt{2}-1)&at\,x=2\end{matrix}\right.$