We observe that \(|x-y|>\frac{1}{4}\)

\(<=>x-y>\frac{1}{4}\) or \(x-y<-\frac{1}{4}\).

A\cap B= shaded region. Then,

\(\displaystyle{P}{\left({A}\cap{B}\right)}={\frac{{\text{area of the sgaded region}}}{{\text{area of the square of side 2}}}}\)

Now, (I)is a right angled triangle with \(base =\frac{7}{4}\), \(altitude=2-\frac{1}{4}=\frac{7}{4}\)

-> Hence area of (I) is \(\displaystyle{\frac{{\text{1}}}{{\text{2}}}}\cdot{\left({\frac{{\text{7}}}{{\text{4}}}}\right)}^{{2}}\)

(II)is also a right angled triangle with:\(\displaystyle{b}{a}{s}{e}{2}-{\frac{{\text{1}}}{{\text{2}}}}={\frac{{\text{3}}}{{\text{2}}}}{\quad\text{and}\quad}{a}{lt}{i}{t}{u}{d}{e}={\frac{{\text{7}}}{{\text{4}}}}-{\frac{{\text{1}}}{{\text{4}}}}={\frac{{\text{6}}}{{\text{4}}}}\)

Hence area of (II) is \(\displaystyle{\frac{{\text{1}}}{{\text{2}}}}\cdot{\frac{{\text{3}}}{{\text{2}}}}\cdot{\frac{{\text{6}}}{{\text{4}}}}={\left({\frac{{\text{3}}}{{\text{2}}}}\right)}^{{2}}\cdot{\frac{{\text{1}}}{{\text{2}}}}\)

\(\displaystyle\therefore{A}{r}{e}{a}\ {o}{f}\ {s}{h}{a}{d}{e}{d}\ {r}{e}{g}{i}{o}{n}={\frac{{\text{1}}}{{\text{2}}}}\cdot{\left({\left\lbrace{\frac{{\text{49}}}{{\text{16}}}}\right\rbrace}+{\frac{{\text{9}}}{{\text{4}}}}\right)}{s}{q}{y}{a}{r}{e}\ {u}{n}{i}{t}{s}={\frac{{\text{85}}}{{\text{32}}}} {u}{n}{i}{t}{s}.\)

\(\displaystyle{P}{\left({A}\cap{B}\right)}={\frac{{\text{85/32}}}{{\text{2*2}}}}={\frac{{\text{85}}}{{\text{128}}}}\)