pythagorean theorem

Relevant wiki: Pythagorean Theorem

Rotate $\triangle AXB$ in its plane $90^\circ$ about $B$ , so that $AB$ falls on $BC$ .

Apply pythagorean theorem on $\triangle XBX'$ , we have

$XX'=\sqrt{3^2+3^2}=3\sqrt{2}$

Apply cosine law on $\triangle XX'C$ ,

$25=49+9(2) -42\sqrt{2} \cos \theta$ $\large \color{#20A900}\implies$ $-42=-42\sqrt{2} \cos \theta$ $\large \color{#20A900}\implies$ $\cos \theta=\dfrac{1}{\sqrt{2}}$ $\large \color{#20A900}\implies$ $\theta=\cos ^{-1} \left(\dfrac{1}{\sqrt{2}}\right)=45^\circ$

$\large \therefore$ $\triangle BXC$ is a right $\triangle$ .

Applying pythagorean theorem on right $\triangle BXC$ , we get

$(BC)^2=3^2+7^2=9+49=58$

$\large \therefore$ $\color{#D61F06}\large \boxed{The~area~of~square~ABCD=58}$