Find the area of the triangle

Note that $(5\sqrt{2})^{2} + (6\sqrt{2})^{2} = 50 + 72 = 122 = (\sqrt{122})^{2}$ , so the given triangle is right-angled with legs length $5\sqrt{2}$ and $6\sqrt{2}$ .

The area of the triangle is therefore $\dfrac{1}{2}(5\sqrt{2})(6\sqrt{2}) = \boxed{30}$ .

Relevant wiki: Heron's Formula

Assuming that we don't know that it is right triangle, use Heron's Formula .

Let $a=5\sqrt{2},b=6\sqrt{2}+\sqrt{122}$ .

$s=\dfrac{5\sqrt{2}+6\sqrt{2}+\sqrt{122}}{2}=\dfrac{11\sqrt{2}+\sqrt{122}}{2}$

$s-a=\dfrac{11\sqrt{2}+\sqrt{122}}{2}-5\sqrt{2}=\dfrac{{11\sqrt{2}+\sqrt{122}}-10\sqrt{2}}{2}=\dfrac{\sqrt{2}+\sqrt{122}}{2}$

$s-b=\dfrac{11\sqrt{2}+\sqrt{122}}{2}-6\sqrt{2}=\dfrac{11\sqrt{2}+\sqrt{122}-12\sqrt{2}}{2}=\dfrac{\sqrt{122}-\sqrt{2}}{2}$

$s-c=\dfrac{11\sqrt{2}+\sqrt{122}}{2}-\sqrt{122}=\dfrac{11\sqrt{2}+\sqrt{122}-2\sqrt{122}}{2}=\dfrac{11\sqrt{2}-\sqrt{122}}{2}$

$s(s-a)=\left(\dfrac{11\sqrt{2}+\sqrt{122}}{2}\right)\left(\dfrac{\sqrt{2}+\sqrt{122}}{2}\right)=\dfrac{11(2)+11\sqrt{2(122)}+\sqrt{2(122)}+122}{4}=\dfrac{144+24\sqrt{61}}{4}$

$(s-b)(s-c)= \left(\dfrac{\sqrt{122}-\sqrt{2}}{2}\right)\left(\dfrac{11\sqrt{2}-\sqrt{122}}{2}\right)=\dfrac{11\sqrt{244}-122-22+\sqrt{244}}{4}=\dfrac{24\sqrt{61}-144}{4}$

$s(s-a)(s-b)(s-c)=\left(\dfrac{144+24\sqrt{61}}{4}\right)\left(\dfrac{24\sqrt{61}-144}{4}\right)=\dfrac{3456\sqrt{61}-20736+35136-3456\sqrt{61}}{16}=900$

$A=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{900}=\color{#D61F06}\boxed{30}$