Sounds Simple

Divide the isosceles triangle into 2 back-to-back congruent right triangles each with hypotenuse length $5$ and base length $\dfrac{6}{2} = 3$ . The shared height will then be $\sqrt{5^{2} - 3^{2}} = 4$ , resulting in a total area of $2 \times \dfrac{3 \times 4}{2} = \boxed{12}$ .

Relevant wiki: Area of Triangles - Heron's Formula

By using the Heron's Formula , we have

$s=\dfrac{5+5+6}{2}=8$

$A=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{8(8-5)(8-5)(8-6)}=\sqrt{144}=\boxed{12}$

Heron's formula

By the formula of Heron's it's easy to solve.

Find the priemeter of tringle .

$\large{P}$ = $\large{5+5+6}$ = $\large{16}$

Area= $\sqrt{P'}({P'-5})({P'-5})({P'-6})$

$\sqrt{8}({3})({3})({2})$ = $\sqrt{144}$ = $\large{12}$ .

P=Priemeter of tringle.

P'=Half of priemeter in tringle.