Area of isosceles triangle is \(2\sqrt { {12} } \) sq cm. What will be the base of a triangle if the height of the triangle is 12 cm.

1
\(\frac{{\sqrt {\left( {11} \right)} }}{3}\) cm
2
\(\frac{{\sqrt {\left( {14} \right)} }}{3}\) cm
3
\(\frac{{\sqrt {\left( {12} \right)} }}{3}\) cm
4
\(\frac{{\sqrt {\left( {10} \right)} }}{3}\) cm

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