The area of an isosceles triangle having base 2cm and the length of one of the equal sides 4cm is

$b=$ 4 cm

$c=$  2 cm

$s=\frac{4+4+2}{2}=\frac{10}{2}=5$  cm

$A=\sqrt{s\times \left(s-a\right)\times \left(s-b\right)\times \left(s-c\right)}$

$=\sqrt{5\times \left(5-4\right)\times \left(5-4\right)\times \left(5-2\right)}$

$=\sqrt{5\times 1\times 1\times 3}=\sqrt{15}{\text{ cm}}^2$