Given,
In an isosceles triangle,
Length of each of two equal sides (a) = 4cm
and length of the base (b) = \(\sqrt[4]2\;cm\)
To find: Area of a triangle (A)
By formula,
Area of an isosceles triangle (A) = \(\frac B4\sqrt{4a^2-b^2}\)
= \(\frac{\sqrt[4]2}4\times\sqrt{4\times(4cm)^2-(\sqrt[4]2cm)^2}\)
= \(\sqrt2cm\times\sqrt{64cm^2-32cm^2}\)
= \(\sqrt2cm\times\sqrt{32cm^2}\)
= \(\sqrt2cm\times\sqrt[4]2cm\)
= 8cm2
Therefore, the Area of the given triangle (A) = 8cm2