Answer
$A(b)=\frac{1}{4}b\cdot \sqrt{64-16b+2b^2}$
Work Step by Step
We know that the perimeter of an isosceles triangle is its base $b$ plus both equal sides $s$: $P=s+s+b$, and we know that the perimeter is 8cm, so $8=2s+b$
We also know that the area of a triangle is $A=\frac{1}{2}bh$. We can calculate its height by using Pythagoras's Theorem using $h$ as the hypotenuse, half of $b$ as one of its sides, and $s$ as its other side:
$h^2=\left(\frac{b}{2}\right)^2+s^2$
since we want everything in terms of $b$, we can use of the earlier equations to represent $s$ in terms of $b$:
$8=2s+b$
$8-b=2s+b-b$
$(8-b )\frac{1}{2}=2s\cdot \frac{1}{2}$
$s=\frac{8-b}{2}$, now we can substitute this into the Pythagoras's Theorem we were using earlier and continue to simplify:
$h^2=\left(\frac{b}{2}\right)^2+(\frac{8-b}{2})^2$
$h^2=\frac{b^2}{2^2}+\frac{(8-b)^2}{2^2}$
$h^2=\frac{b^2}{4}+\frac{64-16b+b^2}{4}$
$h^2=\frac{64-16b+2b^2}{4}$
$\sqrt{h^2}=\sqrt{\frac{64-16b+2b^2}{4}}$
$h=\frac{1}{2}\sqrt{64-16b+2b^2}$ we can replace this into the area equation:
$A=\frac{1}{2}b\cdot \frac{1}{2}\sqrt{64-16b+2b^2}$
$A(b)=\frac{1}{4}b\cdot \sqrt{64-16b+2b^2}$ and that's the area only in function of its base
