Answer
$\frac{\Delta A}{A_0} = 2~\alpha~\Delta T$
Work Step by Step
Let the original area $A_0 = s_0^2$
We can find an expression for the new area after the material expands:
$A = (s_0+\Delta s)^2$
$A = s_0^2+2s_0~\Delta s+(\Delta s)^2$
$A \approx s_0^2+2s_0~\Delta s$
We can find an expression for the ratio $\frac{\Delta A}{A_0}$:
$\Delta A = A-A_0$
$\Delta A = (s_0^2+2s_0~\Delta s)-s_0^2$
$\Delta A = 2s_0~\Delta s$
$\Delta A = 2s_0~(\alpha~\Delta T~s_0)$
$\Delta A = 2s_0^2~\alpha~\Delta T$
$\Delta A = 2~A_0~\alpha~\Delta T$
$\frac{\Delta A}{A_0} = 2~\alpha~\Delta T$
