The area varies

Let $S$ be the length of the shorter side and $L$ the length of the longer side. Then $A = SL$ is the area of the original rectangle, and the area of the new rectangle is

$\left(1 + \dfrac{a}{100}\right)S * \left(1 + \dfrac{a + 5}{100}\right)L = \left(1 + \dfrac{2a + 5}{100} + \dfrac{a(a + 5)}{10000}\right)A$ .

As the area of the new rectangle is $1 + \dfrac{15.5}{100}$ times that of the original, we have that

$1 + \dfrac{2a + 5}{100} + \dfrac{a^{2} + 5a}{10000} = 1 + \dfrac{15.5}{100} \Longrightarrow 100*(2a + 5) + (a^{2} + 5a) = 1550$

$\Longrightarrow a^{2} + 205a - 1050 = 0 \Longrightarrow (a + 210)(a - 5) = 0 \Longrightarrow a = \boxed{5}$ ,

since we require that $a \gt 0$ .