The difference between the semiperimeter and the sides of a ∆ABC are 8 cm,

Let the semi-perimeter of the triangle be s.
Let the sides of the triangle be a, b and c.
Given: s − a = 8, s − b = 7 and s − c = 5      ....(1)

Adding all three equations, we get
3s − (a + b + c) = 8 + 7 + 5
⇒ 3s − (a + b + c) = 20

⇒ 3s − 2s = 20                     $\left(\because s=\frac{a+b+c}{2}\right)$

⇒ s = 20 cm                  ...(2)

∴ By Heron's formula,

Area of $\Delta=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{20(8)(7)(5)} \quad($ from $(1)$ and $(2))$

$=20 \sqrt{14} \mathrm{~cm}^{2}$

Hence, the area of the triangle is $20 \sqrt{14} \mathrm{~cm}^{2}$.