Consider the following for the two (02) items that follow: Let ABC...

Let $AB = c$, $BC = a$, and $AC = b$.

Given that the triangle is right-angled at $B$, we have $a^2 + c^2 = b^2$.

We are given $AB + AC = 3$, which means $c + b = 3 \Rightarrow b = 3 - c$.

Substituting $b$ into the Pythagorean theorem:

$a^2 + c^2 = (3 - c)^2$

$a^2 + c^2 = 9 - 6c + c^2$

$a^2 = 9 - 6c \Rightarrow a = \sqrt{9 - 6c}$

The area of the triangle is $\Delta = \dfrac{1}{2}ac = \dfrac{1}{2}c\sqrt{9 - 6c}$.

To maximize the area, we can maximize its square, $S = \Delta^2$:

$S = \dfrac{1}{4}c^2(9 - 6c) = \dfrac{9}{4}c^2 - \dfrac{3}{2}c^3$

Differentiating $S$ with respect to $c$ and equating to zero:

$\dfrac{dS}{dc} = \dfrac{9}{2}c - \dfrac{9}{2}c^2 = 0$

$\dfrac{9}{2}c(1 - c) = 0$

Since $c > 0$, we get $c = 1$.

For $c = 1$, $\dfrac{d^2S}{dc^2} = \dfrac{9}{2} - 9c = -\dfrac{9}{2} < 0$, which confirms a maximum.

Substituting $c = 1$ into the expression for area:

$\Delta = \dfrac{1}{2}(1)\sqrt{9 - 6(1)} = \dfrac{\sqrt{3}}{2}$ square units.

Answer: $\dfrac{\sqrt{3}}{2}$ square unit