The maximum area of a right angled triangle with hypotenuse $$h$$ is :

Let us denote the two perpendicular sides (legs) of the right-angled triangle by $$a$$ and $$b$$. Because the triangle is right-angled, the hypotenuse is related to the legs by the Pythagoras theorem.

Pythagoras theorem (statement): for a right-angled triangle, $$\text{(hypotenuse)}^{2} = \text{(first leg)}^{2} + \text{(second leg)}^{2}.$$

So we have

$$a^{2} + b^{2} = h^{2}.$$

Next we write the formula for the area of a right-angled triangle.

Area formula (statement): when the two perpendicular sides are $$a$$ and $$b$$, the area $$A$$ is

$$A = \dfrac{1}{2}\, a\, b.$$

Our task is to find the maximum possible value of $$A$$ while the hypotenuse $$h$$ remains fixed. In other words, $$h$$ is a constant and $$a$$, $$b$$ may vary but must satisfy $$a^{2} + b^{2} = h^{2}.$$ We therefore have an optimisation (maximisation) problem with a constraint.

To proceed algebraically, we first express the product $$a b$$ in terms of the constant $$h$$ as far as possible. A very convenient inequality helps us: for any real numbers $$a$$ and $$b,$$ the square of their difference is always non-negative:

$$(a - b)^{2} \ge 0.$$

Expanding the left hand side gives

$$a^{2} - 2 a b + b^{2} \ge 0.$$

Rearranging, we obtain

$$2 a b \le a^{2} + b^{2}.$$

Now we substitute the Pythagoras relation $$a^{2} + b^{2} = h^{2}$$ into this inequality:

$$2 a b \le h^{2}.$$

Dividing both sides by $$2$$, we get an upper bound for the product of the legs:

$$a b \le \dfrac{h^{2}}{2}.$$

Observe that equality in the inequality $$a b \le \dfrac{h^{2}}{2}$$ holds exactly when $$a = b.$$ This fact follows from the condition $$ (a - b)^{2} = 0 \Longrightarrow a = b.$$ Therefore, the product $$a b$$ attains its largest possible value when both legs are equal, i.e. the right-angled triangle is isosceles with

$$a = b.$$

Substituting $$a b_{\text{max}} = \dfrac{h^{2}}{2}$$ into the area formula, we get the maximum area $$A_{\text{max}}$$:

$$A_{\text{max}} = \dfrac{1}{2}\, a b_{\text{max}} = \dfrac{1}{2}\,\left(\dfrac{h^{2}}{2}\right) = \dfrac{h^{2}}{4}.$$

Thus, the greatest possible area of a right-angled triangle whose hypotenuse is $$h$$ is $$\dfrac{h^{2}}{4}.$$

Hence, the correct answer is Option D.