The dimensions of the largest rectangle of side $x$ and $y$ that can be inscribed in the right angled triangle of sides $\mathrm{a}$ and $\mathrm{b}$ is

Let the right-angled triangle have base $a$ and height $b$. The rectangle is inscribed such that one vertex lies on the hypotenuse, and two sides lie along the legs of the triangle. Let the rectangle have width $x$ and height $y$.

Using similar triangles, the small triangle at the top has base $x$ and height $b - y$. The large triangle has base $a$ and height $b$. By similarity:

$\dfrac{x}{a} = \dfrac{b - y}{b}$

This simplifies to $x = \dfrac{a(b - y)}{b} = a - \dfrac{a}{b}y$.

The area $A$ of the rectangle is given by $A = xy = y(a - \dfrac{a}{b}y) = ay - \dfrac{a}{b}y^2$.

To maximize the area, we differentiate $A$ with respect to $y$ and set it to zero:

$\dfrac{dA}{dy} = a - \dfrac{2a}{b}y = 0$

Solving for $y$, we get $y = \dfrac{b}{2}$.

Substituting $y = \dfrac{b}{2}$ into the equation for $x$:

$x = a - \dfrac{a}{b}(\dfrac{b}{2}) = a - \dfrac{a}{2} = \dfrac{a}{2}$.

Thus, the dimensions of the largest rectangle are $x = \dfrac{a}{2}$ and $y = \dfrac{b}{2}$.

Answer: $\dfrac{a}{2}, \dfrac{b}{2}$