Given that PQRS is a square, P = 90^. It is also given that PT = PU.

Given that $PQRS$ is a square, $\angle P = 90^\circ$. It is also given that $PT = PU$.
Therefore, $\triangle PTU$ is a right-angled isosceles triangle.

The sum of angles in a triangle is $180^\circ$.

In a square $PQRS$, the diagonal $QS$ bisects $\angle PQR$ and $\angle PSR$. Thus, $\angle PQS = 45^\circ$.

We have $\angle PTU = 45^\circ$ and $\angle PQS = 45^\circ$.
Since $\angle PTU = \angle PQS$, and these are corresponding angles with transversal $PQ$, we can conclude that $TU \parallel QS$.

Given $PQ = 10$ units. Since $PQRS$ is a square, $PS = PQ = 10$ units.
$\triangle PQS$ is a right-angled triangle at $P$.

The area of a right-angled triangle is $\frac{1}{2} \times \text{base} \times \text{height}$.

Since $PQRS$ is a square, all its sides are equal in length ($QR = RS$) and all its interior angles are $90^\circ$ ($\angle R = 90^\circ$).
Therefore, $\triangle QRS$ has two equal sides ($QR$ and $RS$) and one right angle ($\angle R$), which means it is a right-angled isosceles triangle.

Using the Pythagorean theorem in $\triangle QRS$:

Since $QR = RS = PQ = 10$ units (from the properties of a square and the given information in 6.2):