Jonathan used 4 identical right-angled triangles to form a figure PQRST shown below. PT = 12 cm. The perimeter of the figure is 40 cm. Find the area of PQRST.

1. Find the length of one leg of the identical right-angled triangles.

The line segment $PT$ is divided into two equal segments by $M$.
$PM = MT = 12 \div 2 = 6\text{ cm}$.
Thus, one leg of the identical right-angled triangles is $6\text{ cm}$.

2. Express the perimeter of the figure in terms of the unknown leg.

Let the other leg of the right-angled triangle be $x\text{ cm}$ (which corresponds to $MR$).
The perimeter of $PQRST$ is made up of $PT$, $PQ$, $QR$, $RS$, and $ST$.
Since $\triangle PQR$ and $\triangle TSR$ are identical to $\triangle PMR$, their legs are also $6\text{ cm}$ and $x\text{ cm}$.
$PQ + QR = 6 + x$
$RS + ST = 6 + x$
Perimeter = $PT + (PQ + QR) + (RS + ST) = 12 + (6 + x) + (6 + x) = 24 + 2x$

3. Solve for the unknown leg $x$.

Given the perimeter is $40\text{ cm}$:
$24 + 2x = 40$
$2x = 16$
$x = 8\text{ cm}$

4. Calculate the area of one triangle and the total area of the figure.

Area of one right-angled triangle = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 8 = 24\text{ cm}^2$.
Total area of $PQRST$ = $4 \times 24 = 96\text{ cm}^2$.