In the diagram alongside, AB = 7 cm, AC = 9 cm and B = 38°. Calculate, correct to one decimal digit, the area of ABC.

The Law of Sines states that for a triangle with sides $a, b, c$ and opposite angles $A, B, C$:

$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$

In $\triangle ABC$, we have side $AB = 7$ cm (which is $c$), side $AC = 9$ cm (which is $b$), and angle $\hat{B} = 38^\circ$.
We can set up the ratio to find $\sin \hat{C}$:

$$\frac{AC}{\sin \hat{B}} = \frac{AB}{\sin \hat{C}}$$

Substitute the given values:

$$\frac{9}{\sin 38^\circ} = \frac{7}{\sin \hat{C}}$$

Rearrange to solve for $\sin \hat{C}$:

$$\sin \hat{C} = \frac{7 \times \sin 38^\circ}{9}$$ $$\sin \hat{C} = \frac{7 \times 0.61566}{9}$$ $$\sin \hat{C} \approx 0.47885$$

$$\hat{C} = \arcsin(0.47885)$$ $$\hat{C} \approx 28.6^\circ$$

We must check for a second possible angle for $\hat{C}$ (ambiguous case): $180^\circ - 28.6^\circ = 151.4^\circ$.
If $\hat{C} = 151.4^\circ$, then $\hat{B} + \hat{C} = 38^\circ + 151.4^\circ = 189.4^\circ$. This sum is greater than $180^\circ$, which is not possible for a triangle. Therefore, $\hat{C} \approx 28.6^\circ$ is the only valid angle.

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

$$\hat{A} = 180^\circ - \hat{B} - \hat{C}$$ $$\hat{A} = 180^\circ - 38^\circ - 28.6^\circ$$ $$\hat{A} = 180^\circ - 66.6^\circ$$ $$\hat{A} = 113.4^\circ$$

The area of a triangle can be calculated using the formula:

$$\text{Area} = \frac{1}{2}ab\sin C$$

Using sides $AB$ and $AC$ and the included angle $\hat{A}$:

$$\text{Area} = \frac{1}{2} \times AB \times AC \times \sin \hat{A}$$ $$\text{Area} = \frac{1}{2} \times 7 \text{ cm} \times 9 \text{ cm} \times \sin 113.4^\circ$$ $$\text{Area} = \frac{1}{2} \times 63 \text{ cm}^2 \times 0.9178$$ $$\text{Area} = 31.5 \text{ cm}^2 \times 0.9178$$ $$\text{Area} \approx 28.8967 \text{ cm}^2$$

Rounding to one decimal digit:

$$\text{Area} \approx 28.9 \text{ cm}^2$$

The area of $\triangle ABC$ is $\boxed{\text{28.9 cm}^2}$.