In the diagram, AB = 12 cm, AC = 9 cm and BC = 7 cm. Using as much of the information given below as is necessary (a) express the ratio AC:AB in its simplest form, (b) find the area of triangle ABC.

$$AC = 9 \text{ cm}$$ $$AB = 12 \text{ cm}$$

The ratio $AC:AB$ is $9:12$.

$$9 \div 3 = 3$$ $$12 \div 3 = 4$$

The simplified ratio is $3:4$.

$$\boxed{\text{AC:AB} = \text{3:4}}$$

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

For triangle ABC, with sides AC and AB and included angle A, the formula becomes:

$$\text{Area} = \frac{1}{2} \times AC \times AB \times \sin A$$

$AC = 9 \text{ cm}$, $AB = 12 \text{ cm}$, and $\sin A = 0.58$.

$$\text{Area} = \frac{1}{2} \times 9 \text{ cm} \times 12 \text{ cm} \times 0.58$$

$$\text{Area} = \frac{1}{2} \times 108 \text{ cm}^2 \times 0.58$$ $$\text{Area} = 54 \text{ cm}^2 \times 0.58$$ $$\text{Area} = 31.32 \text{ cm}^2$$ $$\boxed{31.32 \text{ cm}^2}$$