In the composite shape drawn below, DC ⊥ BE, AD = 9 cm, DC = 5 cm and BE = 13 cm. Calculate the area of the whole shape.

The composite shape consists of a rectangle ABCD and a right-angled triangle DCE.
From the given information and diagram:

• For rectangle ABCD: Length $AD = 9 \text{ cm}$ Width $DC = 5 \text{ cm}$
• For right-angled triangle DCE: Height $DC = 5 \text{ cm}$ Base $CE = BE - BC$. Since ABCD is a rectangle, $BC = AD = 9 \text{ cm}$. Therefore, $CE = 13 \text{ cm} - 9 \text{ cm} = 4 \text{ cm}$.

$$\text{Area}_{\text{rectangle}} = \text{length} \times \text{width}$$ $$\text{Area}_{\text{rectangle}} = AD \times DC = 9 \text{ cm} \times 5 \text{ cm} = 45 \text{ cm}^2$$

$$\text{Area}_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height}$$ $$\text{Area}_{\text{triangle}} = \frac{1}{2} \times CE \times DC = \frac{1}{2} \times 4 \text{ cm} \times 5 \text{ cm} = 10 \text{ cm}^2$$

$$\text{Total Area} = \text{Area}_{\text{rectangle}} + \text{Area}_{\text{triangle}}$$ $$\text{Total Area} = 45 \text{ cm}^2 + 10 \text{ cm}^2 = 55 \text{ cm}^2$$

The area of the whole shape is $\boxed{\text{55 cm}^2}$.