In the figure below BC=24cm, < APD = < PCD = < PBA = 90°. Calculate, correct to 2 decimal places, the length AD given that P is the midpoint of BC.

Given that P is the midpoint of BC and BC = 24 cm:

$$BP = PC = \frac{BC}{2} = \frac{24 \text{ cm}}{2} = 12 \text{ cm}$$

Triangle PBA is a right-angled triangle at B ($\angle PBA = 90^\circ$).
We have AB = 5 cm and BP = 12 cm.
Using the Pythagorean theorem ($a^2 + b^2 = c^2$):

$$AP^2 = AB^2 + BP^2$$ $$AP^2 = (5 \text{ cm})^2 + (12 \text{ cm})^2$$ $$AP^2 = 25 \text{ cm}^2 + 144 \text{ cm}^2$$ $$AP^2 = 169 \text{ cm}^2$$ $$AP = \sqrt{169 \text{ cm}^2} = 13 \text{ cm}$$

Triangle PCD is a right-angled triangle at C ($\angle PCD = 90^\circ$).
We have CD = 18 cm and PC = 12 cm.
Using the Pythagorean theorem:

$$PD^2 = PC^2 + CD^2$$ $$PD^2 = (12 \text{ cm})^2 + (18 \text{ cm})^2$$ $$PD^2 = 144 \text{ cm}^2 + 324 \text{ cm}^2$$ $$PD^2 = 468 \text{ cm}^2$$ $$PD = \sqrt{468 \text{ cm}^2}$$

Triangle APD is a right-angled triangle at P ($\angle APD = 90^\circ$).
We have AP = 13 cm and $PD = \sqrt{468}$ cm.
Using the Pythagorean theorem:

$$AD^2 = AP^2 + PD^2$$ $$AD^2 = (13 \text{ cm})^2 + (\sqrt{468 \text{ cm}^2})^2$$ $$AD^2 = 169 \text{ cm}^2 + 468 \text{ cm}^2$$ $$AD^2 = 637 \text{ cm}^2$$ $$AD = \sqrt{637 \text{ cm}^2}$$

$$AD \approx 25.23885 \text{ cm}$$

Rounding to 2 decimal places:

$$AD \approx 25.24 \text{ cm}$$

The length AD is:

$$\text{25.24 cm}$$