Step 1: Identify the properties of $\triangle ABC$. Given that $\angle ABC = 90^\circ$, $\triangle ABC$ is a right-angled triangle. The lengths of the two shorter sides are $AB = 4$ cm and $BC = 3$ cm. Step 2: Calculate the length of the hypotenuse AC using the Pythagorean theorem. In $\triangle ABC$: $$AC^2 = AB^2 + BC^2$$ $$AC^2 = (4\, \text{cm})^2 + (3\, \text{cm})^2$$ $$AC^2 = 16\, \text{cm}^2 + 9\, \text{cm}^2$$ $$AC^2 = 25\, \text{cm}^2$$ $$AC = \sqrt{25\, \text{cm}^2}$$ $$AC = 5\, \text{cm}$$ Step 3: Calculate the area of $\triangle ABC$. The area of a right-angled triangle can be calculated as half the product of its perpendicular sides. $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ Using AB and BC as base and height: $$\text{Area} = \frac{1}{2} \times BC \times AB$$ $$\text{Area} = \frac{1}{2} \times 3\, \text{cm} \times 4\, \text{cm}$$ $$\text{Area} = \frac{1}{2} \times 12\, \text{cm}^2$$ $$\text{Area} = 6\, \text{cm}^2$$ Step 4: Use the area of $\triangle ABC$ to find the length of BD. Given that $\angle BDC = 90^\circ$, BD is the altitude from B to the hypotenuse AC. We can also express the area of $\triangle ABC$ using AC as the base and BD as the height: $$\text{Area} = \frac{1}{2} \times AC \times BD$$ We know the Area is $6\, \text{cm}^2$ and $AC = 5\, \text{cm}$. $$6\, \text{cm}^2 = \frac{1}{2} \times 5\, \text{cm} \times BD$$ Multiply both sides by 2: $$12\, \text{cm}^2 = 5\, \text{cm} \times BD$$ Divide by $5\, \text{cm}$: $$BD = \frac{12\, \text{cm}^2}{5\, \text{cm}}$$ $$BD = 2.4\, \text{cm}$$ The length of BD is $\boxed{\text{2.4 cm}}$. Send me the next one 📸