1. Problem statement: Calculate the area of a right triangle given the lengths of its legs. 2. Formula: The area $A$ of a right triangle with legs $a$ and $b$ is given by $$A = \frac{1}{2}ab$$ 3. Important rule: Multiply the lengths of the two legs and then divide by 2. 4. Part a) Given $a = 15$ cm and $b = 6 \frac{2}{5} = 6 + \frac{2}{5} = \frac{30}{5} + \frac{2}{5} = \frac{32}{5}$ cm. Calculate area: $$A = \frac{1}{2} \times 15 \times \frac{32}{5} = \frac{1}{2} \times \frac{15 \times 32}{5} = \frac{1}{2} \times \frac{480}{5} = \frac{1}{2} \times 96 = 48$$ So, area is 48 cm$^2$. 5. Part b) Given $a = 14 \frac{3}{7} = 14 + \frac{3}{7} = \frac{98}{7} + \frac{3}{7} = \frac{101}{7}$ dm and $b = 14$ dm. Calculate area: $$A = \frac{1}{2} \times \frac{101}{7} \times 14 = \frac{1}{2} \times \frac{101 \times 14}{7} = \frac{1}{2} \times \frac{1414}{7} = \frac{1}{2} \times 202 = 101$$ So, area is 101 dm$^2$. 6. Part c) Given $a = 24$ m and $b = 11 \frac{5}{6} = 11 + \frac{5}{6} = \frac{66}{6} + \frac{5}{6} = \frac{71}{6}$ m. Calculate area: $$A = \frac{1}{2} \times 24 \times \frac{71}{6} = \frac{1}{2} \times \frac{24 \times 71}{6} = \frac{1}{2} \times \frac{1704}{6} = \frac{1}{2} \times 284 = 142$$ So, area is 142 m$^2$. Final answers: a) 48 cm$^2$ b) 101 dm$^2$ c) 142 m$^2$