1. **State the problem:** We need to find the area of a shape composed of a right-angled triangle ABC (right angle at B) and a semicircle with diameter BC. 2. **Given:** - AB = 7.6 cm - AC = 9.5 cm - Triangle ABC is right-angled at B 3. **Find BC using Pythagoras' theorem:** $$BC = \sqrt{AC^2 - AB^2} = \sqrt{9.5^2 - 7.6^2}$$ 4. Calculate: $$BC = \sqrt{90.25 - 57.76} = \sqrt{32.49} = 5.7 \text{ cm (to 3 significant figures)}$$ 5. **Area of triangle ABC:** Since the right angle is at B, AB and BC are perpendicular sides. $$\text{Area}_{\triangle} = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 7.6 \times 5.7$$ 6. Calculate: $$\text{Area}_{\triangle} = \frac{1}{2} \times 7.6 \times 5.7 = 21.66 \text{ cm}^2$$ 7. **Area of the semicircle:** Radius $r = \frac{BC}{2} = \frac{5.7}{2} = 2.85$ cm Area of full circle: $$\pi r^2 = \pi \times 2.85^2 = \pi \times 8.1225$$ Area of semicircle: $$\frac{1}{2} \pi r^2 = \frac{1}{2} \times \pi \times 8.1225 = 4.06125\pi$$ 8. Calculate numerical value: $$4.06125 \times 3.1416 = 12.76 \text{ cm}^2$$ 9. **Total area of the shape:** $$\text{Area}_{total} = \text{Area}_{\triangle} + \text{Area}_{semicircle} = 21.66 + 12.76 = 34.42 \text{ cm}^2$$ 10. **Final answer rounded to 3 significant figures:** $$34.4 \text{ cm}^2$$