1. **State the problem:** We need to find the area of a shape composed of a rectangle ABDE and a right-angled triangle BCD attached along side BD. 2. **Given dimensions:** - Rectangle ABDE has length AE = 8 cm and width AB = 3 cm. - Triangle BCD is right-angled with base BD = 3 cm and hypotenuse BC = 17 cm. 3. **Find the area of the rectangle:** The area of a rectangle is given by the formula: $$\text{Area}_{rectangle} = \text{length} \times \text{width}$$ So, $$\text{Area}_{rectangle} = 8 \times 3 = 24 \text{ cm}^2$$ 4. **Find the height of the triangle BCD:** We know the base BD = 3 cm and hypotenuse BC = 17 cm. Since triangle BCD is right-angled, by Pythagoras theorem: $$BC^2 = BD^2 + CD^2$$ Substitute known values: $$17^2 = 3^2 + CD^2$$ $$289 = 9 + CD^2$$ $$CD^2 = 289 - 9 = 280$$ $$CD = \sqrt{280} = \sqrt{4 \times 70} = 2\sqrt{70} \approx 16.73 \text{ cm}$$ 5. **Calculate the area of triangle BCD:** Area of a triangle is: $$\text{Area}_{triangle} = \frac{1}{2} \times \text{base} \times \text{height}$$ Here base = BD = 3 cm, height = CD \approx 16.73 cm $$\text{Area}_{triangle} = \frac{1}{2} \times 3 \times 16.73 = \frac{3}{2} \times 16.73 = 1.5 \times 16.73 = 25.095 \text{ cm}^2$$ 6. **Find total area of the shape:** $$\text{Area}_{total} = \text{Area}_{rectangle} + \text{Area}_{triangle} = 24 + 25.095 = 49.095 \text{ cm}^2$$ **Final answer:** $$\boxed{49.1 \text{ cm}^2}$$ (rounded to one decimal place)