1. **State the problem:** We have a figure made up of a rectangle, a square, and a right-angled triangle. The area of the square is \(\frac{2}{11}\) of the total area of the figure. We need to find (a) the area of the square and (b) the length of the square. 2. **Calculate the area of the rectangle:** The rectangle has length 15 cm and height 8 cm. $$\text{Area}_{\text{rectangle}} = 15 \times 8 = 120 \text{ cm}^2$$ 3. **Calculate the area of the triangle:** The triangle has base 12 cm and height 1 cm. $$\text{Area}_{\text{triangle}} = \frac{1}{2} \times 12 \times 1 = 6 \text{ cm}^2$$ 4. **Express the total area of the figure:** Let the side length of the square be \(s\) cm. The area of the square is \(s^2\). Total area: $$A = \text{Area}_{\text{rectangle}} + \text{Area}_{\text{square}} + \text{Area}_{\text{triangle}} = 120 + s^2 + 6 = 126 + s^2$$ 5. **Use the given ratio for the square's area:** $$s^2 = \frac{2}{11} A = \frac{2}{11} (126 + s^2)$$ 6. **Solve for \(s^2\):** Multiply both sides by 11: $$11 s^2 = 2 (126 + s^2)$$ $$11 s^2 = 252 + 2 s^2$$ Bring terms to one side: $$11 s^2 - 2 s^2 = 252$$ $$9 s^2 = 252$$ Divide both sides by 9: $$s^2 = \frac{252}{9} = 28$$ 7. **Find the area of the square:** $$\boxed{28 \text{ cm}^2}$$ 8. **Find the length of the square:** $$s = \sqrt{28} = \sqrt{4 \times 7} = 2 \sqrt{7} \approx 5.29 \text{ cm}$$ So the length of the square is approximately \(\boxed{5.29 \text{ cm}}\).