1. **Stating the problem:** Calculate the volume of a rectangular parallelepiped with a base area of 540 dm², where one side of the base is $\frac{3}{5}$ of the other, and the lateral surface area is 1056 dm². 2. **Known formulas:** - Volume of a rectangular parallelepiped: $$V = \text{base area} \times \text{height}$$ - Lateral surface area: $$A_{lat} = 2 \times h \times (a + b)$$ where $a$ and $b$ are the sides of the base and $h$ is the height. 3. **Find the sides of the base:** Let $b$ be the longer side, then the shorter side is $a = \frac{3}{5}b$. Since the base area is $540$ dm²: $$a \times b = 540$$ Substitute $a$: $$\frac{3}{5}b \times b = 540$$ $$\frac{3}{5}b^2 = 540$$ Multiply both sides by $\frac{5}{3}$: $$\cancel{\frac{3}{5}}b^2 \times \frac{5}{3} = 540 \times \frac{5}{3}$$ $$b^2 = 900$$ 4. **Calculate $b$ and $a$:** $$b = \sqrt{900} = 30 \text{ dm}$$ $$a = \frac{3}{5} \times 30 = 18 \text{ dm}$$ 5. **Use lateral surface area to find height $h$:** $$A_{lat} = 2h(a + b) = 1056$$ Substitute $a$ and $b$: $$2h(18 + 30) = 1056$$ $$2h \times 48 = 1056$$ $$96h = 1056$$ Divide both sides by 96: $$\cancel{96}h = \frac{1056}{\cancel{96}}$$ $$h = 11 \text{ dm}$$ 6. **Calculate volume:** $$V = \text{base area} \times h = 540 \times 11 = 5940 \text{ dm}^3$$ **Final answer:** The volume of the parallelepiped is **5940 dm³**.