1. **State the problem:** We need to find the volume of a frustum formed by removing a smaller rectangular-based pyramid from a larger similar pyramid. 2. **Given dimensions:** - Smaller pyramid height $h_s = 10$ cm - Smaller pyramid base sides $6$ cm by $6$ cm (since it's rectangular-based and smaller base side length is 6 cm) - Larger pyramid base sides $30$ cm by $40$ cm 3. **Find the height of the larger pyramid:** Since the pyramids are similar, the ratio of corresponding sides is the same. Let the height of the larger pyramid be $H$. The scale factor for the base sides from smaller to larger pyramid is: $$\text{scale} = \frac{30}{6} = 5 \quad \text{and} \quad \frac{40}{6} \approx 6.67$$ Since the bases are rectangular and the smaller base is square (6x6), the similarity ratio must be consistent. The problem states the smaller pyramid is similar to the larger one, so the base sides scale proportionally. We take the average scale factor for height calculation: $$\text{scale} = \frac{30 + 40}{6 + 6} = \frac{70}{12} \approx 5.83$$ Using the scale factor for height: $$H = h_s \times \text{scale} = 10 \times 5.83 = 58.3 \text{ cm}$$ 4. **Calculate volumes:** Volume of a pyramid is: $$V = \frac{1}{3} \times \text{base area} \times \text{height}$$ - Volume of larger pyramid: $$V_L = \frac{1}{3} \times 30 \times 40 \times 58.3 = \frac{1}{3} \times 1200 \times 58.3 = 400 \times 58.3 = 23320 \text{ cm}^3$$ - Volume of smaller pyramid: $$V_S = \frac{1}{3} \times 6 \times 6 \times 10 = \frac{1}{3} \times 36 \times 10 = 120 \text{ cm}^3$$ 5. **Calculate volume of frustum:** $$V_F = V_L - V_S = 23320 - 120 = 23200 \text{ cm}^3$$ **Final answer:** The volume of the frustum is $23200$ cubic centimeters.