1. **Problem statement:** We have a right frustum ABCDSPQR with upper base rectangle PQRS (5 m by 2 m), lower base rectangle ABCD (15 m by 6 m), and height 12 m. (a) Find the height of pyramid VPQRS. (b) Find the volume of the frustum. 2. **Formulas and rules:** - The frustum is formed by cutting a pyramid with apex V by a plane parallel to the base. - Height of pyramid VPQRS = height of frustum + height of small pyramid VPQRS. - Volume of frustum = volume of large pyramid VABCD - volume of small pyramid VPQRS. - Volume of pyramid = $ \frac{1}{3} \times \text{base area} \times \text{height} $. - Since the frustum is right, linear dimensions scale proportionally. 3. **Find the scale factor:** - Lower base length = 15 m, upper base length = 5 m. - Scale factor for linear dimensions from large base to small base = $ \frac{5}{15} = \frac{1}{3} $. 4. **Find height of small pyramid VPQRS:** - Height of frustum = 12 m. - Let height of small pyramid = $ h $. - Total height of large pyramid = height of frustum + height of small pyramid = $ 12 + h $. - Since linear dimensions scale by $ \frac{1}{3} $, height scales similarly: $$ \frac{h}{12 + h} = \frac{1}{3} $$ - Solve for $ h $: $$ 3h = 12 + h $$ $$ 3h - h = 12 $$ $$ 2h = 12 $$ $$ h = 6 \text{ m} $$ 5. **Find total height of large pyramid:** $$ 12 + 6 = 18 \text{ m} $$ 6. **Calculate volumes:** - Base area of large pyramid = $ 15 \times 6 = 90 \text{ m}^2 $. - Base area of small pyramid = $ 5 \times 2 = 10 \text{ m}^2 $. - Volume of large pyramid: $$ V_{large} = \frac{1}{3} \times 90 \times 18 = \frac{1}{3} \times 1620 = 540 \text{ m}^3 $$ - Volume of small pyramid: $$ V_{small} = \frac{1}{3} \times 10 \times 6 = \frac{1}{3} \times 60 = 20 \text{ m}^3 $$ 7. **Volume of frustum:** $$ V_{frustum} = V_{large} - V_{small} = 540 - 20 = 520 \text{ m}^3 $$ **Final answers:** (a) Height of pyramid VPQRS = 6 m. (b) Volume of frustum = 520 cubic meters.