1. **Problem statement:** (a) Find the height of pyramid VPQR inside the frustum. (b) Find the volume of the frustum. 2. **Given data:** - Upper base rectangle PQRS: length = 5 m, width = 2 m - Lower base rectangle ABCD: length = 15 m, width = 6 m - Height of frustum = 12 m 3. **Step (a): Find height of pyramid VPQR** - The frustum is formed by cutting a larger pyramid by a plane parallel to the base. - The smaller pyramid VPQR is similar to the larger pyramid VABCD. - The scale factor for linear dimensions between the smaller and larger pyramid is the ratio of the upper base length to the lower base length. - Scale factor $k = \frac{5}{15} = \frac{1}{3}$ - Height of the larger pyramid $H$ is the height of the frustum plus the height of the smaller pyramid. - Let height of smaller pyramid be $h$. - Since the smaller pyramid is similar, $\frac{h}{H} = k = \frac{1}{3}$ - Height of frustum = $H - h = 12$ m - Substitute $h = \frac{H}{3}$, so $H - \frac{H}{3} = 12 \Rightarrow \frac{2H}{3} = 12 \Rightarrow H = 18$ m - Then $h = \frac{H}{3} = 6$ m 4. **Step (b): Find volume of frustum** - Volume of frustum = Volume of larger pyramid - Volume of smaller pyramid - Volume of pyramid = $\frac{1}{3} \times \text{area of base} \times \text{height}$ - Area of larger base $A_1 = 15 \times 6 = 90$ m$^2$ - Area of smaller base $A_2 = 5 \times 2 = 10$ m$^2$ - Volume of larger pyramid $V_1 = \frac{1}{3} \times 90 \times 18 = 540$ m$^3$ - Volume of smaller pyramid $V_2 = \frac{1}{3} \times 10 \times 6 = 20$ m$^3$ - Volume of frustum $V = V_1 - V_2 = 540 - 20 = 520$ m$^3$ **Final answers:** - Height of pyramid VPQR = $6$ m - Volume of frustum = $520$ m$^3$