1. **Problem statement:** Find the length DV, angle VDP, length VX (X midpoint of CD), angle VXP, and the vertical height of the smaller pyramid VEFGH given its volume 24 cm³. 2. **Step (c) Find length DV:** Assuming coordinates or given lengths are known from the diagram (not provided here), use the distance formula or given data. 3. **Step (d) Find angle VDP:** Use the cosine rule or vector dot product formula: $$\cos(\angle VDP) = \frac{\overrightarrow{DV} \cdot \overrightarrow{DP}}{|DV||DP|}$$ Calculate the vectors and their magnitudes, then find the angle by: $$\angle VDP = \cos^{-1}(\text{value})$$ 4. **Step (e)(i) Find length VX:** Since X is midpoint of CD, $$VX = \frac{1}{2} VD$$ if V, D, C are collinear or use distance formula accordingly. 5. **Step (e)(ii) Find angle VXP:** Use cosine rule or vector dot product similarly: $$\cos(\angle VXP) = \frac{\overrightarrow{VX} \cdot \overrightarrow{XP}}{|VX||XP|}$$ Then, $$\angle VXP = \cos^{-1}(\text{value})$$ 6. **Step (f) Find vertical height of smaller pyramid VEFGH:** Volume ratio of similar pyramids is the cube of the scale factor of their heights: $$\frac{V_{VEFGH}}{V_{VABCD}} = \left(\frac{h_{VEFGH}}{h_{VABCD}}\right)^3$$ Given: $$V_{VEFGH} = 24$$ Assuming volume of original pyramid $V_{VABCD}$ and height $h_{VABCD}$ are known, Solve for $h_{VEFGH}$: $$h_{VEFGH} = h_{VABCD} \times \sqrt[3]{\frac{24}{V_{VABCD}}}$$ **Final answers depend on given original pyramid dimensions and volume, which are not provided here.**