1. **Problem statement:** (ii) Measure and write down the length of EA and BDPH. b. The net is assembled as a pyramid with apex V and base BDPH. (i) Sketch the cross section VDHI of the pyramid indicating clearly the lengths of VD and DH. (ii) By calculation, determine VX, the height of the pyramid, where X is the center of the base. 2. **Given:** - Pyramid with square base BDPH. - Slant edges length = 5 cm. - Base is square, so BD = DP = PH = HB. 3. **Step (ii) Calculation of VX:** - Let side length of base = s. - Since base is square, diagonal DH = s\sqrt{2}. - VD is slant height = 5 cm. - X is center of base, so DX = \frac{DH}{2} = \frac{s\sqrt{2}}{2}. 4. **Find s:** - From net or problem context, assume s = length of base side (not given explicitly, so assume s). 5. **Use right triangle VDX:** - VX is height (perpendicular from apex to base). - VD is hypotenuse = 5 cm. - DX = \frac{s\sqrt{2}}{2}. 6. **Apply Pythagoras theorem:** $$VX = \sqrt{VD^2 - DX^2} = \sqrt{5^2 - \left(\frac{s\sqrt{2}}{2}\right)^2} = \sqrt{25 - \frac{s^2 \times 2}{4}} = \sqrt{25 - \frac{s^2}{2}}$$ 7. **Without explicit s, VX depends on s.** --- **Cylinder and Hemisphere Surface Area:** 1. Given: - Cylinder height $h=8$ cm - Radius $r=3$ cm - Hemisphere radius $r=3$ cm - Use $\pi=3.14$ 2. Curved surface area of cylinder: $$2\pi rh = 2 \times 3.14 \times 3 \times 8 = 150.72 \text{ cm}^2$$ 3. Surface area of hemisphere: - Surface area of sphere = $4\pi r^2$ - Hemisphere = half sphere surface area plus base circle, but curved surface area is half sphere surface area: $$2\pi r^2 = 2 \times 3.14 \times 3^2 = 56.52 \text{ cm}^2$$ 4. Total surface area of solid paperweight: - Cylinder curved surface + hemisphere curved surface + base of cylinder (circle): - Base area = $\pi r^2 = 3.14 \times 3^2 = 28.26$ cm$^2$ $$\text{Total surface area} = 150.72 + 56.52 + 28.26 = 235.5 \text{ cm}^2$$ --- **Summary:** - Length EA and BDPH depend on measurements (not provided). - Height VX of pyramid: $$VX = \sqrt{25 - \frac{s^2}{2}}$$ where $s$ is base side length. - Curved surface area of cylinder = 150.72 cm$^2$. - Surface area of hemisphere = 56.52 cm$^2$. - Total surface area of paperweight = 235.5 cm$^2$.