Here is the solution to your problem: Task i) The suitable name for the above roof is a rectangular pyramid. This is because its base is a rectangle (with dimensions 12 cm and 8 cm) and its sides meet at a single apex. ii) Step 1: Calculate the length $AV$. Assuming the pyramid is a right pyramid with equal slant edges, and given $BV = CV = DV = 20 \text{ cm}$ from the problem statement and diagram, then: $$AV = 20 \text{ cm}$$ Step 2: Calculate the height $OV$. First, find the length of the diagonal of the base, $AC$. Using the Pythagorean theorem in $\triangle ABC$: $$AC^2 = AB^2 + BC^2$$ $$AC^2 = (12 \text{ cm})^2 + (8 \text{ cm})^2$$ $$AC^2 = 144 \text{ cm}^2 + 64 \text{ cm}^2$$ $$AC^2 = 208 \text{ cm}^2$$ $$AC = \sqrt{208} \text{ cm} = \sqrt{16 \times 13} \text{ cm} = 4\sqrt{13} \text{ cm}$$ The point $O$ is the center of the rectangular base, so $OC$ is half the diagonal $AC$: $$OC = \frac{1}{2} AC = \frac{1}{2} (4\sqrt{13} \text{ cm}) = 2\sqrt{13} \text{ cm}$$ Now, use the Pythagorean theorem in the right-angled triangle $\triangle OVC$ to find the height $OV$: $$OV^2 + OC^2 = CV^2$$ $$OV^2 + (2\sqrt{13} \text{ cm})^2 = (20 \text{ cm})^2$$ $$OV^2 + (4 \times 13) \text{ cm}^2 = 400 \text{ cm}^2$$ $$OV^2 + 52 \text{ cm}^2 = 400 \text{ cm}^2$$ $$OV^2 = 400 \text{ cm}^2 - 52 \text{ cm}^2$$ $$OV^2 = 348 \text{ cm}^2$$ $$OV = \sqrt{348} \text{ cm} = \sqrt{4 \times 87} \text{ cm} = 2