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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.

Chemistry

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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 AVAV.
Assuming the pyramid is a right pyramid with equal slant edges, and given BV=CV=DV=20 cmBV = CV = DV = 20 \text{ cm} from the problem statement and diagram, then:

AV=20 cmAV = 20 \text{ cm}

Step 2: Calculate the height OVOV.
First, find the length of the diagonal of the base, ACAC. Using the Pythagorean theorem in △ABC\triangle ABC:

AC2=AB2+BC2AC^2 = AB^2 + BC^2 AC2=(12 cm)2+(8 cm)2AC^2 = (12 \text{ cm})^2 + (8 \text{ cm})^2 AC2=144 cm2+64 cm2AC^2 = 144 \text{ cm}^2 + 64 \text{ cm}^2 AC2=208 cm2AC^2 = 208 \text{ cm}^2 AC=208 cm=16×13 cm=413 cmAC = \sqrt{208} \text{ cm} = \sqrt{16 \times 13} \text{ cm} = 4\sqrt{13} \text{ cm}

The point OO is the center of the rectangular base, so OCOC is half the diagonal ACAC:

OC=12AC=12(413 cm)=213 cmOC = \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 △OVC\triangle OVC to find the height OVOV:

OV2+OC2=CV2OV^2 + OC^2 = CV^2 OV2+(213 cm)2=(20 cm)2OV^2 + (2\sqrt{13} \text{ cm})^2 = (20 \text{ cm})^2 OV2+(4×13) cm2=400 cm2OV^2 + (4 \times 13) \text{ cm}^2 = 400 \text{ cm}^2 OV2+52 cm2=400 cm2OV^2 + 52 \text{ cm}^2 = 400 \text{ cm}^2 OV2=400 cm2−52 cm2OV^2 = 400 \text{ cm}^2 - 52 \text{ cm}^2 OV2=348 cm2OV^2 = 348 \text{ cm}^2
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