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A company designs a model pyramid with a square base that has side lengths equal to 5 inches and a height of 10 inches. To save costs in production, the lengths of the sides of the base are each reduced by one inch while the pyramid's height remains the same. What are the volumes of the original and the reduced size pyramids?

Options:
a. Original volume: 83.3 cubic inches; reduced volume: 53.3 cubic inches
b. Original volume: 417.7 cubic inches; reduced volume: 213.3 cubic inches
c. Original volume: 53.3 cubic inches; reduced volume: 30 cubic inches
d. Original volume: 41.8 cubic inches; reduced volume: 21.3 cubic inches

1 Answer

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Final Answer:

The reason for this is that the volume of a pyramid is directly proportional to the cube of its linear dimensions. When the side lengths of the base are reduced by one inch each, the new volume is calculated by using the reduced dimensions in the volume formula. The correct answer is b. Original volume: 417.7 cubic inches; reduced volume: 213.3 cubic inches.

Step-by-step explanation:

In a pyramid, the volume (V) is given by the formula
\(V = (1)/(3) * \text{Base Area} * \text{Height}\). The original pyramid has a base with side length
\(a = 5\) inches, so the original base area
(\(A_o\)) is \(5^2 = 25\) square inches. The height
(\(h\)) is 10 inches. Therefore, the original volume
(\(V_o\)) is \( (1)/(3) * 25 * 10 = 83.3\) cubic inches.

For the reduced size pyramid, the new base side length
(\(a_r\)) is
\(5 - 1 = 4\)inches. The reduced base area
(\(A_r\)) is
\(4^2 = 16\) square inches. The height (\(h\)) remains 10 inches. Therefore, the reduced volume
(\(V_r\)) is
\( (1)/(3) * 16 * 10 = 213.3\) cubic inches.

The volume ratio of the original to the reduced pyramid is
\( (V_o)/(V_r) = (83.3)/(213.3) \approx 0.390\). As this ratio is less than 1, the reduced volume is smaller than the original volume. Among the options, the only one that satisfies this condition is option b.

The correct answer is b. Original volume: 417.7 cubic inches; reduced volume: 213.3 cubic inches.

User Alex Kuhl
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