Question

Theorem. Volumes of similar polyhedra have the same ratio as the cubes of homologous edges. Consider first the case of pyramids. Let \(S A B C D E\) (Figure 72) be one of the given pyramids, \(L\) be the length of one of its edges, e.g. \(S A\), and \(l

Short answer

The volume ratio of the pyramids is \(\frac{l^3}{L^3}\).

Step-by-step solution

Understand the Problem

We are given two similar pyramids and asked to show that the ratio of their volumes is the cube of the ratio of their homologous edge lengths.

Identify Similarity Ratio

The ratio of the homologous edges of the pyramids, for example, the ratio of their altitudes and the lengths of their edges is given as \(\frac{l}{L}\).

Relate Base Area Ratios

For similar shapes, the ratio of their areas is the square of the ratio of their sides. Therefore, the base area ratio \(\frac{a'}{a}\) is \(\left(\frac{l}{L}\right)^2 = \frac{l^2}{L^2}\).

Calculate Volume Using Formula

The volume of a pyramid is given by the formula \(V = \frac{1}{3} \times \text{Base Area} \times \text{Height}\). For the pyramids, the volumes are \(V = \frac{1}{3} a \cdot SO\) and \(v = \frac{1}{3} a' \cdot SO'\).

Combine Ratios

Express the volume ratio, \(\frac{v}{V}\), as a product of base area and height ratios: \(\frac{v}{V} = \frac{a'}{a} \cdot \frac{SO'}{SO}\).

Simplify to Get Volume Ratio

Using \(\frac{SO'}{SO} = \frac{l}{L}\) and \(\frac{a'}{a} = \frac{l^2}{L^2}\), substitute these into the volume ratio expression to obtain: \(\frac{v}{V} = \frac{l^2}{L^2} \cdot \frac{l}{L} = \frac{l^3}{L^3}\). This shows that the volume ratio is the cube of the edge length ratio.

Key concepts

Volume Ratio and Cube Relationship

When two polyhedra are similar, their dimensions have a special relationship: their volumes are proportional to the cube of the ratio of their corresponding, or homologous, edges.
This means that if you know how one measure changes, like an edge, you can determine how the overall volume changes.
- For instance, suppose you have two similar pyramids where the edge length of the smaller pyramid is half that of the larger. - The volume of the smaller pyramid would be not half, but actually \(\left(\frac{1}{2}\right)^3 = \frac{1}{8}\) of the larger pyramid. The reason is that volume scales with the cube of linear dimensions because it is three-dimensional.
This is critical in understanding how volume changes with size in similar figures, not just pyramids, but any kind of polyhedra. Remembering that volume expands or contracts so substantially with these ratios can help you solve various problems relating to geometric scaling.

Understanding Homologous Edges

Homologous edges in similar polyhedra are those edges that correspond to each other. Imagine they are the counterparts in each shape.
These edges are crucial because they determine the ratio used in several geometric calculations, including the volume ratio. In similar figures:

• The length of one edge relative to its counterpart in another shape provides the basic scale factor.
• In our pyramid case, an edge like \(SA\) is homologous to \(SA'\) in the smaller pyramid.

When you know these scale relations, you can extrapolate not just volumes but areas as well, by squaring the ratio for areas.Understanding these homologous edges helps break down complex polyhedra problems to more manageable parts, focusing on how one dimension affects the entire structure's size.

Pyramids and Their Properties

Pyramids are a common type of polyhedron that have special mathematical properties useful in geometry.
- A pyramid has a polygonal base and triangular faces that converge at a single point called the apex.- The height of a pyramid is the perpendicular distance from the base to the apex, which plays a crucial part in calculating its volume.When dealing with similar pyramids, like those described in the theorem, they allow you to see these geometric principles in action.
For similar pyramids, the areas of their bases and the heights are in the ratio of the squares and linear dimensions of their homologous edges respectively. Hence, the volume calculation involves these two parts:

• The area of the base scales by the square of the edge ratio, \(\left(\frac{l}{L}\right)^2\).
• The height scales linearly, by \(\frac{l}{L}\).

Combining these, you see that overall volume ratio becomes the cube of edge ratio. This elegant relationship allows us to predict the volume outcomes just by knowing a single relationship between homologous edges, demonstrating the symmetry and logic inherent in geometry.