Solids
In geometry, solids are three-dimensional figures. Cubes, cylinders, spheres, prisms, cones, and so on are all examples of solids.
Similar solids
Similar solids are those solids that have the same structures but the magnitudes of their dimensions may or may not be equal.
Volume
The volume of a solid is defined as the space occupied by a particular solid.
The volume of similar solids is used to measure the dimensions of one solid with the help of another similar solid by using their known dimensions.
Scale Factor
The corresponding dimensions of two similar solids are in the same ratio. This ratio is called the scale ratio or the scale factor.
List of similar solids
1. Pair of Cylinders
2. Pair of Cones
3. Pair of Prisms
4. Pair of Pyramids
5. Pair of Spheres
Steps to find the volume of a similar solid
Step 1: You have to find the ratio of the side lengths or dimensions of the two given similar solids.
Step 2: Find the cube of the ratio obtained from step 1, which is equal to the ratio of the volumes of the two similar solids.
Step 3: Equate the given volume of the solid and the ratio of the volumes of similar solids from Step 2 to find the unknown volume of the similar solid.
The Formulas for the Volumes of Similar Solids
To calculate the volume of similar solids, you need the volume of any one of the solids as well as the dimensions of both solids. Here is an example of how to form the equation when different parameters are given for two similar solids:
\(\frac{Volume~of~solid~A}{Volume~of~solid~B}\) = \((\frac{a}{b})^3\)
\(\frac{Volume~of~solid~A}{Volume~of~solid~B}\) = \((Scale~Factor)^3\)
Where a and b represent any dimensions of the given solids, such as length, width, radius, diameter, height and so on. Also, the scale factor represents the ratio of the corresponding dimensions of the two similar solids.
Solved Examples - The Volume of similar solids Formulas
Example 1: The cones shown below are similar. What is the volume of Cone A?
Round your answer to the nearest tenths.
Solution:
\(\frac{Volume~of~A}{Volume~of~B}\) = \((\frac{Height~of~A}{Height~of~B})^3\)
\(\frac{V}{300}\) = \((\frac{12}{25})^3\) Substitute the values and evaluate.
\(\frac{V}{300}\) = \((\frac{1728}{15625})\)
\(\frac{V}{300}\) x 300 = \(\frac{1728}{15625}\) x 300 Multiply both sides by 300 and simplify.
V \(\approx \) 33.18
The volume of cone A is about 33.18 cubic centimeters.
Example 2: The cylinders shown below are similar. Find the volume of cylinder B.
Solution:
\(\frac{Volume~of~A}{Volume~of~B}\) = \((\frac{Radius~of~A}{Radius~of~B})^3\)
\(\frac{256}{V}\) = \((\frac{4}{8})^3\) Substitute the values and evaluate.
\(\frac{256}{V}\) = \(\frac{64}{512}\)
\(\frac{256}{V}\) = \(\frac{64~÷~64}{512~÷~64}\) or \(\frac{1}{8}\) Divide both sides by 64.
256 \(\bullet\) 8 = V \(\bullet\) 1 Cross multiply and simplify.
V = 2048
The volume of cylinder B is 2048 cubic feet.
Example 3: The pyramids shown below are similar. Find the volume of pyramid A.
Solution:
\(\frac{Volume~of~A}{Volume~of~B}\) = \((\frac{Side ~length ~of ~A}{Side~ length~ of ~B})^3\)
\(\frac{V}{256}\) = \((\frac{3}{8})^3\) Substitute the values and evaluate.
\(\frac{V}{256}\) = \(\frac{27}{512}\)
\(\frac{V}{256}\) x 256 = \(\frac{27}{512}\) x 256 Multiply each side by 256 and simplify.
V = 13.5
The volume of pyramid A is 13.5 cubic inches.
Example 4: The spheres shown below are similar. Find the volume of sphere B.
Solution:
\(\frac{Volume~of~A}{Volume~of~B}\) = \((\frac{Radius~of~A}{Radius~of~B})^3\)
\(\frac{335}{V}\) = \((\frac{24}{5})^3\) Substitute the values and evaluate.
\(\frac{335}{V}\) = \(\frac{13824}{125}\)
355 \(\bullet\) 125 = 13824 \(\bullet\) V Cross multiply to find V
\(\frac{44375}{V}\) = \(\frac{13824~x~V}{V}\) Divide both sides by 13824 and simplify.
V \(\approx\) 3.21
The volume of sphere B is about 3.21 cubic inches.
Example 5: The school aquarium needs an upgrade as the number of fishes have increased. The dimensions of the new aquarium are required to be thrice those of the old aquarium. If one cubic foot of water weighs 62.5 pounds, how many pounds of water will the new aquarium contain?
Solution:
As the dimensions of the new aquarium are thrice those of the old one, the scale factor = 3
\(\frac{Volume~of~the~new~aquarium}{Volume~of~the~old~aquarium}\) = \((scale~factor)^3\)
\(\frac{Volume~of~the~new~aquarium}{1500}\) = \((3)^3\) Substitute the values and evaluate.
\(\frac{Volume~of~the~new~aquarium}{1500}\) x 1500 = 27 x 1500 Multiply both sides by 1500 and simplify.
The volume of the new aquarium = 40500 cubic feet.
The new aquarium holds 40,500 cubic feet of water. To find the weight of water in the aquarium, multiply the volume of the new aquarium by \(\frac{62.5~lb}{1~ft^3}\).
40,500 \(ft^3\) x \(\frac{62.5~lb}{1~ft^3}\)
= 2,531,250 lb
So, the new aquarium will contain 2,531,250 pounds of water.
Two solids that have the same shape are said to be similar solids. Moreover, when two solids are similar, the ratios of their corresponding dimensions are equal.
Step 1: Determine the corresponding dimensions of the given solids.
Step 2: Find the ratio of the lengths of the corresponding dimensions, which is known as the scale factor.
Step 3: Check if the ratios are the same for all the corresponding dimensions – if they are same, the solids are similar.
The surface area of similar solids can be used to find out their volume. Here is the formula for it:
\(\frac{Volume~of~Solid~A}{Volume~of~Solid~B}\) = \(\left( \sqrt{\dfrac{\text{Surface area of solid A}}{\text{Surface area of solid B}}} \right)^{3/2}\)