Composite solids are just a combination of simpler solids.
These solids can be formed by adding two or more prisms, as in the following two examples:
The solid on the left is created by joining three separate cylinders (two of which are the same size). We need to calculate the volume of each component cylinder and add together to determine the volume of this solid.
The solid on the right is created by joining a rectangular prism and a triangular prism. When the newly formed the composite solid is also a prism, it could be easier to use the formula for the volume of a prism. To do this, we would calculate the area of the end (a triangle plus a rectangle) and multiply by the height of the prism.
Composite solids can also be formed by subtracting components, rather than adding, as in the examples below:
The solid on the left is formed by cutting a cylinder from a cube: calculate the volume of the cube and subtract the volume of the cylinder that would fill the circular hole through the prism.
The solid on the right is a truncated pyramid: calculate the volume of the complete pyramid, and subtract the volume of the smaller pyramid that is cut off from the top.
The strategy to find the volume of composite solids is:
If the composite solid is a prism, we may use the formula for the volume of a prism - calculate the area of the end and multiply by the height of the prism.
To determine the volume of the components of a composite solid, we can use the formulas that we have learned previously:
$\text{Volume of any prism }$Volume of any prism | $=$= | $\text{area of base }\times\text{height }$area of base ×height |
$=$= | $A\times h$A×h |
$\text{Volume of a rectangular prism }$Volume of a rectangular prism | $=$= | $\text{area of base }\times\text{height }$area of base ×height |
$=$= | $l\times w\times h$l×w×h |
$\text{Volume of a cube }$Volume of a cube | $=$= | $\text{length }\times\text{length }\times\text{length }$length ×length ×length |
$=$= | $l^3$l3 |
$\text{Volume of a cylinder }$Volume of a cylinder | $=$= | $\text{area of base }\times\text{height }$area of base ×height |
$=$= | $\pi\times\text{radius }^2\times\text{height }$π×radius 2×height | |
$=$= | $\pi r^2\times h$πr2×h |
$\text{Volume of a pyramid }$Volume of a pyramid | $=$= | $\frac{1}{3}\times\text{area of base }\times\text{height }$13×area of base ×height |
$=$= | $\frac{1}{3}\times A\times h$13×A×h |
$\text{Volume of a sphere }$Volume of a sphere | $=$= | $\frac{4}{3}\times\pi\times\text{radius }^3$43×π×radius 3 |
$=$= | $\frac{4}{3}\pi r^3$43πr3 |
Find the volume of the figure shown, correct to two decimal places.
Consider the following hemisphere.
Find the volume of the hemisphere in cubic centimetres.
Round your answer to three decimal places.
What is the capacity in litres?
Round your answer to the nearest litre.
The image below shows the outline of a construction site for the foundations of a large office building.
Find the area of the construction site in square metres.
Prolonged rainstorms bring $0.21$0.21 m of rain to the region and the foundations of the building are flooded.
What is the volume of water in the construction site in cubic metres?
Give your answer as a decimal.