Surface Area and Volume of Composite Solids
Theory
A composite solid is made of prisms and cylinders joined together, or with a part cut out. This Year 10 maths page covers volume by adding and subtracting parts, surface area counting only the faces on the outside, holes, rings and pipes, and practical models with capacity, mass, rain, packing, costs and unknown dimensions. Curriculum: VC2M10M01 / AC9M10M01.
A composite solid is made by joining two or more simple solids, such as rectangular prisms, triangular prisms and cylinders, or by cutting one solid out of another.
Its volume is found by adding the volumes of the parts that are joined and subtracting any part that is cut out, such as a hole or a groove.
A prism whose end is an L, T or other composite shape is still a prism: its volume is the area of that end times the length.
Its surface area is the area of everything a coat of paint would touch. Faces pressed against each other where two parts join are hidden and are not counted.
When a smaller solid sits on a larger face, such as a cylinder standing on a block, the patch it covers is lost but its own top adds the same area back. A hole removes two openings and adds an inside wall.
For a prism with a composite end, with \(A\) the area of the end and \(h\) the length:
| Part | Volume | Surface pieces |
|---|---|---|
| Rectangular prism | \(lwh\) | six rectangles |
| Cylinder | \(\pi r^2 h\) | curved wall \(2\pi rh\), each end \(\pi r^2\) |
| Half-cylinder | \(\dfrac{1}{2}\pi r^2 h\) | curved part \(\pi rh\), two semicircles, flat rectangle \(2r \times h\) |
How to find the volume of a composite solid
- Split the solid into simple parts, or spot a prism with a composite end.
- Find the measurements of each part. Shared lengths and radii are often hidden in the diagram.
- Calculate each volume and add the joined parts; subtract anything cut out.
- Round once at the end, and convert to litres if asked.
How to find the surface area
- List every surface on the outside. Cross out faces that are joined, covered or open.
- Shape on a face: the covered patch and the new top cancel, so add only the side walls of the smaller solid.
- Hole through a solid: subtract the two openings and add the inside wall.
- Calculate each area and add, keeping \(\pi\) exact until the end.
Modelling and unknowns
- Units: change every length to one unit first; \(1\text{ m}^3 = 1000\) L and \(1\text{ cm}^3 = 1\text{ mL}\).
- An unknown dimension: write each part in terms of the pronumeral, add, set equal to the given amount and solve. If the volume is a multiple of \(r^3\), finish with a cube root: \(10r^3 = 270\) gives \(r^3 = 27\) and \(r = 3\).
- A percentage removed: the part cut out is that percentage of the whole volume; set its formula equal to that amount.
- Answer the question: whole blocks, full days or tubs that fit are rounded down; material you must buy is rounded up.
The two parts are joined, so find each volume and add.
| \(V_{\text{base}}\) | \(=\) | \(50 \times 50 \times 20 = 50\,000 \text{ cm}^3\) |
| \(V_{\text{post}}\) | \(=\) | \(\pi \times 10^2 \times 80 = 8000\pi\) |
| \(V\) | \(=\) | \(50\,000 + 8000\pi\) |
| \(\) | \(=\) | \(75\,132.741\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{75\,132.74} \textbf{ cm}^\mathbf{3}\) |
\(1000 \text{ cm}^3 = 1\) L, so that is about \(75.13\) L of concrete.
The hole is a cylinder as tall as the cube. Subtract it:
| \(V\) | \(=\) | \(10^3 - \pi \times 2^2 \times 10\) |
| \(\) | \(=\) | \(1000 - 40\pi\) |
| \(\) | \(=\) | \(874.3362\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{874.34}\text{ cm}^3\) |
A prism is two ends plus a rectangle for every edge of the end. All those rectangles together are (perimeter of the end) \(\times\) length.
| \(\text{one end}\) | \(=\) | \(8 \times 6 - 5 \times 3 = 33\text{ cm}^2\) |
| \(\text{perimeter}\) | \(=\) | \(8 + 3 + 5 + 3 + 3 + 6 = 28\text{ cm}\) |
| \(\text{SA}\) | \(=\) | \(2 \times 33 + 28 \times 10\) |
| \(\) | \(=\) | \(66 + 280\) |
| \(\) | \(=\) | \(\mathbf{346}\text{ cm}^2\) |
The perimeter of an L shape equals the perimeter of the rectangle around it: \(2(8 + 6) = 28\) β
List the surfaces: the outer curved wall, the inner curved wall, and two flat rings (a big circle minus a small circle at each end).
| \(\text{outer wall}\) | \(=\) | \(2\pi(5)(12) = 120\pi\) |
| \(\text{inner wall}\) | \(=\) | \(2\pi(2)(12) = 48\pi\) |
| \(\text{two rings}\) | \(=\) | \(2(\pi \times 5^2 - \pi \times 2^2) = 42\pi\) |
| \(\text{SA}\) | \(=\) | \(120\pi + 48\pi + 42\pi\) |
| \(\) | \(=\) | \(\mathbf{210\pi}\text{ cm}^2\) |
| \(\) | \(\approx\) | \(\mathbf{659.73}\text{ cm}^2\) |
Drilling a hole can increase the surface area: the inside wall (\(48\pi\)) is bigger than the two circles removed (\(8\pi\)).
Common pitfalls
Frequently asked questions
How do you find the volume of a composite solid?
Split it into simple solids, find the volume of each, then add the parts that are joined together and subtract any part that is cut out.
How do you find the surface area of a composite solid?
List every surface on the outside, leave out faces that are joined or hidden, find each area and add them.
Do you include the area where two solids are joined?
No. The two faces pressed together are inside the solid, so neither is part of the surface area.
Does drilling a hole increase or decrease the surface area?
It can do either. You lose the two openings but gain the inside wall of the hole. For a long, narrow hole the inside wall is usually larger, so the surface area goes up.
How do you find the volume of a prism with an L-shaped end?
Find the area of the L-shaped end by adding or subtracting rectangles, then multiply by the length of the prism.
How do you find the mass of a solid from its volume?
Multiply the volume by the density, in matching units. A density in \(\text{g/cm}^3\) times a volume in \(\text{cm}^3\) gives grams; divide by \(1000\) for kilograms.