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Composite 3D figures (surface area and volume)

20 practice questions 2 video lessons Theory + worked examples

Composite Three-Dimensional Figures

Texas Geometry (TEKS) • Standard G.11(C), G.11(D) • Three-Dimensional Measurement

Composite Three-Dimensional Figures is a topic in Three-Dimensional Measurement in the Texas Essential Knowledge and Skills (Geometry, §111.41). It is aligned to Standard G.11(C), G.11(D), which requires students to apply surface area and volume formulas to composite three-dimensional figures.

A composite solid is split into familiar solids whose volumes and surface areas are added or subtracted.

Texas Geometry (TEKS) › Three-Dimensional Measurement › Composite Three-Dimensional Figures  —  Standard G.11(C), G.11(D)

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Practice questions

Every question with a fully worked solution.

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  • Math 10C: Surface Area and Volume of Composite Objects Watch
  • Composite Solids Lesson Watch
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Theory

A composite solid is built from several simple solids. Find its volume by decomposing:

  • Add the volumes of joined solids (a cylinder plus a hemisphere).
  • Subtract the volume of a hollow part or drilled hole.
Break it into pieces you know, compute each, then combine — the same idea as composite 2D areas, one dimension up.
Composite solid A composite solid is built from familiar solids whose volumes are added or subtracted. cylinder + hemisphere: add volumes
A cylinder with a hemisphere on top: add the volumes.
Composite solids Composite solids Composite solids split into familiar solids add joined volumes subtract hollow parts
The composite-solid strategy.

The strategy:

\[V_{\text{total}}=\sum V_{\text{parts}}\ \text{or}\ V_{\text{whole}}-V_{\text{hollow}}\]
add the volumes of the parts, or subtract a hollow part from the whole
Match radii and heights where solids join — a hemisphere on a cylinder shares the radius.

How to find a composite volume

  1. Break the solid into familiar solids.
  2. Compute each volume with its formula.
  3. Add joined parts; subtract hollow parts.
  4. Keep \(\pi\) exact when round solids appear.
Example 1 — Cylinder plus hemisphere
A solid is a cylinder (radius \(3\), height \(8\)) with a hemisphere (radius \(3\)) on top. Find the volume (leave \(\pi\)).
Solution

Add the cylinder and hemisphere volumes.

\(\text{cylinder}\)\(=\)\(\pi(3)^2(8)=72\pi\)
\(\text{hemisphere}\)\(=\)\(\dfrac12\cdot\dfrac43\pi(3)^3=18\pi\)
\(\text{total}\)\(=\)\(72\pi+18\pi=90\pi\)
the total volume is 90 pi
Example 2 — Subtract a hole
A cube of side \(6\) has a cylindrical hole of radius \(1\) and depth \(6\) through it. Find the remaining volume (leave \(\pi\)).
Solution

Subtract the cylinder from the cube.

\(\text{cube}\)\(=\)\(6^3=216\)
\(\text{cylinder}\)\(=\)\(\pi(1)^2(6)=6\pi\)
\(\text{remaining}\)\(=\)\(216-6\pi\)
the remaining volume is 216 minus 6 pi
Example 3 — Cone on a cylinder
A cylinder (radius \(2\), height \(5\)) is topped by a cone (radius \(2\), height \(3\)). Find the volume (leave \(\pi\)).
Solution

Add the two volumes.

\(\text{cylinder}\)\(=\)\(\pi(4)(5)=20\pi\)
\(\text{cone}\)\(=\)\(\dfrac13\pi(4)(3)=4\pi\)
\(\text{total}\)\(=\)\(24\pi\)
the total volume is 24 pi
Example 4 — The strategy
How do you find the volume of a composite solid?
Solution

Break it into familiar solids, find each volume, and add joined parts or subtract hollow parts.

decompose into solids, then add or subtract volumes

Common pitfalls

A hemisphere is half a sphere: \(\dfrac12\cdot\dfrac43\pi r^3\).
Add joined solids; subtract hollow ones. Decide which for each part.
Share the correct radius/height where the solids meet.

Frequently asked questions

How do you find the volume of a composite solid?

Split it into familiar solids, find each volume, and add joined parts or subtract hollow ones.

What is the volume of a hemisphere?

Half a sphere: \(\dfrac12\cdot\dfrac43\pi r^3=\dfrac23\pi r^3\).

When do you subtract volumes?

When part of the solid is hollow or drilled out — subtract that volume from the whole.

How is this like composite 2D area?

The same decompose-and-combine idea, but with volumes of solids instead of areas of shapes.