How to Find Volume and Surface Area of Pyramids and Cone
Read,5 minutes
PYRAMIDS
A pyramid has a polygon base and triangular faces that meet at an apex. The height is the perpendicular distance from the apex to the base.
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Pyramid Volume and Surface Area
If \(B\) is the base area and \(h\) is the height, then \(V = \frac{1}{3}Bh\). For a regular pyramid, \(SA = B + \frac{1}{2}Pl\), where \(P\) is base perimeter and \(l\) is slant height.
CONES
A cone has a circular base and a curved surface that meets at an apex.
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For a cone, \(V = \frac{1}{3}\pi r^2h\) and \(SA = \pi r^2 + \pi rl\). Leave cone answers in terms of \(\pi\) unless an approximation is requested.
Pyramids and Cone
Think of this lesson as more than a rule to memorize. Pyramids and Cone is about three-dimensional measurement, volume, and surface area. A strong student does not rush to the first formula on the page; they pause, identify the structure of the problem, and then choose the tool that matches that structure. That pause is what prevents most mistakes.
Geometry formulas work because they measure a specific feature: length around, space inside, or space enclosed by a solid. Match the question to the measurement first.
Here is the teacher way to approach the topic. First, read the problem slowly and underline the information that is actually given. Next, name the target: are you finding a value, simplifying an expression, comparing two quantities, solving for a variable, or interpreting a graph? Once the target is clear, the calculation becomes much less mysterious because every step has a job.
- Read what is given and what is being asked.
- Choose the rule that connects them.
- Substitute carefully and simplify in small steps.
- Check the final answer against the original question.
A helpful habit is to say what each number represents before using it. For example, if a number is a denominator, a radius, a slope, a common difference, or a coefficient, it should not be treated like an ordinary loose number. Its role tells you where it belongs in the formula. This is especially important on ACT-style questions because many wrong answer choices come from using the right numbers in the wrong places.
Another good habit is to keep the work organized vertically. Write one clean line for substitution, one line for simplifying, and one line for the final answer. If the problem has units, keep the units attached. If the problem has signs, exponents, or parentheses, copy them carefully from one line to the next. Most errors in this topic are not caused by a hard idea; they are caused by dropping a negative sign, combining unlike terms, using the wrong denominator, or skipping a check.
When you finish, ask a quick reasonableness question. Should the answer be positive or negative? Should it be larger or smaller than the original number? Does it fit the graph, table, shape, or equation? Can you plug it back into the original problem? This final check turns the lesson from a procedure into understanding.
On a test, the goal is not to write the longest solution. The goal is to write enough clear work that you can see the structure, avoid traps, and recover quickly if one line goes wrong. Practice the examples below with that mindset: identify the type of problem, choose the matching rule, show the substitution, simplify carefully, and check the answer before moving on.
Free printable Worksheets
Exercises for Pyramids and Cone
1) A square pyramid has base side \(6\) and height \(9\). Find its volume.
2) A cone has radius \(3\) and height \(8\). Find its volume.
3) A rectangular pyramid has base \(10\) by \(6\) and height \(12\). Find volume.
4) A cone has diameter \(10\) and height \(6\). Find volume.
5) A square pyramid has base side \(8\) and slant height \(5\). Find surface area.
6) A cone has radius \(4\) and slant height \(9\). Find surface area.
7) A pyramid has base area \(45\) and height \(10\). Find volume.
8) A cone has volume \(75\pi\) and radius \(5\). Find height.
9) A square pyramid has volume \(192\) and base side \(8\). Find height.
10) A cone has radius \(6\) and slant height \(10\). Find surface area.
11) A square pyramid has base side \(10\), height \(12\), and slant height \(13\). Find volume and surface area.
12) A cone has radius \(5\), height \(12\), and slant height \(13\). Find volume and surface area.
13) A triangular pyramid has base area \(30\) and height \(9\). Find volume.
14) A cone's base circumference is \(14\pi\) and height is \(9\). Find volume.
15) A square pyramid has base side \(12\) and lateral area \(120\). Find total surface area.
16) A cone has total surface area \(80\pi\) and radius \(4\). Find slant height.
17) A pyramid has volume \(45\). If every length is doubled, find the new volume.
18) A cone's radius doubles from \(3\) to \(6\), and height is halved from \(10\) to \(5\). By how much does volume increase?
19) A cone and square pyramid both have height \(12\). Cone base area is \(49\pi\); pyramid base area is \(64\). Find combined volume.
20) A cone has radius \(9\), height \(12\), and slant height \(15\). Find volume and surface area.
1) Base area: \(6^2=36\).
Volume: \(\frac{1}{3}(36)(9)=108\).
Answer: \(\color{red}{108\ \text{cubic units}}\)
2) Use \(V=\frac{1}{3}\pi r^2h\).
Compute \(\frac{1}{3}\pi(3^2)(8)=24\pi\).
Answer: \(\color{red}{24\pi\ \text{cubic units}}\)
3) Base area: \(60\).
Volume: \(\frac{1}{3}(60)(12)=240\).
Answer: \(\color{red}{240\ \text{cubic units}}\)
4) Radius is \(5\).
Volume: \(\frac{1}{3}\pi(5^2)(6)=50\pi\).
Answer: \(\color{red}{50\pi\ \text{cubic units}}\)
5) Base area: \(64\); perimeter: \(32\).
Surface area: \(64+\frac{1}{2}(32)(5)=144\).
Answer: \(\color{red}{144\ \text{square units}}\)
6) Use \(SA=\pi r^2+\pi rl\).
Compute \(16\pi+36\pi=52\pi\).
Answer: \(\color{red}{52\pi\ \text{square units}}\)
7) Use \(V=\frac{1}{3}Bh\).
Compute \(\frac{1}{3}(45)(10)=150\).
Answer: \(\color{red}{150\ \text{cubic units}}\)
8) Set \(75\pi=\frac{1}{3}\pi(5^2)h\).
Solve \(h=9\).
Answer: \(\color{red}{9\ \text{units}}\)
9) Base area: \(64\).
Set \(192=\frac{1}{3}(64)h\).
Solve \(h=9\).
Answer: \(\color{red}{9\ \text{units}}\)
10) Compute \(36\pi+60\pi=96\pi\).
Answer: \(\color{red}{96\pi\ \text{square units}}\)
11) Volume: \(\frac{1}{3}(10^2)(12)=400\).
Surface area: \(100+\frac{1}{2}(40)(13)=360\).
Answer: \(\color{red}{V = 400,\ SA = 360}\)
12) Volume: \(\frac{1}{3}\pi(5^2)(12)=100\pi\).
Surface area: \(25\pi+65\pi=90\pi\).
Answer: \(\color{red}{V = 100\pi,\ SA = 90\pi}\)
13) Compute \(\frac{1}{3}(30)(9)=90\).
Answer: \(\color{red}{90\ \text{cubic units}}\)
14) From \(2\pi r=14\pi\), \(r=7\).
Volume: \(\frac{1}{3}\pi(7^2)(9)=147\pi\).
Answer: \(\color{red}{147\pi\ \text{cubic units}}\)
15) Base area: \(12^2=144\).
Total area: \(144+120=264\).
Answer: \(\color{red}{264\ \text{square units}}\)
16) Set \(80\pi=16\pi+4\pi l\).
Solve \(80=16+4l\), so \(l=16\).
Answer: \(\color{red}{16\ \text{units}}\)
17) Volume scale factor is \(2^3=8\).
New volume: \(45\cdot8=360\).
Answer: \(\color{red}{360\ \text{cubic units}}\)
18) Original volume: \(30\pi\).
New volume: \(60\pi\).
Increase: \(30\pi\).
Answer: \(\color{red}{30\pi\ \text{cubic units}}\)
19) Cone volume: \(\frac{1}{3}(49\pi)(12)=196\pi\).
Pyramid volume: \(\frac{1}{3}(64)(12)=256\).
Combined: \(196\pi+256\).
Answer: \(\color{red}{196\pi + 256\ \text{cubic units}}\)
20) Volume: \(\frac{1}{3}\pi(9^2)(12)=324\pi\).
Surface area: \(81\pi+135\pi=216\pi\).
Answer: \(\color{red}{V = 324\pi,\ SA = 216\pi}\)
Pyramids and Cone Practice Quiz