Three-Dimensional Geometry, Measurement, and Volume

Polyhedra, Cross-Sections, and Solids of Revolution

In 3D geometry, we study shapes that have length, width, and height. Some 3D shapes have flat faces, some have curved surfaces, and some can be created by slicing or rotating 2D shapes.

In this lesson, you will learn how to:

• identify faces, edges, and vertices of polyhedra,
• understand cross-sections made by slicing solids,
• recognize solids of revolution, which are made by rotating 2D shapes,
• connect these ideas to surface area, volume, and visualization of 3D objects.

1. Polyhedra

A polyhedron is a 3D solid made only of flat polygon faces. This means its surfaces are all flat, not curved.

Examples of polyhedra include:

• cubes,
• rectangular prisms,
• triangular prisms,
• pyramids.

Shapes like spheres, cones, and cylinders are not polyhedra because they have curved surfaces.

There are three important parts of a polyhedron:

• Faces: the flat surfaces,
• Edges: the line segments where two faces meet,
• Vertices: the corner points where edges meet.

Example: A cube has:

• 6 faces,
• 12 edges,
• 8 vertices.

A rectangular prism has the same numbers: 6 faces, 12 edges, and 8 vertices.

Euler's Formula

For many polyhedra, the numbers of faces, edges, and vertices are related by:

$$V - E + F = 2$$

where:

• \(V\) = number of vertices,
• \(E\) = number of edges,
• \(F\) = number of faces.

This is a useful way to check whether your counting is correct.

2. Common Polyhedra

Prisms

A prism has:

• two congruent, parallel bases,
• side faces connecting the bases.

The prism is named by the shape of its base. For example:

• a triangular prism has triangular bases,
• a pentagonal prism has pentagonal bases.

Pyramids

A pyramid has:

• one polygon base,
• triangular faces that meet at one point called the apex.

A square pyramid has a square base. A triangular pyramid has a triangular base.

3. Cross-Sections

A cross-section is the 2D shape made when a 3D solid is cut by a plane.

You can think of it like slicing a loaf of bread. Each slice is a cross-section.

The shape of a cross-section depends on:

• the solid being sliced,
• the direction of the slice,
• where the slice passes through the solid.

Examples of cross-sections:

• Slicing a cube parallel to a face gives a square.
• Slicing a rectangular prism parallel to its base gives a rectangle.
• Slicing a cylinder parallel to its base gives a circle.
• Slicing a cone parallel to its base gives a circle.
• Slicing a pyramid or cone in different ways can produce triangles or other shapes.

Cross-sections help us understand the inside of solids and are useful in science, engineering, and design.

Important idea: A cross-section is always a 2D shape, even though it comes from a 3D solid.

4. Solids of Revolution

A solid of revolution is a 3D shape made by rotating a 2D shape around a line called the axis of rotation.

Imagine drawing a shape on paper and spinning it around one side. As it turns, it sweeps out a 3D solid.

Common examples:

• A rectangle rotated around one side forms a cylinder.
• A right triangle rotated around one leg forms a cone.
• A semicircle rotated around its diameter forms a sphere.

These shapes are important because many real-world objects are solids of revolution, such as cans, ice cream cones, and balls.

5. Connecting 2D and 3D Shapes

In geometry, it is very important to connect flat shapes and solid shapes.

• A net shows how the faces of a polyhedron unfold into 2D.
• A cross-section shows the 2D shape inside a slice of a 3D object.
• A solid of revolution shows how a 2D shape can create a 3D object by rotating.

These ideas help you visualize shapes from different points of view.

6. Surface Area and Volume Connections

Knowing the structure of a solid helps when finding its surface area and volume.

For example:

• To find the surface area of a polyhedron, add the areas of all its faces.
• To find the volume of a prism, use:

$$V = Bh$$

where \(B\) is the area of the base and \(h\) is the height.

For a cylinder, which is a solid of revolution, the volume is:

$$V = \pi r^2 h$$

For a cone, the volume is:

$$V = \frac{1}{3}\pi r^2 h$$

Recognizing the type of solid helps you choose the correct formula.

Worked Example 1: Counting Faces, Edges, and Vertices

A triangular prism has 2 triangular bases and 3 rectangular side faces.

Step 1: Count the faces.

There are 2 triangles and 3 rectangles, so:

$$F = 5$$

Step 2: Count the vertices.

Each triangle has 3 vertices, and there are two triangles:

$$V = 6$$

Step 3: Count the edges.

• 3 edges on the top triangle,
• 3 edges on the bottom triangle,
• 3 edges connecting the matching vertices.

So:

$$E = 9$$

Check with Euler's Formula:

$$V - E + F = 6 - 9 + 5 = 2$$

The counts are correct.

Worked Example 2: Identifying a Cross-Section

A cube is sliced by a plane parallel to one of its faces. What is the cross-section?

Reasoning: Each face of a cube is a square. A slice parallel to a face has the same shape as that face.

Answer: The cross-section is a square.

Worked Example 3: Recognizing a Solid of Revolution

A rectangle with height 8 cm and width 3 cm is rotated around one of its longer sides. What solid is formed?

Reasoning: When a rectangle spins around one side, every point on the opposite side moves in a circle.

This creates a cylinder.

The side used as the axis becomes the height, so the cylinder's height is 8 cm. The width becomes the radius, so the radius is 3 cm.

Answer: The solid formed is a cylinder with height 8 cm and radius 3 cm.

Worked Example 4: Using a Volume Formula

A cylinder has radius \(4\) cm and height \(10\) cm. Find its volume.

Use the formula:

$$V = \pi r^2 h$$

Substitute the values:

$$V = \pi (4)^2(10)$$ $$V = \pi (16)(10)$$ $$V = 160\pi$$

So the exact volume is:

$$160\pi\text{ cm}^3$$

If you use \(\pi \approx 3.14\), then:

$$V \approx 160(3.14) = 502.4\text{ cm}^3$$

7. Tips for Success

• When identifying a polyhedron, check that all faces are flat.
• When counting parts, separate faces, edges, and vertices carefully.
• When thinking about a cross-section, imagine the shape of the slice, not the whole solid.
• When thinking about a solid of revolution, ask: “What happens when this 2D shape spins?”
• Use formulas only after you identify the solid correctly.

8. Common Mistakes

• Calling a cylinder or cone a polyhedron. They are not, because they have curved surfaces.
• Mixing up edges and vertices.
• Thinking a cross-section is 3D. It is always 2D.
• Forgetting which side becomes the axis of rotation in a solid of revolution.
• Using the wrong volume formula because the solid was not identified correctly.

Brief Summary

A polyhedron is a 3D shape with flat polygon faces, and its main parts are faces, edges, and vertices. A cross-section is the 2D shape formed when a solid is sliced. A solid of revolution is formed when a 2D shape is rotated around an axis.

These ideas help you understand 3D geometry, choose the correct formulas, and visualize shapes from different views. When you can move between 2D and 3D thinking, geometry becomes much easier.

Surface Area of Prisms and Cylinders

When we talk about surface area, we mean the total area covering the outside of a 3D shape. Imagine wrapping a gift box or covering a can with paper. The amount of paper needed is the surface area.

In this lesson, you will learn how to find the surface area of prisms and cylinders. A helpful idea is to picture the shape as a net, which is a 2D layout of all its faces opened up flat.

Square units are always used for surface area, such as \(\text{cm}^2\), \(\text{m}^2\), or \(\text{in}^2\).

1. What is a prism?

A prism is a 3D shape with:

• two matching, parallel bases, and
• flat side faces connecting the bases.

Examples include rectangular prisms, triangular prisms, and other prisms whose bases are the same shape on top and bottom.

To find the surface area of a prism, add the areas of all faces:

• the two bases
• the lateral faces, which are the side faces

A very useful formula for any prism is:

$$ \text{Surface Area} = 2B + Ph $$

where:

• \(B\) = area of one base
• \(P\) = perimeter of the base
• \(h\) = height of the prism, meaning the distance between the two bases

The term \(2B\) accounts for the two bases. The term \(Ph\) gives the total area of the rectangular side faces.

2. Surface area of a rectangular prism

A rectangular prism has 6 rectangular faces. If its length is \(l\), width is \(w\), and height is \(h\), then:

$$ SA = 2lw + 2lh + 2wh $$

This formula comes from adding the areas of the three pairs of matching faces:

• top and bottom: \(lw\)
• front and back: \(lh\)
• left and right: \(wh\)

3. What is a cylinder?

A cylinder has:

• two matching circular bases, and
• one curved surface around the side.

If you unwrap the curved surface of a cylinder, it forms a rectangle. This is why nets are so helpful.

The net of a cylinder contains:

• two circles
• one rectangle

The rectangle's:

• height is the cylinder height \(h\)
• length is the circumference of the base, which is \(2\pi r\)

So the area of the curved surface, also called the lateral area, is:

$$ 2\pi rh $$

The two circular bases together have area:

$$ 2\pi r^2 $$

So the total surface area of a cylinder is:

$$ SA = 2\pi r^2 + 2\pi rh $$

where:

• \(r\) = radius of the circular base
• \(h\) = height of the cylinder

4. Why nets help

A net lets you see every face of a solid as a flat shape. This makes it easier to decide what areas must be added.

For example:

• a prism's net is made of polygons and rectangles
• a cylinder's net is made of two circles and one rectangle

If you can identify each part of the net, you can find the total surface area step by step.

5. Worked Example 1: Rectangular prism

Find the surface area of a rectangular prism with length \(8\text{ cm}\), width \(3\text{ cm}\), and height \(5\text{ cm}\).

Use the formula:

$$ SA = 2lw + 2lh + 2wh $$

Substitute the values:

$$ SA = 2(8)(3) + 2(8)(5) + 2(3)(5) $$ $$ SA = 48 + 80 + 30 $$ $$ SA = 158 $$

So the surface area is \(158\text{ cm}^2\).

6. Worked Example 2: Triangular prism

A triangular prism has a triangular base with side lengths \(6\text{ cm}\), \(8\text{ cm}\), and \(10\text{ cm}\). The height of the triangle is \(4\text{ cm}\), and the prism length is \(12\text{ cm}\). Find its surface area.

Step 1: Find the area of one triangular base.

$$ B = \frac{1}{2}bh = \frac{1}{2}(8)(4) = 16 $$

So one base has area \(16\text{ cm}^2\).

Step 2: Find the perimeter of the triangular base.

$$ P = 6 + 8 + 10 = 24 $$

Step 3: Use the prism formula.

$$ SA = 2B + Ph $$

Here, the prism height is the distance between the triangular bases, which is \(12\text{ cm}\).

$$ SA = 2(16) + 24(12) $$ $$ SA = 32 + 288 $$ $$ SA = 320 $$

So the surface area is \(320\text{ cm}^2\).

7. Worked Example 3: Cylinder

Find the surface area of a cylinder with radius \(4\text{ m}\) and height \(9\text{ m}\).

Use the formula:

$$ SA = 2\pi r^2 + 2\pi rh $$

Substitute the values:

$$ SA = 2\pi (4^2) + 2\pi (4)(9) $$ $$ SA = 2\pi (16) + 72\pi $$ $$ SA = 32\pi + 72\pi $$ $$ SA = 104\pi $$

So the exact surface area is \(104\pi\text{ m}^2\).

If you want a decimal approximation:

$$ 104\pi \approx 326.7 $$

So the approximate surface area is \(326.7\text{ m}^2\).

8. Worked Example 4: Using a net idea

A cylinder has diameter \(10\text{ cm}\) and height \(7\text{ cm}\). Find its surface area.

Step 1: Find the radius.

$$ r = \frac{10}{2} = 5\text{ cm} $$

Step 2: Find the area of the two circular bases.

$$ 2\pi r^2 = 2\pi (5^2) = 2\pi(25) = 50\pi $$

Step 3: Find the lateral area.

$$ 2\pi rh = 2\pi(5)(7) = 70\pi $$

Step 4: Add them together.

$$ SA = 50\pi + 70\pi = 120\pi $$

So the exact surface area is \(120\pi\text{ cm}^2\).

Approximate value:

$$ 120\pi \approx 377.0 $$

So the approximate surface area is \(377.0\text{ cm}^2\).

9. Common mistakes to avoid

• Forgetting a face: Surface area includes every outside face.
• Mixing up height and slant or side lengths: In prisms, the height in \(2B + Ph\) is the distance between the bases.
• Using diameter instead of radius in a cylinder formula: Remember \(r = \frac{d}{2}\).
• Forgetting the two bases: A cylinder has two circles, not one.
• Wrong units: Surface area must be written in square units.

10. A step-by-step strategy

1. Identify the solid: prism or cylinder.
2. Sketch or imagine its net.
3. Find the area of each base.
4. Find the lateral area.
5. Add all the outside areas.
6. Write the answer in square units.

11. Quick formula list

For any prism:

$$ SA = 2B + Ph $$

For a rectangular prism:

$$ SA = 2lw + 2lh + 2wh $$

For a cylinder:

$$ SA = 2\pi r^2 + 2\pi rh $$

12. Summary

Surface area is the total area on the outside of a 3D object. For prisms, you can add the areas of all faces or use \(SA = 2B + Ph\). For cylinders, use the two circular bases and the curved side to get \(SA = 2\pi r^2 + 2\pi rh\).

Always think about the net of the solid. If you can picture the shape opened flat, it becomes much easier to see what areas need to be added together.

Surface Area of Pyramids and Cones

When we talk about surface area, we mean the total area covering the outside of a 3D shape. If you could unwrap the shape into flat pieces, the surface area would be the sum of all those pieces.

In this lesson, you will learn how to find the surface area of pyramids and cones. A very important idea is knowing the difference between height and slant height, because using the wrong one gives the wrong answer.

1. Surface Area of a Pyramid

A pyramid has:

• one base, which can be a square, rectangle, triangle, or another polygon,
• triangular faces that meet at one point called the vertex.

To find the total surface area of a pyramid, add:

• the area of the base, and
• the areas of all the triangular side faces.

The side faces together are called the lateral area.

So:

$$\text{Surface Area} = \text{Base Area} + \text{Lateral Area}$$

For a regular pyramid (where the base is a regular polygon and all side faces match), the lateral area can be found using:

$$\text{Lateral Area} = \frac{1}{2}Pl$$

where:

• \(P\) = perimeter of the base
• \(l\) = slant height of the pyramid

Then the total surface area is:

$$\text{Surface Area} = B + \frac{1}{2}Pl$$

where \(B\) is the area of the base.

Height vs. Slant Height in a Pyramid

The height of a pyramid is the straight-line distance from the vertex down to the center of the base, measured perpendicular to the base.

The slant height is the distance from the vertex to the midpoint of a base edge, measured along the triangular face.

For surface area, you use the slant height, not the vertical height, because the triangular faces are tilted.

2. Surface Area of a Cone

A cone has:

• one circular base,
• one curved surface that wraps around to a point.

Its surface area also has two parts:

• the area of the circular base,
• the curved surface area, also called the lateral area.

The formula for the lateral area of a cone is:

$$\text{Lateral Area} = \pi rl$$

where:

• \(r\) = radius of the base
• \(l\) = slant height

The base area is:

$$\text{Base Area} = \pi r^2$$

So the total surface area of a cone is:

$$\text{Surface Area} = \pi r^2 + \pi rl$$

You can also factor this as:

$$\text{Surface Area} = \pi r(r+l)$$

Height vs. Slant Height in a Cone

The height of a cone goes straight from the tip down to the center of the circular base.

The slant height goes from the tip to a point on the edge of the base, along the outside surface.

For the surface area of a cone, use the slant height. The curved surface stretches along the slanted side, not straight down.

If you know the height and radius but not the slant height, you can find it using the Pythagorean theorem:

$$l^2 = r^2 + h^2$$ $$l = \sqrt{r^2 + h^2}$$

This works because the radius, height, and slant height form a right triangle.

Steps for Finding Surface Area

For a pyramid:

1. Find the area of the base.
2. Find the perimeter of the base.
3. Use the slant height to find the lateral area: \(\frac{1}{2}Pl\).
4. Add base area and lateral area.
5. Write the answer in square units.

For a cone:

1. Find the radius of the base.
2. If needed, find the slant height using \(l=\sqrt{r^2+h^2}\).
3. Find the base area: \(\pi r^2\).
4. Find the lateral area: \(\pi rl\).
5. Add them and write the answer in square units.

Worked Example 1: Square Pyramid

A square pyramid has base side length \(6\text{ cm}\) and slant height \(5\text{ cm}\). Find its total surface area.

Step 1: Find the base area.

$$B = 6 \times 6 = 36\text{ cm}^2$$

Step 2: Find the perimeter of the base.

$$P = 4 \times 6 = 24\text{ cm}$$

Step 3: Find the lateral area.

$$\text{Lateral Area} = \frac{1}{2}Pl = \frac{1}{2}(24)(5) = 60\text{ cm}^2$$

Step 4: Add base area and lateral area.

$$\text{Surface Area} = 36 + 60 = 96\text{ cm}^2$$

Answer: The total surface area is \(96\text{ cm}^2\).

Worked Example 2: Cone with Given Slant Height

A cone has radius \(4\text{ m}\) and slant height \(7\text{ m}\). Find its total surface area.

Step 1: Find the base area.

$$\pi r^2 = \pi(4^2) = 16\pi$$

Step 2: Find the lateral area.

$$\pi rl = \pi(4)(7) = 28\pi$$

Step 3: Add the areas.

$$\text{Surface Area} = 16\pi + 28\pi = 44\pi\text{ m}^2$$

If you want a decimal approximation:

$$44\pi \approx 138.2\text{ m}^2$$

Answer: The total surface area is \(44\pi\text{ m}^2\), or about \(138.2\text{ m}^2\).

Worked Example 3: Cone When Height Is Given Instead of Slant Height

A cone has radius \(3\text{ cm}\) and height \(4\text{ cm}\). Find its total surface area.

Step 1: Find the slant height.

$$l = \sqrt{r^2 + h^2} = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5\text{ cm}$$

Step 2: Find the base area.

$$\pi r^2 = \pi(3^2) = 9\pi$$

Step 3: Find the lateral area.

$$\pi rl = \pi(3)(5) = 15\pi$$

Step 4: Add the areas.

$$\text{Surface Area} = 9\pi + 15\pi = 24\pi\text{ cm}^2$$

Approximate value:

$$24\pi \approx 75.4\text{ cm}^2$$

Answer: The total surface area is \(24\pi\text{ cm}^2\), or about \(75.4\text{ cm}^2\).

Worked Example 4: Pyramid with Rectangular Base

A pyramid has a rectangular base measuring \(8\text{ cm}\) by \(5\text{ cm}\). Suppose the two faces attached to the 8 cm sides have slant height \(6\text{ cm}\), and the two faces attached to the 5 cm sides have slant height \(7\text{ cm}\). Find the total surface area.

This pyramid is not a regular pyramid, so we find each group of triangles separately.

Step 1: Find the base area.

$$B = 8 \times 5 = 40\text{ cm}^2$$

Step 2: Find the area of the two triangles with base 8 cm.

Area of one triangle:

$$\frac{1}{2}(8)(6)=24\text{ cm}^2$$

Area of two such triangles:

$$2 \times 24 = 48\text{ cm}^2$$

Step 3: Find the area of the two triangles with base 5 cm.

Area of one triangle:

$$\frac{1}{2}(5)(7)=17.5\text{ cm}^2$$

Area of two such triangles:

$$2 \times 17.5 = 35\text{ cm}^2$$

Step 4: Add everything.

$$\text{Surface Area} = 40 + 48 + 35 = 123\text{ cm}^2$$

Answer: The total surface area is \(123\text{ cm}^2\).

Common Mistakes to Avoid

• Using height instead of slant height. For lateral area of pyramids and cones, use slant height.
• Forgetting the base. Total surface area includes the base unless the question says otherwise.
• Mixing up perimeter and area. A pyramid formula uses the perimeter of the base, not the base area, in the lateral area part.
• Forgetting square units. Surface area is measured in units such as \(\text{cm}^2\), \(\text{m}^2\), or \(\text{in}^2\).
• Not checking whether the pyramid is regular. If the triangular faces are not all the same, you may need to find each face area separately.

Quick Formula Review

• Regular pyramid: $$\text{Surface Area} = B + \frac{1}{2}Pl$$
• Cone: $$\text{Surface Area} = \pi r^2 + \pi rl$$
• Slant height of a cone: $$l = \sqrt{r^2+h^2}$$

Summary

To find the surface area of a pyramid or cone, add the base area and the lateral area. The most important idea is to use the slant height for the side surfaces, not the vertical height.

For a regular pyramid, use \(B + \frac{1}{2}Pl\). For a cone, use \(\pi r^2 + \pi rl\). If a cone gives you the height instead of the slant height, first use the Pythagorean theorem to find the slant height.

Volume Formulas Overview

Volume tells us how much space a three-dimensional object takes up. You can think of it as the amount of space inside a solid. Volume is measured in cubic units, such as cm^3"), m^3"), or in^3").

In this lesson, you will learn the main volume formulas for prisms, cylinders, pyramids, cones, and spheres. You will also learn why tapered solids like pyramids and cones use a one-third factor in their formulas.

A helpful idea for almost every volume problem is this:

$$\text{Volume} = \text{area of base} \times \text{height}$$

This works directly for solids with the same cross-section all the way through, like prisms and cylinders. For tapered solids, like pyramids and cones, we use a related formula with an extra factor of \frac13").

1. Volume of a Prism

A prism has two matching parallel bases, and the shape stays the same all the way through. Examples include rectangular prisms and triangular prisms.

The general formula is:

$$V = Bh$$

where:

• V") = volume
• B") = area of the base
• h") = height of the prism, meaning the distance between the two bases

For a rectangular prism, the base area is length times width, so:

$$V = lwh$$

2. Volume of a Cylinder

A cylinder is like a prism with circular bases. Since the base is a circle, its area is:

$$B = \pi r^2$$

So the volume formula becomes:

$$V = Bh = \pi r^2 h$$

where r") is the radius of the circular base and h") is the height.

3. Volume of a Pyramid

A pyramid has one base and triangular faces that meet at a point called the apex. Because the shape narrows as it rises, it does not fill space as quickly as a prism with the same base and height.

The formula is:

$$V = \frac13 Bh$$

This means a pyramid has one-third the volume of a prism with the same base area and height.

Why the \(\frac13\) factor?

If you compare a prism and a pyramid that have the same base and height, the pyramid is tapered. Its cross-sections get smaller as you move upward. Because of that taper, it only holds one-third as much volume.

This is not just a coincidence. In geometry, it can be shown that three matching pyramids can be rearranged to fill a prism with the same base area and height, depending on the shape. So:

$$\text{Volume of pyramid} = \frac13(\text{base area})(\text{height})$$

4. Volume of a Cone

A cone is like a pyramid with a circular base. Since the base area is \pi r^2"), we use the pyramid idea with the one-third factor:

$$V = \frac13 \pi r^2 h$$

A cone has one-third the volume of a cylinder with the same base radius and height.

5. Volume of a Sphere

A sphere is a perfectly round three-dimensional shape, like a ball. Its volume formula is different from the others because it does not have a base and height in the same way.

The formula is:

$$V = \frac43 \pi r^3$$

where r") is the radius of the sphere.

Notice that the radius is cubed. That makes sense because volume is a three-dimensional measurement.

Important Notes About Measurements

• Always use the same unit for all measurements before calculating.
• Volume answers must be written in cubic units.
• Be careful to use the correct height. In volume formulas, height means the perpendicular distance from base to top.
• For circles and spheres, make sure you know whether you are given the radius or the diameter. Remember: $$d = 2r$$

Volume Formulas Summary Table

• Prism: $$V = Bh$$
• Rectangular prism: $$V = lwh$$
• Cylinder: $$V = \pi r^2 h$$
• Pyramid: $$V = \frac13 Bh$$
• Cone: $$V = \frac13 \pi r^2 h$$
• Sphere: $$V = \frac43 \pi r^3$$

Worked Example 1: Rectangular Prism

A rectangular prism has length 6 cm"), width 4 cm"), and height 3 cm"). Find its volume.

Use the formula:

$$V = lwh$$

Substitute the values:

$$V = 6 \cdot 4 \cdot 3 = 72$$

So the volume is:

$$\boxed{72\text{ cm}^3}$$

Worked Example 2: Cylinder

A cylinder has radius 5 m") and height 8 m"). Find its volume.

Use the formula:

$$V = \pi r^2 h$$

Substitute the values:

$$V = \pi (5)^2 (8)$$ $$V = \pi (25)(8) = 200\pi$$

So the exact volume is:

$$\boxed{200\pi\text{ m}^3}$$

If you want an approximate decimal value:

$$200\pi \approx 628.3$$

So the volume is about 628.3 m^3").

Worked Example 3: Pyramid

A pyramid has a square base with side length 9 cm") and height 12 cm"). Find its volume.

First find the base area:

$$B = 9 \cdot 9 = 81$$

Now use the pyramid formula:

$$V = \frac13 Bh$$ $$V = \frac13 (81)(12)$$ $$V = \frac13 (972) = 324$$

So the volume is:

$$\boxed{324\text{ cm}^3}$$

Worked Example 4: Cone and Sphere

Part A: Cone

A cone has radius 3 cm") and height 10 cm"). Find its volume.

Use the formula:

$$V = \frac13 \pi r^2 h$$ $$V = \frac13 \pi (3)^2 (10)$$ $$V = \frac13 \pi (9)(10) = \frac13 (90\pi) = 30\pi$$

So the volume is:

$$\boxed{30\pi\text{ cm}^3}$$

Part B: Sphere

A sphere has radius 4 cm"). Find its volume.

Use the formula:

$$V = \frac43 \pi r^3$$ $$V = \frac43 \pi (4)^3$$ $$V = \frac43 \pi (64) = \frac{256}{3}\pi$$

So the exact volume is:

$$\boxed{\frac{256}{3}\pi\text{ cm}^3}$$

An approximate decimal value is:

$$\frac{256}{3}\pi \approx 268.1$$

So the volume is about 268.1 cm^3").

How to Choose the Right Formula

1. Identify the solid.
2. Find the measurements you need, such as radius, height, length, or base area.
3. Choose the correct formula.
4. Substitute carefully.
5. Check that your answer is in cubic units.

Common Mistakes to Avoid

• Forgetting the \frac13") in pyramids and cones
• Using diameter instead of radius in circle formulas
• Mixing units, such as centimeters and meters
• Writing square units instead of cubic units
• Confusing slant height with vertical height

Final Summary

Volume measures how much space is inside a three-dimensional object. Prisms and cylinders use V = Bh") because their cross-sections stay the same all the way through.

Pyramids and cones are tapered, so they use V = \frac13 Bh") or V = \frac13\pi r^2 h"), which shows they have one-third the volume of a matching prism or cylinder. Spheres use the special formula V = \frac43\pi r^3").

If you learn to identify the solid, find the correct measurements, and choose the right formula, you will be able to solve many volume problems with confidence.

Square-Cube Law in Dimensional Scaling

When a 3D shape is made larger or smaller, its measurements do not all change in the same way.

If the length of a shape changes by some scale factor, then its surface area and volume change by different factors. This idea is called the Square-Cube Law.

This law is very important in geometry because it helps us predict what happens when we enlarge or shrink solids such as cubes, rectangular prisms, cylinders, and spheres.

Key idea: if every linear measurement of a solid is multiplied by a scale factor of \(k\), then:

• Lengths are multiplied by \(k\)
• Surface areas are multiplied by \(k^2\)
• Volumes are multiplied by \(k^3\)

So the Square-Cube Law can be written as:

$$ \text{New surface area} = \text{Original surface area} \times k^2 $$ $$ \text{New volume} = \text{Original volume} \times k^3 $$

This happens because surface area is measured in square units and volume is measured in cubic units.

For example, if a side length doubles, then each area measurement becomes \(2 \times 2 = 4\) times as large, and each volume measurement becomes \(2 \times 2 \times 2 = 8\) times as large.

Why does area use a square?

Area measures a flat surface. A rectangle has area found by multiplying two lengths:

$$ A = l \times w $$

If both length and width are multiplied by \(k\), then the new area is:

$$ A' = (kl)(kw) = k^2lw $$

So the area becomes \(k^2\) times the original area.

Why does volume use a cube?

Volume measures the space inside a 3D object. For a rectangular prism, volume is:

$$ V = l \times w \times h $$

If all three dimensions are multiplied by \(k\), then the new volume is:

$$ V' = (kl)(kw)(kh) = k^3lwh $$

So the volume becomes \(k^3\) times the original volume.

Important note: the scale factor must apply to every linear dimension. If only one measurement changes, then you cannot use the full Square-Cube Law.

Main teaching points

• A linear scale factor compares side lengths.
• If side lengths scale by \(k\), then area scales by \(k^2\).
• If side lengths scale by \(k\), then volume scales by \(k^3\).
• Doubling is scale factor \(2\), halving is scale factor \(\frac{1}{2}\), tripling is scale factor \(3\).
• Surface area grows faster than length, but volume grows even faster than surface area.

This means large objects can hold much more inside compared to how much outer surface they have.

For example:

• If \(k=2\): area changes by \(2^2=4\), volume changes by \(2^3=8\)
• If \(k=3\): area changes by \(3^2=9\), volume changes by \(3^3=27\)
• If \(k=\frac{1}{2}\): area changes by \(\left(\frac{1}{2}\right)^2=\frac{1}{4}\), volume changes by \(\left(\frac{1}{2}\right)^3=\frac{1}{8}\)

Worked Example 1: Enlarging a cube

A cube has side length \(4\) cm. A larger cube is made by doubling every side length.

Step 1: Find the scale factor.

The side length is doubled, so \(k=2\).

Step 2: Find how the surface area changes.

$$ k^2 = 2^2 = 4 $$

The surface area becomes 4 times the original.

Step 3: Find how the volume changes.

$$ k^3 = 2^3 = 8 $$

The volume becomes 8 times the original.

Check with actual values:

Original surface area of cube:

$$ 6s^2 = 6(4^2)=6(16)=96\text{ cm}^2 $$

New side length:

$$ 8\text{ cm} $$

New surface area:

$$ 6(8^2)=6(64)=384\text{ cm}^2 $$

Compare:

$$ 384 \div 96 = 4 $$

Original volume:

$$ s^3 = 4^3 = 64\text{ cm}^3 $$

New volume:

$$ 8^3 = 512\text{ cm}^3 $$

Compare:

$$ 512 \div 64 = 8 $$

So doubling the side length makes the surface area 4 times as large and the volume 8 times as large.

Worked Example 2: Shrinking a rectangular prism

A rectangular prism is reduced so that every side length is halved.

Step 1: Identify the scale factor.

$$ k=\frac{1}{2} $$

Step 2: Find the surface area scale factor.

$$ k^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} $$

The new surface area is one quarter of the original.

Step 3: Find the volume scale factor.

$$ k^3 = \left(\frac{1}{2}\right)^3 = \frac{1}{8} $$

The new volume is one eighth of the original.

If the original volume was \(320\text{ cm}^3\), then the new volume is:

$$ 320 \times \frac{1}{8} = 40\text{ cm}^3 $$

If the original surface area was \(200\text{ cm}^2\), then the new surface area is:

$$ 200 \times \frac{1}{4} = 50\text{ cm}^2 $$

Worked Example 3: Finding a new measurement from a scale factor

A cylinder is enlarged by a linear scale factor of \(3\). Its original surface area is \(40\text{ cm}^2\), and its original volume is \(25\text{ cm}^3\).

Find the new surface area and new volume.

Surface area:

$$ 40 \times 3^2 = 40 \times 9 = 360\text{ cm}^2 $$

Volume:

$$ 25 \times 3^3 = 25 \times 27 = 675\text{ cm}^3 $$

So the enlarged cylinder has:

• Surface area \(=360\text{ cm}^2\)
• Volume \(=675\text{ cm}^3\)

Worked Example 4: Finding the scale factor from volume

A solid is enlarged, and its volume becomes \(64\) times the original volume. What is the linear scale factor? How much does its surface area change?

Step 1: Use the volume rule.

$$ k^3 = 64 $$

We need the number whose cube is \(64\).

$$ k=4 $$

Step 2: Use the area rule.

$$ k^2 = 4^2 = 16 $$

So:

• The linear scale factor is \(4\)
• The surface area becomes \(16\) times the original

How to solve Square-Cube Law problems

1. Find the linear scale factor \(k\).
2. For surface area, use \(k^2\).
3. For volume, use \(k^3\).
4. Multiply the original area or volume by that scale factor.
5. Make sure your units are correct: area in square units, volume in cubic units.

Common mistakes to avoid

• Mistake 1: multiplying area by \(k\) instead of \(k^2\)
• Mistake 2: multiplying volume by \(k^2\) instead of \(k^3\)
• Mistake 3: forgetting that the scale factor must apply to all lengths
• Mistake 4: mixing up square units and cubic units

Quick comparison table

• Length scales by \(k\)
• Area scales by \(k^2\)
• Volume scales by \(k^3\)

If a shape is made bigger, volume increases the fastest. If a shape is made smaller, volume decreases the fastest.

This is why even a small increase in side length can cause a very large increase in how much a solid can hold.

Brief summary

The Square-Cube Law tells us how measurements change when a 3D shape is scaled.

If every length is multiplied by \(k\), then surface area is multiplied by \(k^2\) and volume is multiplied by \(k^3\).

Remember: length uses 1 factor of \(k\), area uses 2 factors, and volume uses 3 factors. This helps you solve scaling problems quickly and correctly.

Density and Mass Modeling connects geometry to real life. In many situations, we do not just want to know the volume of a solid. We also want to know how heavy it is or how much mass it has. To do that, we use density.

For example, a block of wood and a block of metal can have the same size, but the metal block is much heavier. That happens because metal has a greater density than wood. In this lesson, you will learn how to combine volume and density to find mass in three-dimensional problems.

1. What is density?

Density tells us how much mass is packed into a certain amount of space.

The basic formula is:

$$\text{Density} = \frac{\text{Mass}}{\text{Volume}}$$

This formula can also be rearranged to solve for mass or volume:

$$\text{Mass} = \text{Density} \times \text{Volume}$$ $$\text{Volume} = \frac{\text{Mass}}{\text{Density}}$$

These three formulas are the main tools for density and mass modeling.

2. Units matter

When working with density, the units must match. Common units are:

• Volume in cubic centimeters: \(\text{cm}^3\)
• Volume in cubic meters: \(\text{m}^3\)
• Mass in grams: \(\text{g}\)
• Mass in kilograms: \(\text{kg}\)
• Density in \(\text{g/cm}^3\) or \(\text{kg/m}^3\)

For example:

• If density is in \(\text{g/cm}^3\), volume should be in \(\text{cm}^3\), and mass will come out in grams.
• If density is in \(\text{kg/m}^3\), volume should be in \(\text{m}^3\), and mass will come out in kilograms.

If the units do not match, convert them before calculating.

3. Finding volume of common solids

Since mass depends on volume, you must first know how to calculate the volume of the solid.

• Rectangular prism: $$V = lwh$$
• Cube: $$V = s^3$$
• Cylinder: $$V = \pi r^2 h$$
• Triangular prism: $$V = (\text{area of triangular base}) \times \text{length}$$

Once you find the volume, multiply by density to get mass.

4. The modeling idea

In real-world problems, a solid object is often made from a material such as aluminum, steel, plastic, or concrete. Each material has its own density. A mass model uses:

1. the shape of the object,
2. its dimensions,
3. the formula for volume, and
4. the density of the material.

This lets us estimate the mass of the object.

5. Step-by-step method

Use this process whenever you solve a density and mass problem:

1. Identify the shape of the solid.
2. Find its volume using the correct volume formula.
3. Check the units of volume and density.
4. Use $$\text{Mass} = \text{Density} \times \text{Volume}$$
5. Write the answer with the correct units.

Worked Example 1: Rectangular prism

A metal block is shaped like a rectangular prism. Its length is \(8\text{ cm}\), width is \(5\text{ cm}\), and height is \(3\text{ cm}\). The metal has density \(7.8\text{ g/cm}^3\). Find the mass of the block.

Step 1: Find the volume.

$$V = lwh = 8 \times 5 \times 3 = 120\text{ cm}^3$$

Step 2: Use density to find mass.

$$\text{Mass} = \text{Density} \times \text{Volume}$$ $$\text{Mass} = 7.8 \times 120 = 936\text{ g}$$

Answer: The mass of the block is \(936\text{ g}\).

Worked Example 2: Cylinder

A solid plastic cylinder has radius \(4\text{ cm}\) and height \(10\text{ cm}\). The plastic has density \(1.2\text{ g/cm}^3\). Find the mass of the cylinder. Give your answer to the nearest tenth if needed.

Step 1: Find the volume of the cylinder.

$$V = \pi r^2 h$$ $$V = \pi (4)^2(10) = 160\pi\text{ cm}^3$$ $$V \approx 502.7\text{ cm}^3$$

Step 2: Find the mass.

$$\text{Mass} = 1.2 \times 502.7 \approx 603.2\text{ g}$$

Answer: The mass is about \(603.2\text{ g}\).

Worked Example 3: Unit conversion

A concrete cube has side length \(0.5\text{ m}\). Concrete has density \(2400\text{ kg/m}^3\). Find the mass of the cube.

Step 1: Find the volume.

$$V = s^3 = (0.5)^3 = 0.125\text{ m}^3$$

Step 2: Use the density formula.

$$\text{Mass} = 2400 \times 0.125 = 300\text{ kg}$$

Answer: The mass of the concrete cube is \(300\text{ kg}\).

Notice that the density was given in \(\text{kg/m}^3\), and the volume was in \(\text{m}^3\), so the mass came out in kilograms.

Worked Example 4: A composite solid

A solid object is made by joining two rectangular prisms of the same material.

• Prism A: \(6\text{ cm} \times 4\text{ cm} \times 2\text{ cm}\)
• Prism B: \(3\text{ cm} \times 4\text{ cm} \times 2\text{ cm}\)
• Density of the material: \(2.5\text{ g/cm}^3\)

Find the total mass.

Step 1: Find the volume of each prism.

$$V_A = 6 \times 4 \times 2 = 48\text{ cm}^3$$ $$V_B = 3 \times 4 \times 2 = 24\text{ cm}^3$$

Step 2: Find the total volume.

$$V_{\text{total}} = 48 + 24 = 72\text{ cm}^3$$

Step 3: Find the mass.

$$\text{Mass} = 2.5 \times 72 = 180\text{ g}$$

Answer: The total mass is \(180\text{ g}\).

6. Scaling and how size affects mass

When a solid is enlarged or reduced, its volume changes, and so does its mass.

If all dimensions are multiplied by the same scale factor, the volume changes by the cube of that factor.

For example, if the side lengths of a solid are doubled, then:

$$\text{New volume} = 2^3 = 8 \text{ times the old volume}$$

If the material stays the same, the density does not change. That means the mass also becomes 8 times as large.

This is an important idea in measurement and modeling: mass depends on volume, and volume grows quickly when dimensions increase.

Example of scaling

A small cube has side length \(2\text{ cm}\) and mass \(40\text{ g}\). A larger cube is made of the same material, but each side length is \(3\) times as long. Find the mass of the larger cube.

Since the side length is multiplied by \(3\), the volume is multiplied by:

$$3^3 = 27$$

Because the material is the same, the mass also multiplies by \(27\).

$$40 \times 27 = 1080\text{ g}$$

Answer: The larger cube has mass \(1080\text{ g}\).

7. Common mistakes to avoid

• Using the wrong volume formula. Always identify the solid first.
• Forgetting units. A number without units is incomplete in measurement problems.
• Mixing units. Do not use \(\text{cm}^3\) with \(\text{kg/m}^3\) unless you convert first.
• Confusing mass and density. Density is not the same as weight or mass. It is mass per unit volume.
• Ignoring composite shapes. If a solid is made of several parts, find each volume and add them.

8. Quick check: which formula do I use?

• If you know mass and volume, find density with $$\text{Density} = \frac{\text{Mass}}{\text{Volume}}$$
• If you know density and volume, find mass with $$\text{Mass} = \text{Density} \times \text{Volume}$$
• If you know mass and density, find volume with $$\text{Volume} = \frac{\text{Mass}}{\text{Density}}$$

9. Why this matters

Density and mass modeling are used in construction, engineering, shipping, manufacturing, and science. Workers need to know how heavy concrete blocks are, how much metal is in a machine part, or whether a container can safely hold an object.

By combining geometry and density, you can solve practical problems that involve real objects and materials.

Summary

Density tells how much mass is in a certain volume. The key formula for most problems is $$\text{Mass} = \text{Density} \times \text{Volume}$$. To use it, first find the volume of the 3D shape, make sure the units match, and then multiply by the density. For larger or smaller versions of the same object, mass changes in the same way as volume.