Two- and Three-Dimensional Measurement

Area of Polygons

When we find the area of a shape, we are finding how much surface is inside it. Area is measured in square units, such as square centimeters \\(cm^2\\), square meters \\(m^2\\), or square inches \\(in^2\\).

In this lesson, you will learn how to find the area of three important polygons: parallelograms, triangles, and trapezoids. You will also see why the formulas work by breaking shapes apart and comparing them to rectangles.

A polygon is a closed shape made of straight sides. Many area formulas come from shapes you already know, especially the rectangle.

Remember: the area of a rectangle is

$$A = l \times w$$

This means length times width. We will use this idea to build the formulas for other polygons.

Important idea: base and height

For many polygons, area depends on the base and the height.

• Base: a side chosen to be the bottom of the shape
• Height: the perpendicular distance from the base to the opposite side or opposite vertex

The height must make a right angle with the base. This is very important. A slanted side is not always the height.

1. Area of a Parallelogram

A parallelogram is a four-sided shape with two pairs of parallel sides.

If you look at a parallelogram, it may seem different from a rectangle. But if you cut a triangular piece from one side and move it to the other side, the shape becomes a rectangle.

That means a parallelogram has the same area as a rectangle with the same base and height.

So the formula for the area of a parallelogram is

$$A = b \times h$$

where \\(b\\) is the base and \\(h\\) is the height.

Why this works: rearranging the parallelogram does not change its area. It simply turns into a rectangle, and rectangle area is base times height.

Example 1: Parallelogram

Find the area of a parallelogram with base \\(8\\) cm and height \\(5\\) cm.

Use the formula:

$$A = b \times h$$

Substitute the values:

$$A = 8 \times 5 = 40$$

So the area is \\(40 \, cm^2\\).

Watch out: if a parallelogram has a slanted side of \\(6\\) cm, but the height is \\(5\\) cm, you use the height, not the slanted side, unless the slanted side is also perpendicular to the base.

2. Area of a Triangle

A triangle can be connected to another triangle of the same size to make a parallelogram.

Since the area of the parallelogram is \\(b \times h\\), one triangle is half of that.

So the formula for the area of a triangle is

$$A = \frac{1}{2}bh$$

where \\(b\\) is the base and \\(h\\) is the height.

Why this works: two matching triangles form a parallelogram. One triangle is half the area of the whole parallelogram.

Example 2: Triangle

Find the area of a triangle with base \\(10\\) m and height \\(7\\) m.

Use the formula:

$$A = \frac{1}{2}bh$$

Substitute:

$$A = \frac{1}{2}(10)(7)$$

Multiply:

$$A = \frac{1}{2}(70) = 35$$

So the area is \\(35 \, m^2\\).

Tip: It does not matter which side you choose as the base, as long as you use the height that goes with that base.

3. Area of a Trapezoid

A trapezoid is a four-sided shape with one pair of parallel sides. These parallel sides are called the bases.

Let the two bases be \\(b_1\\) and \\(b_2\\), and let the height be \\(h\\).

If you put two matching trapezoids together, they make a parallelogram. The base of that parallelogram is the sum of the two trapezoid bases, \\(b_1 + b_2\\). Its height is still \\(h\\).

The area of the parallelogram is

$$A = (b_1 + b_2)h$$

One trapezoid is half of that, so the formula is

$$A = \frac{1}{2}(b_1 + b_2)h$$

You can also think of this as:

$$A = \left(\frac{b_1 + b_2}{2}\right)h$$

This means: find the average of the two bases, then multiply by the height.

Example 3: Trapezoid

Find the area of a trapezoid with bases \\(6\\) in and \\(10\\) in, and height \\(4\\) in.

Use the formula:

$$A = \frac{1}{2}(b_1 + b_2)h$$

Substitute:

$$A = \frac{1}{2}(6 + 10)(4)$$

Add inside the parentheses:

$$A = \frac{1}{2}(16)(4)$$

Multiply:

$$A = 8 \times 4 = 32$$

So the area is \\(32 \, in^2\\).

How these formulas are connected

• A rectangle has area \\(l \times w\\).
• A parallelogram has area \\(b \times h\\) because it can be rearranged into a rectangle.
• A triangle has area \\(\frac{1}{2}bh\\) because it is half of a parallelogram.
• A trapezoid has area \\(\frac{1}{2}(b_1+b_2)h\\) because two trapezoids make a parallelogram.

This is called geometric decomposition. It means breaking shapes apart or rearranging them to understand and find area.

Using rectangle framing

Another way to understand area is by placing a shape inside a rectangle. Then you can subtract the extra parts.

For example, a triangle can fit inside a rectangle. If the rectangle has the same base and height as the triangle, the triangle takes up exactly half of the rectangle.

A trapezoid can also be seen inside a rectangle, with triangles on the sides that can be removed. This helps explain why the area depends on the average of the two bases.

Example 4: Comparing shapes and choosing the correct height

A triangle has base \\(12\\) cm and height \\(9\\) cm. One slanted side is \\(10\\) cm. Find the area.

Even though the triangle has a slanted side of \\(10\\) cm, the formula uses the height, which is \\(9\\) cm.

Use the formula:

$$A = \frac{1}{2}bh$$

Substitute:

$$A = \frac{1}{2}(12)(9)$$

Multiply:

$$A = \frac{1}{2}(108) = 54$$

So the area is \\(54 \, cm^2\\).

Common mistakes to avoid

• Do not use a slanted side as the height unless it is perpendicular to the base.
• Do not forget the \\(\frac{1}{2}\\) in triangle and trapezoid formulas.
• Use square units in your final answer.
• Add both bases for a trapezoid before multiplying by \\(\frac{1}{2}\\) and the height.
• Check that base and height match each other.

Steps for solving area problems

1. Identify the type of polygon.
2. Find the base or bases and the height.
3. Choose the correct formula.
4. Substitute the numbers carefully.
5. Solve.
6. Write the answer in square units.

Formulas to remember

• Parallelogram: $$A = bh$$
• Triangle: $$A = \frac{1}{2}bh$$
• Trapezoid: $$A = \frac{1}{2}(b_1+b_2)h$$

Brief Summary

The area of a polygon tells how much space is inside the shape. For parallelograms, multiply base by height. For triangles, take half of base times height. For trapezoids, take half of the sum of the two bases times the height.

These formulas make sense because the shapes can be rearranged, matched, or framed with rectangles. Understanding why the formulas work will help you remember them and use them correctly.

Area and Perimeter of Composite Figures

A composite figure is a shape made by putting together two or more simple shapes, such as rectangles, squares, and triangles.

When you work with composite figures, you usually need to find one of two things:

• Area: how much space is inside the figure
• Perimeter: the total distance around the outside edge of the figure

Composite figures may look tricky at first, but the main idea is simple: break the figure into smaller shapes you already know how to measure.

For 7th Grade, the most common shapes you will split a composite figure into are:

• Rectangles
• Squares
• Triangles

Important formulas

• Rectangle area: \(A = l \times w\)
• Square area: \(A = s \times s = s^2\)
• Triangle area: $$A = \frac{1}{2}bh$$
• Perimeter of any figure: add all outside side lengths

How to find the area of a composite figure

1. Look at the figure carefully.
2. Split it into smaller familiar shapes.
3. Find the area of each smaller shape.
4. Add the areas together.

How to find the perimeter of a composite figure

1. Find the length of every outside edge.
2. Do not include sides inside the figure where shapes touch.
3. Add the outside side lengths.

Very important difference

• For area, you add the areas of all the parts.
• For perimeter, you add only the lengths around the outside.

Students often make mistakes when they accidentally include an inside line in the perimeter or forget a missing side length. Always check whether a side is on the outside boundary.

Finding missing side lengths

Sometimes a composite figure does not show every side length. You can often find a missing length by using what you know about opposite sides of rectangles or by subtracting.

For example, if a long horizontal side is 12 units and part of it is 5 units, then the remaining part is:

\(12 - 5 = 7\) units

This idea is used often in L-shaped figures.

Worked Example 1: Composite figure made of two rectangles

Suppose a figure is made from:

• A large rectangle with length 10 cm and width 4 cm
• A small rectangle attached on top with length 3 cm and width 2 cm

Step 1: Find the area of each rectangle.

Large rectangle:

\(A = 10 \times 4 = 40\text{ cm}^2\)

Small rectangle:

\(A = 3 \times 2 = 6\text{ cm}^2\)

Step 2: Add the areas.

$$40 + 6 = 46\text{ cm}^2$$

Area of the composite figure: \(46\text{ cm}^2\)

Step 3: Find the perimeter.

To find perimeter, add only the outside edges. Suppose the small rectangle sits on top of the large one, and its bottom side is attached, so that bottom side is inside the figure and should not be counted.

The outside sides are:

• Bottom of large rectangle: 10 cm
• Left side of large rectangle: 4 cm
• Right side of large rectangle: 4 cm
• Top exposed parts of the large rectangle: total \(10 - 3 = 7\) cm
• Two sides of the small rectangle: \(2 + 2 = 4\) cm
• Top of small rectangle: 3 cm

Add them:

$$10 + 4 + 4 + 7 + 4 + 3 = 32\text{ cm}$$

Perimeter of the composite figure: \(32\text{ cm}\)

Worked Example 2: An L-shaped figure

An L-shape can be split into two rectangles.

Suppose the full height on the left is 9 m, the bottom length is 8 m, the top horizontal part is 3 m, and the vertical drop on the right from the top to the inner corner is 4 m.

Step 1: Find missing side lengths.

The lower right vertical part is:

\(9 - 4 = 5\) m

The inner horizontal part is:

\(8 - 3 = 5\) m

Step 2: Split into rectangles and find area.

One way is to split the L-shape into:

• A left rectangle: \(3 \text{ m} \times 9 \text{ m}\)
• A bottom rectangle: \(5 \text{ m} \times 5 \text{ m}\)

Left rectangle area:

\(3 \times 9 = 27\text{ m}^2\)

Bottom rectangle area:

\(5 \times 5 = 25\text{ m}^2\)

Total area:

$$27 + 25 = 52\text{ m}^2$$

Area of the L-shape: \(52\text{ m}^2\)

Step 3: Find the perimeter.

Add the outside sides:

• Top: 3 m
• Right upper side: 4 m
• Inner horizontal side: 5 m
• Right lower side: 5 m
• Bottom: 8 m
• Left side: 9 m

$$3 + 4 + 5 + 5 + 8 + 9 = 34\text{ m}$$

Perimeter of the L-shape: \(34\text{ m}\)

Worked Example 3: Rectangle with a triangle on top

Suppose a composite figure is made of:

• A rectangle with length 12 cm and width 5 cm
• A triangle on top of the rectangle with base 12 cm and height 4 cm

Step 1: Find the area of the rectangle.

\(A = 12 \times 5 = 60\text{ cm}^2\)

Step 2: Find the area of the triangle.

$$A = \frac{1}{2}bh = \frac{1}{2}(12)(4) = 24\text{ cm}^2$$

Step 3: Add the areas.

$$60 + 24 = 84\text{ cm}^2$$

Area of the composite figure: \(84\text{ cm}^2\)

Step 4: Find the perimeter.

For perimeter, do not count the triangle's base because it is attached to the top of the rectangle inside the figure.

Suppose the two slanted sides of the triangle are each 5 cm. Then the outside edges are:

• Bottom of rectangle: 12 cm
• Left side of rectangle: 5 cm
• Right side of rectangle: 5 cm
• Two slanted sides of triangle: \(5 + 5\) cm

$$12 + 5 + 5 + 5 + 5 = 32\text{ cm}$$

Perimeter of the composite figure: \(32\text{ cm}\)

Worked Example 4: Using subtraction to find area

Sometimes a composite figure is easier to solve by thinking of it as a large shape with a piece missing.

Suppose an L-shaped figure fits inside a rectangle that is 11 ft by 8 ft. A small rectangle measuring 4 ft by 3 ft is missing from one corner.

Step 1: Find the area of the large rectangle.

\(11 \times 8 = 88\text{ ft}^2\)

Step 2: Find the area of the missing rectangle.

\(4 \times 3 = 12\text{ ft}^2\)

Step 3: Subtract.

$$88 - 12 = 76\text{ ft}^2$$

Area of the composite figure: \(76\text{ ft}^2\)

This method works well when the figure is almost a rectangle but has one part cut out.

Tips for solving composite figure problems

• Draw lines to split the figure into smaller shapes.
• Label every side length you know.
• Find missing lengths before calculating area or perimeter.
• Use square units for area, such as \(\text{cm}^2\), \(\text{m}^2\), or \(\text{ft}^2\).
• Use regular units for perimeter, such as cm, m, or ft.
• Check that you did not count any inside edges in the perimeter.

Common mistakes to avoid

• Adding side lengths to find area
• Adding areas to find perimeter
• Counting shared inside sides as part of the perimeter
• Forgetting to find a missing side length
• Writing the wrong units

Quick check questions

1. If a figure is made from two rectangles, what do you do to find the total area?
2. When finding perimeter, do you include the shared side inside the figure?
3. What units should area have?
4. What units should perimeter have?

Answers

1. Find each rectangle's area and add them.
2. No, only outside edges count.
3. Square units, like \(\text{cm}^2\).
4. Linear units, like cm.

Summary

To find the area of a composite figure, split it into simple shapes, find each area, and add them together. To find the perimeter, add only the lengths around the outside edge.

Always look carefully for missing side lengths and for edges that are inside the figure. If you stay organized and work one part at a time, composite figures become much easier to solve.

Circle Anatomy and Pi

Circles are all around us: wheels, coins, clocks, plates, and lids are all shaped like circles. To measure circles, we need to know the names of their parts and how those parts are connected.

In this lesson, you will learn the anatomy of a circle and the meaning of c0 (pi). You will also learn how to use pi to find the circumference, which is the distance around a circle.

1. Parts of a Circle

A circle is the set of all points that are the same distance from one fixed point. That fixed point is called the center.

• Center: the middle point of the circle
• Radius: a line segment from the center to the circle
• Diameter: a line segment that goes across the circle through the center
• Circumference: the distance all the way around the circle

The radius and diameter are closely related. The diameter is always twice the radius.

$$d = 2r$$

That also means the radius is half the diameter.

$$r = \frac{d}{2}$$

2. What Is Pi?

Pi, written as \(\pi\), is a special number that appears in every circle. It compares the circumference of a circle to its diameter.

No matter how big or small the circle is, the ratio stays the same:

$$\pi = \frac{C}{d}$$

This means:

$$C = \pi d$$

Since the diameter is twice the radius, we can also write:

$$C = 2\pi r$$

Both formulas find the circumference. Use the first one if you know the diameter. Use the second one if you know the radius.

3. Exact Answers and Approximate Answers

Pi is an irrational number, which means its decimal goes on forever without repeating. For 7th Grade math, we usually use:

• \(\pi\) for an exact answer
• \(3.14\) for an approximate answer

For example, if the diameter is 8 units, the circumference is:

$$C = \pi d = \pi(8) = 8\pi$$

This is the exact form.

If you want a decimal approximation, use \(\pi \approx 3.14\):

$$C \approx 3.14 \times 8 = 25.12$$

So the circumference is \(8\pi\) units exactly, or about 25.12 units.

4. How to Find Circumference

To find the circumference of a circle:

1. Decide whether you know the radius or the diameter.
2. Choose the correct formula.
3. Substitute the value into the formula.
4. Simplify for an exact answer or multiply by \(3.14\) for an approximate answer.
5. Include units in your final answer.

5. Worked Examples

Example 1: Find the circumference when the diameter is known

A circle has a diameter of 10 cm. Find its circumference.

Use the formula:

$$C = \pi d$$

Substitute \(d = 10\):

$$C = \pi(10) = 10\pi$$

Exact answer: \(10\pi\) cm

Approximate answer:

$$C \approx 3.14 \times 10 = 31.4$$

Approximate circumference: 31.4 cm

Example 2: Find the circumference when the radius is known

A circle has a radius of 7 m. Find its circumference.

Use the formula:

$$C = 2\pi r$$

Substitute \(r = 7\):

$$C = 2\pi(7) = 14\pi$$

Exact answer: \(14\pi\) m

Approximate answer:

$$C \approx 2 \times 3.14 \times 7 = 43.96$$

Approximate circumference: 43.96 m

Example 3: Find the circumference when you must first find the radius

A circle has a diameter of 18 in. Find its circumference using the radius formula.

First find the radius:

$$r = \frac{d}{2} = \frac{18}{2} = 9$$

Now use:

$$C = 2\pi r$$ $$C = 2\pi(9) = 18\pi$$

Exact answer: \(18\pi\) in

Approximate answer:

$$C \approx 2 \times 3.14 \times 9 = 56.52$$

Approximate circumference: 56.52 in

Example 4: Find the diameter from the circumference

A circle has a circumference of \(37.68\) ft. Find the diameter.

Use the formula:

$$C = \pi d$$

Solve for \(d\):

$$d = \frac{C}{\pi}$$

Substitute \(C = 37.68\) and use \(\pi \approx 3.14\):

$$d \approx \frac{37.68}{3.14} = 12$$

The diameter is 12 ft.

6. Common Mistakes to Avoid

• Mixing up radius and diameter: Remember, the diameter is twice the radius.
• Using the wrong formula: If you know diameter, use \(C = \pi d\). If you know radius, use \(C = 2\pi r\).
• Forgetting units: Always write cm, m, in, ft, or whatever unit is given.
• Rounding too early: Keep \(\pi\) in the calculation until the end if possible.

7. Quick Check

Try these on your own:

• A circle has radius 5 cm. What is its circumference in exact form?
• A circle has diameter 16 mm. What is its approximate circumference?
• A circle has circumference 62.8 m. About what is its diameter?

Answers:

• \(C = 2\pi(5) = 10\pi\) cm
• \(C \approx 3.14 \times 16 = 50.24\) mm
• \(d \approx 62.8 \div 3.14 = 20\) m

8. Summary

A circle has important parts: the center, radius, diameter, and circumference. The diameter is twice the radius, and the radius is half the diameter.

Pi, written as \(\pi\), is the constant ratio of a circles circumference to its diameter. You can find circumference with either of these formulas:

$$C = \pi d$$

or

$$C = 2\pi r$$

Use \(\pi\) for exact answers and \(3.14\) when you need an approximate decimal answer. Understanding circle anatomy and pi helps you measure real-world circular objects accurately.

Area of Circles and Sectors

In this lesson, you will learn how to find the area of a circle and the area of a sector. A circle is a flat shape with all points the same distance from its center. A sector is a slice of a circle, like a piece of pizza.

Knowing how to find these areas helps in real life. For example, you might need to find the area of a round garden, a circular table top, or part of a circular track.

1. Review of circle parts

• Radius: the distance from the center of the circle to the edge.
• Diameter: the distance across the circle through the center.
• Sector: a part of the circle made by two radii and the curved edge.
• Semicircle: half of a circle.

The diameter is always twice the radius.

$$d = 2r$$

So if you know the diameter, you can find the radius by dividing by 2.

$$r = \frac{d}{2}$$

2. Area of a full circle

The formula for the area of a circle is:

$$A = \pi r^2$$

Here:

• \(A\) means area
• \(r\) means radius
• \(\pi\) is a special number that is about \(3.14\)

To find the area of a circle:

1. Find the radius.
2. Square the radius, which means multiply it by itself.
3. Multiply by \(\pi\), or use \(3.14\) if needed.
4. Write the answer in square units, like \(cm^2\), \(m^2\), or \(in^2\).

Example 1: Finding the area of a circle

A circle has radius \(5\) cm. Find its area.

Use the formula:

$$A = \pi r^2$$

Substitute \(r = 5\):

$$A = \pi (5)^2$$ $$A = 25\pi$$

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

$$A \approx 3.14 \times 25 = 78.5$$

The area is \(25\pi\, cm^2\) or about \(78.5\, cm^2\).

Important reminder: Do not square \(\pi\). Only the radius is squared.

3. When the diameter is given

Sometimes a problem gives the diameter instead of the radius. Remember to divide by 2 first.

Example 2: Finding area from the diameter

A circle has diameter \(12\) m. Find its area.

First find the radius:

$$r = \frac{12}{2} = 6$$

Now use the area formula:

$$A = \pi r^2$$ $$A = \pi (6)^2$$ $$A = 36\pi$$

Using \(\pi \approx 3.14\):

$$A \approx 3.14 \times 36 = 113.04$$

The area is \(36\pi\, m^2\) or about \(113.04\, m^2\).

4. Area of a sector

A sector is only part of a circle, so its area is only part of the whole area.

To find the area of a sector, use this idea:

$$\text{sector area} = \frac{\text{central angle}}{360} \times \pi r^2$$

The central angle is the angle at the center of the circle. A full circle has \(360^\circ\).

This formula works because the sector is a fraction of the whole circle. For example:

• \(180^\circ\) is half of \(360^\circ\), so it is half the area.
• \(90^\circ\) is one quarter of \(360^\circ\), so it is one quarter of the area.
• \(60^\circ\) is \(\frac{60}{360} = \frac{1}{6}\) of the full area.

Example 3: Finding the area of a sector

A circle has radius \(8\) cm. A sector has a central angle of \(90^\circ\). Find the area of the sector.

Start with the sector formula:

$$\text{sector area} = \frac{90}{360} \times \pi (8)^2$$

Simplify:

$$\frac{90}{360} = \frac{1}{4}$$ $$\text{sector area} = \frac{1}{4} \times \pi \times 64$$ $$\text{sector area} = 16\pi$$

Using \(\pi \approx 3.14\):

$$\text{sector area} \approx 16 \times 3.14 = 50.24$$

The area of the sector is \(16\pi\, cm^2\) or about \(50.24\, cm^2\).

5. Area of a semicircle

A semicircle is half of a circle, so its area is:

$$\text{semicircle area} = \frac{1}{2}\pi r^2$$

You can also think of it as a sector with a central angle of \(180^\circ\):

$$\frac{180}{360} \times \pi r^2 = \frac{1}{2}\pi r^2$$

Example 4: Finding the area of a semicircle

A semicircle has diameter \(14\) in. Find its area.

First find the radius:

$$r = \frac{14}{2} = 7$$

Now use the semicircle formula:

$$\text{area} = \frac{1}{2}\pi (7)^2$$ $$\text{area} = \frac{1}{2}\pi (49)$$ $$\text{area} = 24.5\pi$$

Using \(\pi \approx 3.14\):

$$\text{area} \approx 24.5 \times 3.14 = 76.93$$

The area of the semicircle is \(24.5\pi\, in^2\) or about \(76.93\, in^2\).

6. Steps for solving area problems

• Read carefully to see if the problem gives radius or diameter.
• If diameter is given, divide by 2 to get the radius.
• Use \(A = \pi r^2\) for a full circle.
• Use $$\frac{\text{angle}}{360} \times \pi r^2$$ for a sector.
• Use $$\frac{1}{2}\pi r^2$$ for a semicircle.
• Always include square units in your answer.

7. Common mistakes to avoid

• Using the diameter as the radius: if the problem gives diameter, divide by 2 first.
• Forgetting to square the radius: \(r^2\) means \(r \times r\).
• Using the whole circle formula for a sector: multiply by the fraction \(\frac{\text{angle}}{360}\).
• Forgetting units: area must be in square units.

8. Quick check

Try these on your own:

• A circle has radius \(3\) cm. What is its area?
• A circle has diameter \(10\) m. What is its area?
• A sector has radius \(6\) in and central angle \(120^\circ\). What is its area?
• A semicircle has radius \(4\) ft. What is its area?

Answers:

• \(9\pi\, cm^2\) or about \(28.26\, cm^2\)
• \(25\pi\, m^2\) or about \(78.5\, m^2\)
• \(12\pi\, in^2\) or about \(37.68\, in^2\)
• \(8\pi\, ft^2\) or about \(25.12\, ft^2\)

Summary

The area of a full circle is found using $$A = \pi r^2$$. A sector is part of a circle, so its area is a fraction of the full area: $$\frac{\text{angle}}{360} \times \pi r^2$$. A semicircle is half of a circle, so its area is $$\frac{1}{2}\pi r^2$$.

Always make sure you know the radius, square it, and use the correct fraction for the part of the circle you are finding. With practice, circle and sector area problems become much easier.

Surface Area of Prisms and Cylinders

When we talk about surface area, we mean the total area of all the outside faces of a 3-dimensional object.

Imagine covering a box or a can completely with paper. The amount of paper you need is the surface area.

In this lesson, you will learn how to find the surface area of rectangular prisms, triangular prisms, and cylinders by thinking about their nets.

What is a net?

A net is a flat 2-dimensional drawing that shows all the faces of a 3-dimensional shape opened up.

If you can find the area of each part of the net and add them together, you can find the surface area.

Important idea: Surface area is always measured in square units, such as square centimeters \,\(cm^2\), square meters \,\(m^2\), or square inches \,\(in^2\).

1. Surface Area of a Rectangular Prism

A rectangular prism is a box-shaped solid. It has:

• 6 rectangular faces
• 3 pairs of equal faces

If a rectangular prism has length \,\(l\), width \,\(w\), and height \,\(h\), then its faces are:

• 2 faces with area \,\(lw\)
• 2 faces with area \,\(lh\)
• 2 faces with area \,\(wh\)

So the surface area formula is:

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

You can also write it as:

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

How this connects to the net: When you unfold the prism, you see all 6 rectangles. Add the area of all 6 rectangles.

Example 1: Rectangular Prism

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

Step 1: Use the formula.

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

Step 2: Substitute the values.

$$SA = 2(8 \cdot 3) + 2(8 \cdot 5) + 2(3 \cdot 5)$$

Step 3: Multiply.

$$SA = 2(24) + 2(40) + 2(15)$$ $$SA = 48 + 80 + 30$$

Step 4: Add.

$$SA = 158$$

The surface area is \(158\,cm^2\).

2. Surface Area of a Triangular Prism

A triangular prism has:

• 2 triangular bases
• 3 rectangular side faces

To find the surface area, add:

• the area of both triangles
• the area of all 3 rectangles

This gives the formula:

$$SA = 2B + Ph$$

Here:

• \(B\) = area of one triangular base
• \(P\) = perimeter of the triangular base
• \(h\) = length of the prism

Why does this work?

The two triangular ends give \,\(2B\).

The three rectangles wrap around the prism. Their total area is the same as:

$$\text{perimeter of triangle} \times \text{length of prism}$$

That is \,\(Ph\).

Finding the area of a triangle

Remember, the area of a triangle is:

$$B = \frac{1}{2}bh$$

Here, for the triangle, \,\(b\) is the triangle's base and \,\(h\) is the triangle's height.

Be careful: in a prism problem, the letter \,\(h\) might also be used for the prism's length. Always check what each measurement means.

Example 2: Triangular Prism

A triangular prism has a triangular base with side lengths \,\(6\,cm\), \,\(8\,cm\), and \,\(10\,cm\). The triangle has a base of \,\(8\,cm\) and a height of \,\(6\,cm\). The prism length is \,\(12\,cm\).

Find the surface area.

Step 1: Find the area of one triangular base.

$$B = \frac{1}{2}bh$$ $$B = \frac{1}{2}(8)(6) = 24$$

So one triangle has area \,\(24\,cm^2\).

Step 2: Find the perimeter of the triangular base.

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

Step 3: Use the surface area formula.

$$SA = 2B + Ph$$ $$SA = 2(24) + 24(12)$$

Step 4: Calculate.

$$SA = 48 + 288 = 336$$

The surface area is \(336\,cm^2\).

3. Surface Area of a Cylinder

A cylinder has:

• 2 circular bases
• 1 curved surface around the side

If you unwrap the curved surface, it becomes a rectangle in the net.

What are the dimensions of that rectangle?

• Its height is the cylinder's height
• Its width is the distance around the circle, which is the circumference

The circumference of a circle is:

$$C = 2\pi r$$

So the area of the curved surface is:

$$2\pi r h$$

The two circular bases each have area:

$$\pi r^2$$

So both circles together have area:

$$2\pi r^2$$

Now add the parts:

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

Here:

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

Example 3: Cylinder

Find the surface area of a cylinder with radius \,\(4\,cm\) and height \,\(9\,cm\).

Step 1: Use the formula.

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

Step 2: Substitute.

$$SA = 2\pi (4^2) + 2\pi (4)(9)$$

Step 3: Simplify.

$$SA = 2\pi (16) + 2\pi (36)$$ $$SA = 32\pi + 72\pi$$ $$SA = 104\pi$$

This is the exact answer.

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

$$SA \approx 104(3.14) = 326.56$$

The surface area is \(104\pi\,cm^2\) or about \(326.56\,cm^2\).

4. Using Nets to Understand Surface Area

Nets help you see where the formulas come from instead of just memorizing them.

• A rectangular prism net shows 6 rectangles.
• A triangular prism net shows 2 triangles and 3 rectangles.
• A cylinder net shows 2 circles and 1 rectangle.

If you ever forget a formula, draw or imagine the net and add the areas of all the parts.

5. Common Mistakes to Avoid

• Forgetting a face: Surface area includes all outside surfaces.
• Mixing up area and perimeter: Area is measured in square units. Perimeter is measured in units.
• Using the diameter instead of the radius in a cylinder: The formula uses \,\(r\), not diameter.
• For triangular prisms, forgetting both triangles: You need \,\(2B\), not just \,\(B\).
• Not labeling units: Always write square units like \,\(cm^2\).

Example 4: A Problem That Combines Careful Thinking

A rectangular prism has length \,\(10\,m\), width \,\(4\,m\), and height \,\(2\,m\). Find its surface area.

Step 1: Write the formula.

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

Step 2: Substitute.

$$SA = 2(10 \cdot 4) + 2(10 \cdot 2) + 2(4 \cdot 2)$$

Step 3: Multiply.

$$SA = 2(40) + 2(20) + 2(8)$$ $$SA = 80 + 40 + 16$$

Step 4: Add.

$$SA = 136$$

The surface area is \(136\,m^2\).

Quick Check: Which formula should I use?

• Rectangular prism: $$SA = 2lw + 2lh + 2wh$$
• Triangular prism: $$SA = 2B + Ph$$
• Cylinder: $$SA = 2\pi r^2 + 2\pi rh$$

Steps for Solving Surface Area Problems

1. Identify the solid.
2. Picture or draw its net.
3. Find the area of each part of the net.
4. Add all the areas together.
5. Write the answer in square units.

Summary

Surface area is the total area on the outside of a 3-dimensional figure. Nets help us unfold a solid into flat shapes so we can add their areas.

For a rectangular prism, add the areas of 6 rectangles. For a triangular prism, add 2 triangles and 3 rectangles. For a cylinder, add 2 circles and the rectangle made by the curved surface.

If you understand the net, the formula makes sense.

Lesson: Volume of Prisms and Cylinders

When we talk about volume, we mean the amount of space inside a 3-dimensional object. Volume tells us how much a shape can hold.

For example, volume can describe how much water fits in a bottle, how much air is inside a room, or how much space is inside a box.

Volume is measured in cubic units. If the measurements are in centimeters, the volume will be in cubic centimeters, written as \(cm^3\). If the measurements are in meters, the volume will be in \(m^3\).

What are prisms and cylinders?

A prism is a solid shape with two matching, parallel bases that are polygons. The sides connect the bases. Examples include rectangular prisms and triangular prisms.

A cylinder is a solid shape with two matching, parallel circular bases connected by a curved surface.

The key idea for both prisms and cylinders is the same:

Volume = area of the base × height

In math symbols, this is:

$$V = Bh$$

Here:

• \(V\) = volume
• \(B\) = area of the base
• \(h\) = height of the solid

The height is the distance from one base straight to the other base. It is not always a side slanting across the shape.

Why does this formula work?

Imagine stacking many thin layers of the same base shape on top of each other. Each layer has the same area. The number of layers depends on the height. So the total space inside is the base area multiplied by the height.

This is why both prisms and cylinders use the same idea: find the area of the base, then multiply by the height.

Volume of a Rectangular Prism

A rectangular prism has a rectangular base. Since the area of a rectangle is length × width, the volume formula becomes:

$$V = lwh$$

Here:

• \(l\) = length
• \(w\) = width
• \(h\) = height

Worked Example 1: Rectangular Prism

Find the volume of a rectangular prism with length 8 cm, width 3 cm, and height 5 cm.

Use the formula:

$$V = lwh$$

Substitute the values:

$$V = 8 \times 3 \times 5$$ $$V = 24 \times 5 = 120$$

So the volume is:

$$120\;cm^3$$

Volume of Any Prism

Not all prisms have rectangular bases. Some have triangular bases or other polygon bases. The rule stays the same:

$$V = Bh$$

First find the area of the base, then multiply by the prism's height.

Worked Example 2: Triangular Prism

A triangular prism has a triangular base with base 6 m and triangle height 4 m. The prism height is 10 m. Find the volume.

Step 1: Find the area of the triangular base.

$$B = \frac{1}{2}bh$$

For the triangle:

$$B = \frac{1}{2}(6)(4) = 12$$

So the base area is \(12\;m^2\).

Step 2: Multiply by the prism height.

$$V = Bh$$ $$V = 12 \times 10 = 120$$

So the volume is:

$$120\;m^3$$

Important: In this example, there are two different heights.

• The triangle's height is used to find the area of the base.
• The prism's height is used to find the volume.

Be careful not to mix them up.

Volume of a Cylinder

A cylinder has a circular base. So we first find the area of the circle:

$$B = \pi r^2$$

Then multiply by the height:

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

Here:

• \(r\) = radius of the circular base
• \(h\) = height of the cylinder
• \(\pi\) is about \(3.14\)

Worked Example 3: Cylinder

Find the volume of a cylinder with radius 4 cm and height 9 cm.

Use the formula:

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

Substitute the values:

$$V = \pi (4)^2(9)$$ $$V = \pi (16)(9)$$ $$V = 144\pi$$

If using \(\pi \approx 3.14\):

$$V \approx 144(3.14) = 452.16$$

So the volume is:

Exact form: \(144\pi\;cm^3\)

Approximate form: \(452.16\;cm^3\)

How to Find Volume Step by Step

1. Identify the solid: prism or cylinder.
2. Find the area of the base.
3. Find the height of the solid.
4. Multiply base area by height.
5. Write the answer in cubic units.

Worked Example 4: A Problem with Mixed Information

A cylinder has diameter 10 in and height 7 in. Find its volume.

Be careful: the formula needs radius, not diameter.

Step 1: Find the radius.

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

Step 2: Use the volume formula.

$$V = \pi r^2 h$$ $$V = \pi (5)^2(7)$$ $$V = \pi (25)(7) = 175\pi$$

If using \(\pi \approx 3.14\):

$$V \approx 175(3.14) = 549.5$$

So the volume is:

Exact form: \(175\pi\;in^3\)

Approximate form: \(549.5\;in^3\)

Common Mistakes to Avoid

• Forgetting cubic units: Volume must be written in units like \(cm^3\), \(m^3\), or \(in^3\).
• Using the wrong measurement for height: The height is the distance between the two bases.
• Using diameter instead of radius in a cylinder: Always check whether you need to divide by 2.
• Not finding the base area first: In prisms and cylinders, volume comes from base area × height.
• Mixing up area and volume: Area is measured in square units, but volume is measured in cubic units.

Quick Check

• If a prism has base area \(15\;cm^2\) and height \(6\;cm\), then \(V = 15 \times 6 = 90\;cm^3\).
• If a cylinder has radius \(3\;m\) and height \(5\;m\), then \(V = \pi (3)^2(5) = 45\pi\;m^3\).

Summary

The main rule for both prisms and cylinders is:

$$V = Bh$$

This means volume = area of the base × height.

For a rectangular prism, you can also use:

$$V = lwh$$

For a cylinder, use:

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

Always find the base area carefully, use the correct height, and write your answer in cubic units.

Dimensional Change and Scale Factors

When a shape is enlarged or reduced, its measurements do not all change in the same way. A scale factor tells how much each length changes.

For example, if every side length is multiplied by 2, the scale factor is 2. If every side length is multiplied by \(\frac{1}{2}\), the scale factor is \(\frac{1}{2}\).

In this lesson, you will learn an important pattern:

• Lengths change by the scale factor.
• Areas change by the square of the scale factor.
• Volumes change by the cube of the scale factor.

This idea helps us predict what happens to perimeter, area, surface area, and volume when figures are resized.

1. What is a scale factor?

A scale factor compares the new measurement to the original measurement.

$$\text{Scale factor} = \frac{\text{new length}}{\text{original length}}$$

If a rectangle has a side of 4 cm and the matching side in a larger copy is 12 cm, then the scale factor is

$$\frac{12}{4} = 3$$

That means every length in the new figure is 3 times the original.

2. How one-dimensional measures change

Measurements like side length, height, width, radius, and perimeter are one-dimensional. They are based on length.

If the scale factor is \(k\), then each length is multiplied by \(k\).

Perimeter also changes by \(k\) because perimeter is the total of side lengths.

For example, if a triangle is enlarged by a scale factor of 4, then:

• each side becomes 4 times as long,
• the perimeter becomes 4 times as large.

3. How area changes

Area is a two-dimensional measurement. It depends on two lengths, such as length and width.

If both dimensions are multiplied by the scale factor \(k\), then the area is multiplied by \(k^2\).

$$\text{New area} = \text{original area} \times k^2$$

Why does this happen? Think about a rectangle.

If the original area is

$$A = l \times w$$

and both dimensions are scaled by \(k\), then the new area is

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

$$A' = k^2A$$

So if the scale factor doubles, the area does not just double. It becomes \(2^2 = 4\) times as large.

4. How volume changes

Volume is a three-dimensional measurement. It depends on three lengths, such as length, width, and height.

If all dimensions are multiplied by the scale factor \(k\), then the volume is multiplied by \(k^3\).

$$\text{New volume} = \text{original volume} \times k^3$$

For a rectangular prism,

$$V = lwh$$

After scaling by \(k\), the new volume is

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

$$V' = k^3V$$

So if the scale factor is 2, the volume becomes \(2^3 = 8\) times as large.

5. Surface area also follows the area rule

Surface area measures the total area covering the outside of a 3D figure. Since surface area is made of areas of faces, it changes by the square of the scale factor.

$$\text{New surface area} = \text{original surface area} \times k^2$$

So for 3D figures:

• lengths change by \(k\),
• surface area changes by \(k^2\),
• volume changes by \(k^3\).

6. Important patterns to remember

• If the scale factor is 2, area becomes \(2^2 = 4\) times as large, and volume becomes \(2^3 = 8\) times as large.
• If the scale factor is 3, area becomes \(3^2 = 9\) times as large, and volume becomes \(3^3 = 27\) times as large.
• If the scale factor is \(\frac{1}{2}\), area becomes \(\left(\frac{1}{2}\right)^2 = \frac{1}{4}\), and volume becomes \(\left(\frac{1}{2}\right)^3 = \frac{1}{8}\).

This means shrinking a figure has a big effect too.

7. Worked Examples

Example 1: Finding the new area of a rectangle

A rectangle has an area of \(18\text{ cm}^2\). It is enlarged by a scale factor of 2. What is the new area?

Step 1: Use the area rule.

$$\text{new area} = 18 \times 2^2$$

Step 2: Square the scale factor.

$$2^2 = 4$$

Step 3: Multiply.

$$18 \times 4 = 72$$

Answer: The new area is \(72\text{ cm}^2\).

Example 2: Finding the new volume of a prism

A rectangular prism has a volume of \(30\text{ cm}^3\). It is enlarged by a scale factor of 3. What is the new volume?

Step 1: Use the volume rule.

$$\text{new volume} = 30 \times 3^3$$

Step 2: Cube the scale factor.

$$3^3 = 27$$

Step 3: Multiply.

$$30 \times 27 = 810$$

Answer: The new volume is \(810\text{ cm}^3\).

Example 3: Working backward from area change

A figure is enlarged, and its area becomes 9 times the original area. What is the scale factor?

Step 1: Area changes by \(k^2\).

$$k^2 = 9$$

Step 2: Find the number whose square is 9.

$$k = 3$$

Answer: The scale factor is 3.

Example 4: Shrinking a cube

A cube has a volume of \(64\text{ in}^3\). Its side lengths are reduced by a scale factor of \(\frac{1}{2}\). What is the new volume?

Step 1: Use the volume rule.

$$\text{new volume} = 64 \times \left(\frac{1}{2}\right)^3$$

Step 2: Cube the scale factor.

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

Step 3: Multiply.

$$64 \times \frac{1}{8} = 8$$

Answer: The new volume is \(8\text{ in}^3\).

8. A quick way to decide what power to use

You can match the type of measurement to the number of dimensions:

• 1-dimensional measurement: use \(k\)
• 2-dimensional measurement: use \(k^2\)
• 3-dimensional measurement: use \(k^3\)

This leads to these rules:

• Perimeter changes by \(k\)
• Area changes by \(k^2\)
• Surface area changes by \(k^2\)
• Volume changes by \(k^3\)

9. Common mistakes to avoid

• Mistake: Multiplying area by the scale factor instead of squaring it.
  If the scale factor is 4, area changes by \(4^2 = 16\), not 4.
• Mistake: Multiplying volume by the scale factor instead of cubing it.
  If the scale factor is 2, volume changes by \(2^3 = 8\), not 2.
• Mistake: Forgetting that shrinking uses fractions.
  If the scale factor is \(\frac{1}{3}\), then area changes by \(\frac{1}{9}\) and volume changes by \(\frac{1}{27}\).

10. Summary

When a figure is resized, lengths change by the scale factor. Areas and surface areas change faster because they depend on two dimensions, so they change by the square of the scale factor. Volumes change even faster because they depend on three dimensions, so they change by the cube of the scale factor.

Always ask yourself: Is this measurement 1D, 2D, or 3D? Then use \(k\), \(k^2\), or \(k^3\).