Two-Dimensional Geometry, Area, and Perimeter

Area of Parallelograms and Triangles Review

In this lesson, you will review how to find the area of parallelograms and triangles. You will also learn how to use the area formulas to find a missing base or missing height.

Area tells how much space is inside a flat shape. We measure area in square units, such as square centimeters \\(cm^2\\), square meters \\(m^2\\), or square inches \\(in^2\\).

When finding area, it is very important to use the base and the height. The height is the perpendicular distance, which means it makes a right angle with the base. A slanted side is not the height unless it forms a right angle with the base.

1. Area of a Parallelogram

A parallelogram is a four-sided figure with two pairs of opposite sides that are parallel. To find its area, multiply the base by the height.

Formula:

$$A = bh$$

where:

• \\(A\\) = area
• \\(b\\) = base
• \\(h\\) = height

Even if a parallelogram is slanted, the formula is still the same. What matters is using the correct height, not the length of the slanted side.

2. Area of a Triangle

A triangle covers half the area of a parallelogram with the same base and height. That is why the triangle formula includes \\(\\).

Formula:

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

where:

• \\(A\\) = area
• \\(b\\) = base
• \\(h\\) = height

Just like with parallelograms, the height must go straight up and down, or straight across, to make a right angle with the base.

3. Choosing the Correct Height

One common mistake is using a side length instead of the height. Remember:

• The base can often be any side you choose.
• The height must match that base.
• The height forms a right angle with the base.

If the figure is tilted, the height may be drawn inside or outside the shape. That is okay. As long as it is perpendicular to the base, it can be used in the formula.

4. Finding a Missing Base or Height

Sometimes you know the area and one measurement, and you need to find the missing base or height. To do this, substitute the known values into the formula and solve the equation.

For a parallelogram:

$$A = bh$$

If you know \\(A\\) and \\(b\\), then

$$h = \frac{A}{b}$$

If you know \\(A\\) and \\(h\\), then

$$b = \frac{A}{h}$$

For a triangle:

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

If you know the area and one measurement, solve carefully. You may need to multiply both sides by 2 first.

For example, if \\(A = \frac{1}{2}bh\\), then:

$$2A = bh$$

From there, divide by the known base or height to find the missing value.

Worked Example 1: Find the area of a parallelogram

A parallelogram has a base of 9 cm and a height of 4 cm. Find the area.

Use the formula:

$$A = bh$$

Substitute the values:

$$A = 9 \cdot 4$$ $$A = 36$$

The area is 36 square centimeters, or \\(36\,cm^2\\).

Worked Example 2: Find the area of a triangle

A triangle has a base of 12 m and a height of 5 m. Find the area.

Use the formula:

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

Substitute the values:

$$A = \frac{1}{2}(12)(5)$$ $$A = 6 \cdot 5$$ $$A = 30$$

The area is 30 square meters, or \\(30\,m^2\\).

Worked Example 3: Find a missing height in a parallelogram

A parallelogram has an area of \\(54\,in^2\\) and a base of 6 in. Find the height.

Start with the formula:

$$A = bh$$

Substitute what you know:

$$54 = 6h$$

Now divide both sides by 6:

$$h = \frac{54}{6} = 9$$

The height is 9 inches.

Worked Example 4: Find a missing base in a triangle

A triangle has an area of \\(40\,ft^2\\) and a height of 5 ft. Find the base.

Use the triangle formula:

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

Substitute the known values:

$$40 = \frac{1}{2}b(5)$$

Simplify the right side:

$$40 = \frac{5}{2}b$$

Now solve for \\(b\\). One way is to multiply both sides by 2:

$$80 = 5b$$

Then divide by 5:

$$b = 16$$

The base is 16 feet.

Tips to Remember

• Parallelogram area: \\(A = bh\\)
• Triangle area: \\(A = \frac{1}{2}bh\\)
• Always use the height that is perpendicular to the base.
• Include square units in your final area answer.
• If finding a missing value, write an equation and solve step by step.

Common Mistakes

• Using a slanted side instead of the height
• Forgetting the \\(\\) in the triangle formula
• Leaving off units
• Writing units like cm instead of \\(cm^2\\) for area

Quick Check for Yourself

1. A parallelogram has base 11 cm and height 3 cm. What is its area?
2. A triangle has base 14 in and height 6 in. What is its area?
3. A parallelogram has area \\(72\,m^2\\) and height 8 m. What is its base?
4. A triangle has area \\(24\,yd^2\\) and base 8 yd. What is its height?

Answers

1. \\(33\,cm^2\\)
2. \\(42\,in^2\\)
3. 9 m
4. 6 yd

Summary

To find the area of a parallelogram, multiply the base by the height. To find the area of a triangle, multiply the base by the height and then divide by 2.

If you know the area and one measurement, you can use the formula to solve for the missing base or height. Always make sure the height is perpendicular to the base, and remember to label area in square units.

Area of Trapezoids and Kites

In geometry, area tells how much space is covered inside a flat shape. We measure area in square units, such as square centimeters \,\((cm^2)\) or square meters \,\((m^2)\).

In this lesson, you will learn how to find the area of two important shapes: trapezoids and kites. We will also see why the formulas work by breaking the shapes into simpler parts.

Understanding where formulas come from helps you remember them and use them correctly. That is especially helpful when problems look different from the examples you have seen before.

1. Area of a Trapezoid

A trapezoid is a quadrilateral with one pair of parallel sides. The parallel sides are called the bases. The distance straight up and down between the bases is called the height.

The area formula for a trapezoid is:

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

Here:

• \(b_1\) and \(b_2\) are the lengths of the two bases
• \(h\) is the height

You can think of this formula as: average of the bases, then multiply by the height.

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

Why does this formula make sense?

Imagine placing two matching trapezoids together to make a parallelogram. The new shape has base \,\((b_1+b_2)\) and height \,\(h\). The area of that parallelogram is:

$$ (b_1+b_2)h $$

Since the parallelogram is made of two equal trapezoids, one trapezoid has half that area:

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

Important reminder: The height is not just a slanted side. The height must be the perpendicular distance between the two bases.

2. Worked Example: Trapezoid

Find the area of a trapezoid with bases 10 cm and 6 cm, and height 4 cm.

Step 1: Write the formula.

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

Step 2: Substitute the values.

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

Step 3: Simplify.

$$A = \frac{1}{2}(16)(4)$$ $$A = 8 \cdot 4$$ $$A = 32$$

Answer: The area is 32 \(cm^2\).

3. Another Way to Understand Trapezoid Area

You can also break a trapezoid into simpler shapes, such as a rectangle and one or two triangles. Then you add the areas of those parts.

This method is useful when you want to see how the formula connects to shapes you already know well.

For example, if a trapezoid has bases 12 and 8, with height 5, you could think of it as:

• a rectangle with width 8 and height 5
• plus two small triangles whose combined extra base length is \(12-8=4\)

The rectangle area is:

$$8 \cdot 5 = 40$$

The two triangles together have total base 4 and height 5, so their combined area is:

$$\frac{1}{2}(4)(5)=10$$

Total area:

$$40+10=50$$

This matches the trapezoid formula:

$$A=\frac{1}{2}(12+8)(5)=\frac{1}{2}(20)(5)=50$$

4. Area of a Kite

A kite is a quadrilateral with two pairs of equal side lengths. The equal sides are next to each other, not opposite each other.

To find the area of a kite, we use its diagonals. A diagonal is a segment connecting opposite vertices.

The area formula for a kite is:

$$A = \frac{1}{2}d_1d_2$$

Here:

• \(d_1\) is the length of one diagonal
• \(d_2\) is the length of the other diagonal

Why does this formula work?

The diagonals of a kite divide it into triangles. In many kites, one diagonal splits the other into two equal parts, and the diagonals meet at right angles. This creates smaller right triangles inside the kite.

If you put the pieces together, the total area becomes half the product of the diagonals:

$$A = \frac{1}{2}d_1d_2$$

This is similar to the area formula for some other diagonal-based shapes, but for a kite you should focus on using the two diagonals.

5. Worked Example: Kite

Find the area of a kite with diagonals 14 cm and 10 cm.

Step 1: Write the formula.

$$A = \frac{1}{2}d_1d_2$$

Step 2: Substitute the values.

$$A = \frac{1}{2}(14)(10)$$

Step 3: Simplify.

$$A = \frac{1}{2}(140)$$ $$A = 70$$

Answer: The area is 70 \(cm^2\).

6. Worked Example: Finding a Missing Height in a Trapezoid

A trapezoid has area 54 square meters. Its bases are 7 m and 11 m. Find the height.

Step 1: Start with the formula.

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

Step 2: Substitute what you know.

$$54 = \frac{1}{2}(7+11)h$$

Step 3: Simplify inside the parentheses.

$$54 = \frac{1}{2}(18)h$$ $$54 = 9h$$

Step 4: Solve for \(h\).

$$h = \frac{54}{9} = 6$$

Answer: The height is 6 m.

7. Worked Example: Finding a Missing Diagonal in a Kite

A kite has area 96 square inches. One diagonal is 12 in. Find the other diagonal.

Step 1: Use the formula.

$$A = \frac{1}{2}d_1d_2$$

Step 2: Substitute known values.

$$96 = \frac{1}{2}(12)(d_2)$$

Step 3: Simplify.

$$96 = 6d_2$$

Step 4: Solve.

$$d_2 = \frac{96}{6} = 16$$

Answer: The other diagonal is 16 in.

8. Common Mistakes to Avoid

• Using a slanted side as the height in a trapezoid. The height must be perpendicular to the bases.
• Forgetting to add both bases in the trapezoid formula.
• Forgetting the \(\frac{1}{2}\) in both formulas.
• Using side lengths instead of diagonals for a kite. The kite formula needs diagonals, not the outside sides.
• Forgetting square units in the final answer.

9. Quick Check: Which Formula Do I Use?

• If the shape is a trapezoid, use:

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

• If the shape is a kite, use:

$$A = \frac{1}{2}d_1d_2$$

10. Strategy for Solving Area Problems

1. Identify the shape.
2. Write the correct formula.
3. Check which measurements are needed.
4. Substitute carefully.
5. Solve step by step.
6. Write the answer with square units.

11. Brief Summary

The area of a trapezoid is found by averaging the two bases and multiplying by the height:

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

The area of a kite is found by multiplying the diagonals and taking half:

$$A = \frac{1}{2}d_1d_2$$

Both formulas can be understood by breaking the shapes into simpler parts. When you understand how a formula is built, it becomes much easier to remember and use correctly.

Anatomy of a Circle

A circle is a very important shape in geometry. You see circles in wheels, clocks, coins, pizzas, and rings. To understand circles well, you need to know the names of their parts and how those parts are connected.

In this lesson, you will learn the main parts of a circle: the center, radius, diameter, chord, arc, and circumference. You will also learn the relationships between these parts.

1. What is a circle?

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

If point O is the center, then every point on the circle is the same distance from O.

2. The center

The center is the middle point of the circle. It is usually labeled with a capital letter, such as O.

The center is important because many other parts of the circle are described using it.

3. Radius

A radius is a line segment from the center of the circle to any point on the circle.

If the center is O and A is a point on the circle, then segment OA is a radius.

All radii in the same circle have the same length.

If the radius is represented by the letter r, then that length tells you how far the edge of the circle is from the center.

4. Diameter

A diameter is a line segment that goes from one side of the circle to the other side through the center.

The diameter is the longest chord in a circle because it passes through the center.

If the diameter is represented by d and the radius is r, then:

$$d = 2r$$

and

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

This means the diameter is always twice the radius.

5. Chord

A chord is a line segment with both endpoints on the circle.

A chord does not have to go through the center. If it does go through the center, then it is a diameter.

So, every diameter is a chord, but not every chord is a diameter.

6. Arc

An arc is a part of the circle's edge. It is a curved section of the circle between two points.

For example, if points A and B are on the circle, then the curved part from A to B is an arc.

Arcs can be short or long. A smaller curved part is called a minor arc, and a larger curved part is called a major arc. In 8th Grade, it is most important to understand that an arc is simply a curved part of the circle.

7. Circumference

The circumference is the total distance around the circle. It is like the perimeter of a circle.

The formula for circumference is:

$$C = 2\pi r$$

Because the diameter is twice the radius, the circumference can also be written as:

$$C = \pi d$$

Here, \(\pi\) is a special number that is about \(3.14\).

8. Relationships between circle parts

The parts of a circle are connected in important ways.

• Center to edge gives a radius.
• Edge to edge through the center gives a diameter.
• Edge to edge without needing the center gives a chord.
• Curved part of the edge gives an arc.
• Entire distance around gives the circumference.

The most important number relationship is:

$$d = 2r$$

This means:

• If you know the radius, multiply by 2 to get the diameter.
• If you know the diameter, divide by 2 to get the radius.

You can then use either one to find the circumference:

$$C = 2\pi r \quad \text{or} \quad C = \pi d$$

9. How to identify parts of a circle

When looking at a diagram, ask yourself these questions:

1. Is the point in the middle? If yes, it is the center.
2. Does the segment go from the center to the circle? If yes, it is a radius.
3. Does the segment go across the circle through the center? If yes, it is a diameter.
4. Does the segment connect two points on the circle? If yes, it is a chord.
5. Is it a curved piece of the circle? If yes, it is an arc.
6. Is it the whole distance around the circle? If yes, it is the circumference.

10. Worked Examples

Example 1: Find the diameter from the radius

A circle has radius \(6\) cm. What is its diameter?

Use the relationship:

$$d = 2r$$

Substitute \(r = 6\):

$$d = 2(6) = 12$$

Answer: The diameter is 12 cm.

Example 2: Find the radius from the diameter

A circle has diameter \(18\) inches. What is its radius?

Use the relationship:

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

Substitute \(d = 18\):

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

Answer: The radius is 9 inches.

Example 3: Find the circumference using the radius

A circle has radius \(5\) m. Find the circumference.

Use the formula:

$$C = 2\pi r$$

Substitute \(r = 5\):

$$C = 2\pi(5) = 10\pi$$

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

$$C \approx 10(3.14) = 31.4$$

Answer: The circumference is \(10\pi\) m, or about 31.4 m.

Example 4: Name the parts of a circle

Suppose a circle has center O. Point A and point B are on the circle.

• Segment OA goes from the center to the circle.
• Segment AB connects two points on the circle.
• The curved part from A to B is shown.

What is each part called?

• OA is a radius.
• AB is a chord.
• The curved part from A to B is an arc.

If segment AB also passed through the center O, then AB would be a diameter.

11. Common mistakes to avoid

• Mixing up radius and diameter: Remember, the diameter is twice the radius.
• Thinking every chord is a diameter: Only chords that pass through the center are diameters.
• Confusing arc and chord: An arc is curved, but a chord is straight.
• Forgetting circumference is the whole way around: It is the circle's perimeter.

12. Quick review

• Center: the middle point of the circle
• Radius: from center to circle
• Diameter: across the circle through the center
• Chord: straight segment connecting two points on the circle
• Arc: curved part of the circle
• Circumference: total distance around the circle

Key formulas:

$$d = 2r$$

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

$$C = 2\pi r$$

$$C = \pi d$$

Summary

The anatomy of a circle is about knowing the names and meanings of its parts. The radius goes from the center to the edge, the diameter goes across the circle through the center, and the circumference is the distance around the circle. A chord connects two points on the circle, and an arc is a curved part of the circle. Understanding these parts helps you describe circles clearly and solve geometry problems correctly.

Lesson: Circumference and Pi

When you measure around a circle, you are finding its circumference. Circumference is the distance all the way around the outside of a circle.

In this lesson, you will learn what pi means, how it connects to circles, and how to use formulas to find circumference. You will also learn when to give an exact answer using \(\pi\), and when to give an approximate answer using a decimal like \(3.14\).

1. Important circle parts

Before finding circumference, it helps to know two important measurements in a circle.

• Radius: the distance from the center of the circle to the edge.
• Diameter: the distance across the circle through the center.

The diameter is always twice the radius.

$$d = 2r$$

And the radius is half the diameter.

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

2. What is pi?

For every circle, if you divide the circumference by the diameter, you get the same number. That number is called pi, written as \(\pi\).

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

This means the ratio of circumference to diameter is always constant for all circles.

Pi is an irrational number, which means its decimal goes on forever without repeating. In 8th grade, you will usually use:

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

A common approximation is:

$$\pi \approx 3.14$$

3. Circumference formulas

Since \(\pi = \frac{C}{d}\), multiplying both sides by \(d\) gives the formula for circumference:

$$C = \pi d$$

Because \(d = 2r\), you can also write the formula as:

$$C = 2\pi r$$

So there are two correct circumference formulas:

• $$C = \pi d$$
• $$C = 2\pi r$$

Use \(C = \pi d\) when you know the diameter.

Use \(C = 2\pi r\) when you know the radius.

4. Exact answers and approximate answers

Sometimes a problem asks for the exact circumference. That means you should leave your answer in terms of \(\pi\).

For example, if the diameter is 8 units:

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

The exact answer is \(8\pi\) units.

If the problem asks for an approximation, replace \(\pi\) with \(3.14\):

$$C \approx 8(3.14) = 25.12$$

The approximate answer is 25.12 units.

5. Worked Examples

Example 1: Find the circumference when the diameter is given

A circle has diameter \(10\) cm. Find the circumference exactly and approximately.

Step 1: Use the formula \(C = \pi d\).

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

Exact answer: \(10\pi\) cm

Step 2: Approximate using \(\pi \approx 3.14\).

$$C \approx 10(3.14) = 31.4$$

Approximate answer: \(31.4\) cm

Example 2: Find the circumference when the radius is given

A circle has radius \(7\) m. Find the circumference exactly and approximately.

Step 1: Use the formula \(C = 2\pi r\).

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

Exact answer: \(14\pi\) m

Step 2: Approximate using \(\pi \approx 3.14\).

$$C \approx 14(3.14) = 43.96$$

Approximate answer: \(43.96\) m

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

A circle has radius \(4.5\) in. Find the circumference approximately.

Step 1: Find the diameter.

$$d = 2r = 2(4.5) = 9$$

Step 2: Use \(C = \pi d\).

$$C = \pi(9) = 9\pi$$

Step 3: Approximate.

$$C \approx 9(3.14) = 28.26$$

Approximate answer: \(28.26\) in

You could also have used \(C = 2\pi r\):

$$C = 2\pi(4.5) = 9\pi$$

Both methods give the same result.

Example 4: Find the missing measurement

The circumference of a circle is \(18.84\) ft. Find the diameter.

Step 1: Start with the formula \(C = \pi d\).

$$18.84 = 3.14d$$

Step 2: Divide both sides by \(3.14\).

$$d = \frac{18.84}{3.14} = 6$$

Answer: The diameter is \(6\) ft.

If needed, you could then find the radius:

$$r = \frac{d}{2} = \frac{6}{2} = 3$$

6. Common mistakes to avoid

• Mixing up radius and diameter: Remember, diameter is twice the radius.
• Using the wrong formula: If you know the radius, use \(C = 2\pi r\). If you know the diameter, use \(C = \pi d\).
• Forgetting units: Circumference is a length, so use units like cm, m, or inches.
• Rounding too early: Keep \(\pi\) in the calculation until the end for a more accurate approximation.
• Confusing circumference with area: Circumference is the distance around a circle, not the space inside it.

7. Quick check idea

If the diameter of a circle is about \(5\), the circumference should be a little more than \(15\), because:

$$\pi \cdot 5 \approx 3.14 \cdot 5 = 15.7$$

This helps you decide whether your answer makes sense.

8. Summary

The circumference of a circle is the distance around it. The number pi, written as \(\pi\), is the constant ratio of circumference to diameter.

Use these formulas to find circumference:

• $$C = \pi d$$
• $$C = 2\pi r$$

Leave answers in terms of \(\pi\) for exact answers, and use \(\pi \approx 3.14\) for approximate answers. Always check whether the problem gives the radius or the diameter before choosing your formula.

Area of a Circle

In this lesson, you will learn how to find the area of a circle. Area tells us how much space is inside a shape. For a circle, we use a special formula that involves the radius and the number .

A circle is different from shapes like rectangles and triangles because it has no corners or straight sides. Even so, we can still find its area using a pattern that mathematicians discovered.

By the end of this lesson, you should be able to:

• identify the radius and diameter of a circle,
• use the formula for area of a circle,
• solve problems when given the radius or diameter,
• understand why the formula is \(A = \pi r^2\).

1. Important parts of a circle

Before finding area, lets review two important parts of a circle.

• Radius: the distance from the center of the circle to the edge.
• Diameter: the distance across the circle through the center.

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. The formula for area of a circle

The formula for the area of a circle is:

$$A = \pi r^2$$

In this formula:

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

The expression \(r^2\) means radius radius.

So if the radius is 5 cm, then:

$$r^2 = 5^2 = 25$$

3. Why does the formula work?

One way to understand the formula is to imagine cutting a circle into many thin equal slices, like pieces of a pizza.

If you rearrange those slices by putting some up and some down, the shape starts to look like a parallelogram.

The more slices you make, the closer the rearranged shape gets to a parallelogram.

In that rearranged shape:

• the height is about the radius, \(r\),
• the base is about half the circumference, which is \(\pi r\).

Since area of a parallelogram is base  height, we get:

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

This is why the area formula for a circle is \(A = \pi r^2\).

4. Steps for finding the area of a circle

1. Find the radius.
2. Square the radius: \(r^2\).
3. Multiply by \(\pi\).
4. Write the answer in square units, such as \(cm^2\), \(m^2\), or \(in^2\).

5. Worked Examples

Example 1: Find the area when the radius is given

A circle has radius \(4\) cm. Find its area.

Step 1: Write the formula.

$$A = \pi r^2$$

Step 2: Substitute \(r = 4\).

$$A = \pi(4^2)$$

Step 3: Square the radius.

$$A = \pi(16) = 16\pi$$

Step 4: Write an exact answer or an approximate answer.

Exact form: \(16\pi \text{ cm}^2\)

Approximate form: $$16 \times 3.14 = 50.24$$

So the area is about \(50.24 \text{ cm}^2\).

Example 2: Find the area when the diameter is given

A circle has diameter \(10\) m. Find its area.

Step 1: Find the radius.

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

Step 2: Use the area formula.

$$A = \pi r^2$$

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

Step 3: Approximate using \(\pi \approx 3.14\).

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

So the area is \(25\pi \text{ m}^2\), or about \(78.5 \text{ m}^2\).

Example 3: A word problem

A circular garden has radius \(7\) ft. How much ground does it cover?

This asks for the area of the garden.

Use the formula:

$$A = \pi r^2$$

Substitute \(r = 7\):

$$A = \pi(7^2) = 49\pi$$

Approximate:

$$A \approx 49 \times 3.14 = 153.86$$

So the garden covers about \(153.86 \text{ ft}^2\).

Example 4: Find the radius from the area

The area of a circle is \(154 \text{ cm}^2\). Find the radius. Use \(\pi \approx 3.14\).

Start with the formula:

$$A = \pi r^2$$

Substitute the known area:

$$154 = 3.14r^2$$

Divide both sides by \(3.14\):

$$r^2 = \frac{154}{3.14} \approx 49.04$$

Take the square root:

$$r \approx 7$$

So the radius is about \(7\) cm.

6. Common mistakes to avoid

• Using the diameter instead of the radius: The formula uses \(r\), not \(d\). If you are given the diameter, divide by 2 first.
• Forgetting to square the radius: \(\pi r^2\) does not mean \(\pi \times r \times 2\). It means \(\pi \times r \times r\).
• Mixing up area and circumference: Area measures space inside the circle. Circumference measures distance around the circle.
• Forgetting square units: Area must be written in units like \(cm^2\) or \(m^2\).

7. Quick check

Try these on your own:

• A circle has radius \(3\) in. What is its area?
• A circle has diameter \(12\) cm. What is its area?
• A circular rug has radius \(2.5\) m. How much floor space does it cover?

Answers:

• \(A = 9\pi \text{ in}^2 \approx 28.26 \text{ in}^2\)
• Radius \(= 6\), so \(A = 36\pi \text{ cm}^2 \approx 113.04 \text{ cm}^2\)
• \(A = \pi(2.5)^2 = 6.25\pi \text{ m}^2 \approx 19.625 \text{ m}^2\)

8. Summary

The area of a circle tells us how much space is inside it. To find it, use the formula $$A = \pi r^2$$ where \(r\) is the radius.

If you are given the diameter, divide by 2 first to get the radius. Then square the radius, multiply by \(\pi\), and write your answer in square units.

Remember: the formula works because a circle can be cut into slices and rearranged into a shape that looks like a parallelogram with base \(\pi r\) and height \(r\).

Area of Composite Figures

A composite figure is a shape made by joining two or more simple shapes together. These simple shapes are usually rectangles, triangles, parallelograms, trapezoids, or parts of circles such as semicircles.

To find the area of a composite figure, we do not need one new formula. Instead, we break the figure into smaller shapes we already know, find the area of each part, and then combine the results.

Sometimes we add areas because the figure is made of parts joined together. Sometimes we subtract areas because a piece has been cut out of a larger shape.

Why this works: Area measures how much flat space is inside a figure. If a figure is built from smaller parts, the total area is the sum of the areas of those parts, as long as the parts do not overlap.

Main idea: Look for familiar shapes inside the composite figure.

• Rectangles
• Squares
• Triangles
• Parallelograms
• Trapezoids
• Semicircles or quarter circles

Here are some area formulas you may need:

• Rectangle: \(A = lw\)
• Square: \(A = s^2\)
• Triangle: \(A = \frac{1}{2}bh\)
• Parallelogram: \(A = bh\)
• Trapezoid: \(A = \frac{1}{2}(b_1+b_2)h\)
• Circle: \(A = \pi r^2\)
• Semicircle: \(A = \frac{1}{2}\pi r^2\)

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

Steps for finding the area of a composite figure

1. Look at the whole figure carefully.
2. Break it into familiar shapes.
3. Label all known side lengths.
4. Find any missing lengths if needed.
5. Use the correct area formula for each part.
6. Add or subtract the areas.
7. Write the answer in square units.

Tip: Draw lines on the figure to split it into easier pieces. A vertical or horizontal line often helps.

Tip: If a shape has a curved part, check whether it is a full circle, semicircle, or quarter circle.

Tip: Be careful with the radius and diameter of a circle. The diameter goes all the way across the circle, and the radius is half of that.

Example 1: Composite figure made of two rectangles

Suppose an L-shaped figure can be split into:

• Rectangle A: length \(8\) cm and width \(3\) cm
• Rectangle B: length \(4\) cm and width \(5\) cm

Find the area of each rectangle.

Rectangle A:

$$A_1 = 8 \times 3 = 24$$

Rectangle B:

$$A_2 = 4 \times 5 = 20$$

Add the areas:

$$A_{total} = 24 + 20 = 44$$

The area of the composite figure is \(44\;cm^2\).

What to notice: This problem uses addition because the two rectangles together make the whole figure.

Example 2: A large rectangle with a smaller rectangle cut out

A figure is a \(12\) m by \(9\) m rectangle with a \(4\) m by \(3\) m rectangular corner removed.

First find the area of the large rectangle:

$$A_{large} = 12 \times 9 = 108$$

Now find the area of the missing rectangle:

$$A_{cutout} = 4 \times 3 = 12$$

Subtract the missing part:

$$A_{total} = 108 - 12 = 96$$

The area of the composite figure is \(96\;m^2\).

What to notice: We subtract because part of the large shape is missing.

Finding missing side lengths

Sometimes a side length is not given directly. You can often find it by using lengths that line up.

For example, if the full width of a figure is \(15\) cm and one part is \(6\) cm wide, then the missing width is:

$$15 - 6 = 9$$

This step is very important. A correct method can still give a wrong answer if a missing side length is found incorrectly.

Example 3: Rectangle and triangle

A composite figure is made from a rectangle and a triangle on top of it.

• The rectangle has length \(10\) ft and width \(6\) ft.
• The triangle has base \(10\) ft and height \(4\) ft.

First find the area of the rectangle:

$$A_{rect} = 10 \times 6 = 60$$

Now find the area of the triangle:

$$A_{tri} = \frac{1}{2}(10)(4) = 20$$

Add the two areas:

$$A_{total} = 60 + 20 = 80$$

The area of the composite figure is \(80\;ft^2\).

What to notice: The triangle and rectangle share the same base length, which makes the problem easier.

Example 4: Rectangle with a semicircle

A figure is made of a rectangle with a semicircle attached to one side.

• The rectangle is \(12\) in by \(8\) in.
• The semicircle has diameter \(8\) in, so its radius is \(4\) in.

First find the area of the rectangle:

$$A_{rect} = 12 \times 8 = 96$$

Now find the area of the semicircle:

$$A_{semi} = \frac{1}{2}\pi r^2 = \frac{1}{2}\pi (4^2) = \frac{1}{2}\pi (16) = 8\pi$$

Using \(\pi \approx 3.14\):

$$A_{semi} \approx 8(3.14) = 25.12$$

Add the areas:

$$A_{total} = 96 + 25.12 = 121.12$$

The area of the composite figure is about \(121.12\;in^2\).

What to notice: The diameter was given, but the circle formula uses the radius, so we had to divide by \(2\).

Common mistakes to avoid

• Using perimeter instead of area. Area counts space inside the figure, not distance around it.
• Forgetting to square the units. Write answers like \(cm^2\), not just \(cm\).
• Adding when you should subtract. If a part is cut out, subtract its area.
• Using the diameter as the radius. For circles, always check which one you have.
• Not finding missing side lengths first. Make sure each smaller shape has the measurements you need.

How to decide whether to add or subtract

• If the composite figure is made by joining pieces, add their areas.
• If the composite figure is made by removing a piece from a larger shape, subtract.
• Some problems need both adding and subtracting.

Strategy for difficult figures

If a figure looks confusing, try more than one way to split it. There is often more than one correct method.

For example, an L-shape might be split into:

• Two rectangles by drawing one line, or
• A large rectangle minus one small rectangle.

If both methods use correct measurements, they should give the same final area.

Quick practice thinking

Ask yourself these questions:

• What simple shapes do I see?
• Do I need to add, subtract, or both?
• Are all needed lengths given?
• Is there a circle part, and if so, do I know the radius?
• Did I label my answer with square units?

Summary

A composite figure is made from smaller familiar shapes. To find its area, break it apart, find the area of each piece, and then add or subtract as needed.

Always use the correct formulas, find missing lengths carefully, and check whether a curved part is a semicircle or another part of a circle. Finish by writing the answer in square units.

Area of Shaded Regions

Sometimes a shape is partly shaded and partly unshaded. To find the area of the shaded region, we usually find the area of the whole shape first and then subtract the area of the part that is not shaded.

This idea is often called subtracting areas. It is very useful when one shape is inside another shape, or when a piece has been cut out of a larger figure.

Main idea:

$$\text{Area of shaded region} = \text{Area of whole shape} - \text{Area of unshaded part}$$

To solve these problems, you need to know the area formulas for common 2D shapes.

• Rectangle: \(A = l \times w\)
• Square: \(A = s^2\)
• Triangle: \(A = \frac{1}{2}bh\)
• Circle: \(A = \pi r^2\)

In shaded-region questions, the steps are usually the same.

1. Identify the outer shape.
2. Identify the inner shape or cut-out part.
3. Find the area of each shape.
4. Subtract: outer area minus inner area.
5. Write the correct square units, such as \(\text{cm}^2\), \(\text{m}^2\), or \(\text{in}^2\).

Important reminder: Area measures the amount of surface inside a shape. That is why area is always written in square units.

Let’s look at some examples, starting with simple shapes and moving to harder ones.

Example 1: Rectangle with a smaller rectangle inside

A large rectangle has length \(12\text{ cm}\) and width \(8\text{ cm}\). Inside it is a smaller unshaded rectangle with length \(5\text{ cm}\) and width \(3\text{ cm}\). Find the shaded area.

Step 1: Find the area of the large rectangle.

$$A = l \times w = 12 \times 8 = 96\text{ cm}^2$$

Step 2: Find the area of the smaller rectangle.

$$A = l \times w = 5 \times 3 = 15\text{ cm}^2$$

Step 3: Subtract.

$$\text{Shaded area} = 96 - 15 = 81\text{ cm}^2$$

Answer: The shaded area is \(81\text{ cm}^2\).

Example 2: Square with a circular cut-out

A square has side length \(10\text{ m}\). Inside it is an unshaded circle with radius \(3\text{ m}\). Find the shaded area in terms of \(\pi\).

Step 1: Find the area of the square.

$$A = s^2 = 10^2 = 100\text{ m}^2$$

Step 2: Find the area of the circle.

$$A = \pi r^2 = \pi(3)^2 = 9\pi\text{ m}^2$$

Step 3: Subtract.

$$\text{Shaded area} = 100 - 9\pi\text{ m}^2$$

Answer: The shaded area is \(100 - 9\pi\text{ m}^2\).

If you want a decimal answer, use \(\pi \approx 3.14\).

$$9\pi \approx 9(3.14) = 28.26$$

$$100 - 28.26 = 71.74\text{ m}^2$$

So the shaded area is about \(71.74\text{ m}^2\).

Example 3: Circle with a smaller circle inside

A large circle has radius \(7\text{ cm}\). Inside it is a smaller unshaded circle with radius \(4\text{ cm}\). Find the shaded area in terms of \(\pi\).

This kind of shape is like a ring. We still use the same idea: big area minus small area.

Step 1: Find the area of the large circle.

$$A = \pi r^2 = \pi(7)^2 = 49\pi\text{ cm}^2$$

Step 2: Find the area of the small circle.

$$A = \pi r^2 = \pi(4)^2 = 16\pi\text{ cm}^2$$

Step 3: Subtract.

$$\text{Shaded area} = 49\pi - 16\pi = 33\pi\text{ cm}^2$$

Answer: The shaded area is \(33\pi\text{ cm}^2\).

Notice that because both shapes are circles, both areas have \(\pi\), so subtracting is simple.

Example 4: Rectangle with a triangular cut-out

A rectangle measures \(14\text{ in}\) by \(9\text{ in}\). A triangular piece inside it is unshaded. The triangle has base \(6\text{ in}\) and height \(4\text{ in}\). Find the shaded area.

Step 1: Find the area of the rectangle.

$$A = l \times w = 14 \times 9 = 126\text{ in}^2$$

Step 2: Find the area of the triangle.

$$A = \frac{1}{2}bh = \frac{1}{2}(6)(4) = 12\text{ in}^2$$

Step 3: Subtract.

$$\text{Shaded area} = 126 - 12 = 114\text{ in}^2$$

Answer: The shaded area is \(114\text{ in}^2\).

Tips for Success

• Read carefully: Make sure you know which part is shaded and which part is not.
• Use the correct formula: Different shapes need different area formulas.
• Check the measurements: For circles, be sure you know whether the number given is the radius or the diameter.
• Subtract in the right order: Always do larger area minus smaller area.
• Include units: Your final answer must be in square units.

Common Mistakes

• Forgetting to subtract and giving only the area of the outer shape.
• Using perimeter formulas instead of area formulas.
• Using the diameter as the radius in a circle problem.
• Forgetting to square the radius in \(\pi r^2\).
• Leaving off square units.

Quick Check

Try these on your own:

1. A square has side length \(9\text{ cm}\). Inside it is a smaller square with side length \(4\text{ cm}\). What is the shaded area?
2. A circle has radius \(8\text{ m}\). Inside it is a smaller circle with radius \(5\text{ m}\). What is the shaded area in terms of \(\pi\)?
3. A rectangle is \(15\text{ ft}\) by \(10\text{ ft}\). A triangle inside it has base \(8\text{ ft}\) and height \(5\text{ ft}\). What is the shaded area?

Quick Check Answers

1. Outer square: \(9^2 = 81\text{ cm}^2\). Inner square: \(4^2 = 16\text{ cm}^2\). Shaded area: \(81 - 16 = 65\text{ cm}^2\).
2. Outer circle: \(\pi(8^2) = 64\pi\text{ m}^2\). Inner circle: \(\pi(5^2) = 25\pi\text{ m}^2\). Shaded area: \(64\pi - 25\pi = 39\pi\text{ m}^2\).
3. Outer rectangle: \(15 \times 10 = 150\text{ ft}^2\). Inner triangle: \(\frac{1}{2}(8)(5) = 20\text{ ft}^2\). Shaded area: \(150 - 20 = 130\text{ ft}^2\).

Summary

To find the area of a shaded region, find the area of the whole figure and subtract the area of the part that is not shaded. This works for rectangles, squares, triangles, circles, and many mixed shapes. Always use the correct area formula, subtract carefully, and include square units in your final answer.