Areas Related to Circles

Perimeter and Area Fundamentals of Circles

Before working with sectors, segments, and more complex circle shapes, it is important to understand the two basic measurements of a whole circle: its perimeter and its area.

The perimeter of a circle is called its circumference. It tells us the distance all the way around the circle. The area of a circle tells us how much space is inside the circle.

These two ideas are different, so it is very important not to mix up their formulas.

• Circumference = distance around the circle
• Area = amount of space inside the circle

1. Important parts of a circle

To use circle formulas correctly, you need to know the meanings of radius and diameter.

• The radius, written as \(r\), is the distance from the center of the circle to any point on the circle.
• The diameter, written as \(d\), is the distance across the circle through the center.

The diameter is always twice the radius:

So if you know one of them, you can find the other:

2. The number \(\pi\)

Circle formulas use the special number \(\pi\) (pi). Pi is the ratio of the circumference of a circle to its diameter. It is an irrational number, which means its decimal goes on forever without repeating.

In school math, we usually use:

• Exact form: \(\pi\)
• Approximate decimal form: \(3.14\) or \(3.1416\)

When a question says leave your answer in terms of \(\pi\), do not change \(\pi\) into a decimal. When a question asks for a decimal answer, substitute a decimal approximation for \(\pi\).

For example:

• Exact: \(12\pi\)
• Approximate: \(12\pi \approx 37.7\)

3. Formula for circumference

The circumference of a circle can be found in two equivalent ways:

or

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

4. Formula for area

The area of a circle is found using:

This means you square the radius, then multiply by \(\pi\).

Be careful: the area formula uses radius, not diameter. If you are given the diameter, divide by 2 first to get the radius.

5. Units matter

Circumference is a length, so its units are just units like:

• cm
• m
• in

Area measures surface, so its units are squared:

• cm²
• m²
• in²

If you forget the squared units for area, your answer is not complete.

6. Common mistakes to avoid

• Using diameter in the area formula instead of radius
• Forgetting to square the radius in \(A = \pi r^2\)
• Confusing circumference and area formulas
• Changing exact answers into decimals when the question says to leave answers in terms of \(\pi\)
• Using the wrong units

Worked Example 1: Find circumference from radius

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

Step 1: Choose the correct formula.

Step 2: Substitute \(r = 5\).

This is the exact answer.

If a decimal approximation is needed:

Answer: \(10\pi\) cm, or about \(31.4\) cm.

Worked Example 2: Find area from radius

A circle has radius \(7\) m. Find its area.

Step 1: Use the area formula.

Step 2: Substitute \(r = 7\).

This is the exact answer.

For a decimal approximation:

Answer: \(49\pi\) m², or about \(153.86\) m².

Worked Example 3: Find circumference and area from diameter

A circle has diameter \(12\) in. Find both the circumference and the area.

Step 1: Find the radius.

Step 2: Find circumference using \(C = \pi d\).

Approximate value:

Step 3: Find area using \(A = \pi r^2\).

Approximate value:

Answer:

• Circumference: \(12\pi\) in, or about \(37.68\) in
• Area: \(36\pi\) in², or about \(113.04\) in²

Worked Example 4: Decide whether to use exact or approximate form

A circular garden has radius \(9\) ft. Find the area:

1. In terms of \(\pi\)
2. As a decimal rounded to the nearest tenth

Step 1: Use the area formula.

So the exact answer is:

Step 2: Convert to a decimal.

Rounded to the nearest tenth:

Answer:

• Exact form: \(81\pi\) ft²
• Approximate form: \(254.3\) ft²

7. How to tell which formula to use

Ask yourself what the problem wants.

• If it asks for the distance around the circle, use circumference.
• If it asks for the space inside the circle, use area.

You can also use this quick guide:

• Around the circle → \(C = 2\pi r\) or \(C = \pi d\)
• Inside the circle → \(A = \pi r^2\)

8. Exact answers vs decimal answers

In mathematics, exact answers are often preferred unless the question asks for an approximation.

For example, if the radius is \(4\) cm:

• Circumference exact: \(2\pi(4) = 8\pi\) cm
• Circumference approximate: \(8\pi \approx 25.1\) cm
• Area exact: \(\pi(4^2) = 16\pi\) cm²
• Area approximate: \(16\pi \approx 50.2\) cm²

Exact answers are more precise because they keep \(\pi\) instead of rounding it.

9. Final check strategy

After solving a circle problem, check these questions:

1. Did I use the correct formula?
2. Did I use radius or diameter correctly?
3. Did I square the radius for area?
4. Did I keep \(\pi\) if the answer should be exact?
5. Did I write the correct units?

Brief Summary

The circumference of a circle is the distance around it, and the area is the space inside it. Use \(C = 2\pi r\) or \(C = \pi d\) for circumference, and use \(A = \pi r^2\) for area. Always pay attention to whether the problem gives radius or diameter, and be careful to tell the difference between exact answers in terms of \(\pi\) and decimal approximations.

Deriving and Calculating Arc Length

When we measure around a whole circle, we are finding its circumference. But sometimes we only want the length of part of the circle. That curved part is called an arc.

In this lesson, you will learn how to derive the arc length formula from the circumference of a circle, and then how to use it to solve problems.

This idea is very useful when working with sectors, curved edges, and shapes made from parts of circles.

1. Review: Circumference of a Circle

The circumference of a circle is the total distance around it.

The formula for circumference is:

where:

• \(C\) = circumference

• \(r\) = radius of the circle

• \(\pi\) is approximately \(3.14\)

If you know the whole distance around the circle, then an arc is just a fraction of that total distance.

2. What Is an Arc?

An arc is a portion of the circle’s circumference.

The size of the arc depends on the size of the central angle that cuts it off. A central angle is an angle with its vertex at the center of the circle.

For example:

• A \(90^\circ\) central angle cuts off one-quarter of the circle.

• A \(180^\circ\) central angle cuts off half of the circle.

• A \(360^\circ\) central angle is the whole circle.

So the arc length must be the same fraction of the circumference as the central angle is of \(360^\circ\).

3. Deriving the Arc Length Formula

A full circle has:

• angle measure \(360^\circ\)

• circumference \(2\pi r\)

If the central angle is \(\theta\) degrees, then the arc is only \(\frac{\theta}{360}\) of the full circle.

So the arc length, which we will call \(L\), is:

This is the formula for arc length when the angle is given in degrees.

Arc Length Formula:

where:

• \(L\) = arc length

• \(\theta\) = central angle in degrees

• \(r\) = radius

4. Why the Formula Makes Sense

The formula works because it uses a simple idea:

part of the circle = same part of the circumference

If the angle is half of \(360^\circ\), then the arc length is half of the circumference.

If the angle is one-fourth of \(360^\circ\), then the arc length is one-fourth of the circumference.

This is why the fraction \(\frac{\theta}{360}\) is so important.

5. Steps for Finding Arc Length

1. Identify the radius \(r\).

2. Identify the central angle \(\theta\).

3. Use the formula $$L = \frac{\theta}{360} \cdot 2\pi r$$

4. Simplify your answer.

5. If needed, give a decimal approximation.

6. Worked Examples

Example 1: Find the arc length of a \(90^\circ\) sector with radius \(8\text{ cm}\).

Step 1: Write the formula.

Step 2: Substitute the values.

Step 3: Simplify.

Answer:

Approximate decimal:

This makes sense because a \(90^\circ\) angle is one-quarter of a full circle, so the arc length should be one-quarter of the circumference.

Example 2: Find the arc length of a \(120^\circ\) sector with radius \(15\text{ m}\).

Step 1: Use the formula.

Step 2: Substitute.

Step 3: Simplify the fraction.

Now simplify:

Answer:

Approximate decimal:

Example 3: Find the arc length when the radius is \(7\text{ cm}\) and the central angle is \(225^\circ\).

This example is a little harder because the angle is greater than \(180^\circ\), but the same method still works.

Step 1: Write the formula.

Step 2: Substitute.

Step 3: Simplify.

Answer:

Approximate decimal:

Example 4: A circle has diameter \(20\text{ cm}\). Find the length of an arc with central angle \(72^\circ\).

This problem is different because it gives the diameter instead of the radius.

Step 1: Find the radius.

Step 2: Use the arc length formula.

Step 3: Simplify.

Answer:

Approximate decimal:

7. Common Mistakes to Avoid

• Using the diameter instead of the radius
  In the formula \(2\pi r\), \(r\) means radius. If you are given the diameter, divide by 2 first.

• Forgetting to use the fraction of the circle
  Do not just find the whole circumference unless the angle is \(360^\circ\).

• Mixing up area and arc length
  Arc length is a distance around the circle, so the unit is just \(\text{cm}\), \(\text{m}\), and so on. It is not square units.

• Not simplifying the angle fraction
  Simplifying \(\frac{\theta}{360}\) often makes the calculation easier.

8. Quick Check: Does Your Answer Make Sense?

After solving, ask yourself:

• Is the arc length less than the full circumference?

• If the angle is small, is the arc length also small?

• If the angle is half of \(360^\circ\), is the arc length half of the circumference?

These checks can help you catch mistakes.

9. Summary

An arc is part of a circle’s circumference. To find its length, first remember that the whole circle measures \(360^\circ\) and has circumference \(2\pi r\).

The arc length formula is:

This formula works because the arc is the same fraction of the circumference as the central angle is of the full \(360^\circ\).

Whenever you solve an arc length problem, identify the radius, identify the central angle, substitute into the formula, and simplify carefully.

Area of a Sector of a Circle

A sector is a pie-shaped part of a circle. It is formed by two radii and the arc between them.

To find the area of a sector, we use the idea that the sector is only a fraction of the whole circle. If we know what fraction of the full circle the sector takes up, we can find its area.

This lesson will show you how to find the area of a sector using its central angle and radius.

1. Review: Area of a Circle

Before finding the area of a sector, remember the formula for the area of a full circle:

Here,

• \(A\) is the area

• \(r\) is the radius

• \(\pi \approx 3.14\) or \(\frac{22}{7}\) in some questions

A full circle has an angle of \(360^\circ\). A sector has only part of that full turn.

2. Formula for the Area of a Sector

If the central angle of the sector is \(\theta\), then the sector is \(\frac{\theta}{360}\) of the full circle.

So the formula for the area of a sector is:

This formula means:

• Find the area of the whole circle

• Multiply by the fraction \(\frac{\theta}{360}\)

3. What Each Part Means

• \(\theta\) is the central angle of the sector, measured in degrees

• \(r\) is the radius of the circle

• \(360\) is used because a full circle is \(360^\circ\)

If the angle is small, the sector is a small part of the circle. If the angle is large, the sector is a larger part of the circle.

4. Steps for Finding the Area of a Sector

1. Write down the radius and the central angle.

2. Find the area of the whole circle using \(\pi r^2\).

3. Find the fraction of the circle using \(\frac{\theta}{360}\).

4. Multiply the circle’s area by that fraction.

5. Simplify your answer. Round only if the question asks you to.

5. Worked Examples

Example 1: A simple sector

Find the area of a sector with radius \(7\text{ cm}\) and central angle \(90^\circ\).

Step 1: Use the formula

Step 2: Substitute the values

In decimal form,

Answer: \(\frac{49\pi}{4}\text{ cm}^2\) or about \(38.5\text{ cm}^2\)

Why this makes sense: A \(90^\circ\) sector is one-quarter of a full circle, so its area should be one-quarter of the circle’s area.

Example 2: A sector that is not a simple fraction

Find the area of a sector with radius \(10\text{ cm}\) and central angle \(135^\circ\).

Step 1: Write the formula

Step 2: Substitute

Simplify the fraction:

Now multiply:

In decimal form,

Answer: \(37.5\pi\text{ cm}^2\) or about \(117.8\text{ cm}^2\)

Example 3: Finding area when \(\pi = 3.14\) is used

A sector has radius \(6\text{ m}\) and central angle \(200^\circ\). Find its area using \(\pi = 3.14\).

Step 1: Use the formula

Step 2: Simplify

Answer: \(62.8\text{ m}^2\)

Example 4: Working backward

The area of a sector is \(24\pi\text{ cm}^2\). The radius of the circle is \(12\text{ cm}\). Find the central angle.

Use the sector area formula:

Cancel \(\pi\) from both sides:

Now solve for \(\theta\):

Multiply both sides by \(5\):

Answer: \(60^\circ\)

6. Common Mistakes to Avoid

• Using diameter instead of radius: If you are given the diameter, divide by 2 first.

• Forgetting to square the radius: In \(\pi r^2\), only the radius is squared.

• Using the wrong fraction: The fraction must be \(\frac{\theta}{360}\), not \(\frac{360}{\theta}\).

• Rounding too early: Keep exact values like \(\pi\) as long as possible.

• Wrong units: Area is always in square units, such as \(\text{cm}^2\), \(\text{m}^2\), or \(\text{mm}^2\).

7. Quick Check Questions

Try these on your own:

1. Find the area of a sector with radius \(8\text{ cm}\) and angle \(45^\circ\).

2. Find the area of a sector with radius \(14\text{ m}\) and angle \(120^\circ\).

3. A sector has area \(16\pi\text{ cm}^2\) and radius \(8\text{ cm}\). Find the central angle.

8. Summary

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

To find the area of a sector, use:

The key idea is proportional reasoning: if the angle is a certain fraction of \(360^\circ\), then the area is the same fraction of the full circle’s area.

Always check that you are using the radius, squaring it correctly, and writing the final answer in square units.

Area of a Segment of a Circle

When a circle is cut by a chord, it creates a region between the chord and the arc. This region is called a segment of a circle.

To find the area of a segment, we do not use a new formula from memory. Instead, we build it from ideas you already know: the area of a sector and the area of a triangle.

The key idea is:

$$\text{Area of segment} = \text{Area of sector} - \text{Area of triangle}$$

This lesson will show you what a segment is, when to subtract, how to find each part, and how to solve problems step by step.

1. Important circle parts

• Radius: a line from the center of the circle to the edge.
• Chord: a straight line joining two points on the circle.
• Arc: a curved part of the circle.
• Sector: the region between two radii and an arc, like a slice of pizza.
• Segment: the region between a chord and its arc.

If you draw two radii to the ends of a chord, you create:

• a sector, and
• a triangle inside that sector.

The segment is the part of the sector that is left after removing the triangle.

2. Main formula

If the central angle is \(\theta\) and the radius is \(r\), then:

$$\text{Area of segment} = \text{Area of sector} - \text{Area of triangle}$$

The area of the sector is:

$$\text{Area of sector} = \frac{\theta}{360} \times \pi r^2$$

The triangle is formed by the two radii and the chord.

In many 10th Grade problems, the triangle area is found using:

• the standard triangle formula \(\frac{1}{2} \times \text{base} \times \text{height}\), or
• a known special triangle if the angle makes it easy.

So, in full:

$$\text{Area of segment} = \frac{\theta}{360} \times \pi r^2 - \text{Area of triangle}$$

3. How to solve these questions

1. Identify the radius and the central angle.
2. Find the area of the sector.
3. Find the area of the triangle inside the sector.
4. Subtract:

$$\text{segment} = \text{sector} - \text{triangle}$$

4. When do we subtract?

For the minor segment, which is the smaller region between the chord and the shorter arc, we usually subtract the triangle from the sector.

$$\text{minor segment} = \text{minor sector} - \text{triangle}$$

If a problem asks for the major segment, which is the larger region, then:

$$\text{major segment} = \text{area of circle} - \text{minor segment}$$

Always read the question carefully to see which segment is required.

5. Worked Example 1: Simple segment with a right triangle

A circle has radius \(6\) cm. A chord creates a central angle of \(90^\circ\). Find the area of the minor segment.

Step 1: Find the sector area

$$\text{Area of sector} = \frac{90}{360} \times \pi \times 6^2$$

$$= \frac{1}{4} \times \pi \times 36$$

$$= 9\pi \text{ cm}^2$$

Step 2: Find the triangle area

The triangle is made from two radii of length \(6\) cm with a right angle between them.

So it is a right triangle with base \(6\) cm and height \(6\) cm.

$$\text{Area of triangle} = \frac{1}{2} \times 6 \times 6 = 18 \text{ cm}^2$$

Step 3: Subtract

$$\text{Area of segment} = 9\pi - 18$$

Using \(\pi \approx 3.14\):

$$9\pi - 18 \approx 28.26 - 18 = 10.26 \text{ cm}^2$$

Answer: The area of the segment is \(9\pi - 18\text{ cm}^2\), or about \(10.26\text{ cm}^2\).

6. Worked Example 2: Segment from an equilateral triangle setup

A circle has radius \(8\) cm and central angle \(60^\circ\). Find the area of the minor segment.

Step 1: Sector area

$$\text{Area of sector} = \frac{60}{360} \times \pi \times 8^2$$

$$= \frac{1}{6} \times 64\pi$$

$$= \frac{32\pi}{3} \text{ cm}^2$$

Step 2: Triangle area

The triangle has two sides of length \(8\) cm and included angle \(60^\circ\). In this special case, the triangle is equilateral, so all sides are \(8\) cm.

The area of an equilateral triangle of side \(8\) cm is:

$$\text{Area} = \frac{\sqrt{3}}{4} \times 8^2 = \frac{\sqrt{3}}{4} \times 64 = 16\sqrt{3} \text{ cm}^2$$

Step 3: Subtract

$$\text{Area of segment} = \frac{32\pi}{3} - 16\sqrt{3}$$

Approximate value:

$$\frac{32\pi}{3} - 16\sqrt{3} \approx 33.51 - 27.71 = 5.80 \text{ cm}^2$$

Answer: The area of the segment is \(\frac{32\pi}{3} - 16\sqrt{3}\text{ cm}^2\), or about \(5.80\text{ cm}^2\).

7. Worked Example 3: Finding the major segment

A circle has radius \(5\) cm. A chord subtends an angle of \(120^\circ\) at the center. Find the area of the major segment.

Step 1: Find the minor sector area

$$\text{Area of sector} = \frac{120}{360} \times \pi \times 5^2$$

$$= \frac{1}{3} \times 25\pi = \frac{25\pi}{3}$$

Step 2: Find the triangle area

The triangle has sides \(5\) cm and \(5\) cm. Draw a perpendicular from the center to the chord. This splits the triangle into two right triangles.

The whole triangle has base:

$$2\times 5\sin 60^\circ = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}$$

Its height is:

$$5\cos 60^\circ = 5 \times \frac{1}{2} = 2.5$$

So:

$$\text{Area of triangle} = \frac{1}{2} \times 5\sqrt{3} \times 2.5 = \frac{25\sqrt{3}}{4}$$

Step 3: Find the minor segment

$$\text{Minor segment} = \frac{25\pi}{3} - \frac{25\sqrt{3}}{4}$$

Step 4: Find the whole circle area

$$\text{Area of circle} = \pi r^2 = 25\pi$$

Step 5: Find the major segment

$$\text{Major segment} = 25\pi - \left(\frac{25\pi}{3} - \frac{25\sqrt{3}}{4}\right)$$

$$= \frac{50\pi}{3} + \frac{25\sqrt{3}}{4}$$

Approximate value:

$$\frac{50\pi}{3} + \frac{25\sqrt{3}}{4} \approx 52.36 + 10.83 = 63.19 \text{ cm}^2$$

Answer: The area of the major segment is \(\frac{50\pi}{3} + \frac{25\sqrt{3}}{4}\text{ cm}^2\), or about \(63.19\text{ cm}^2\).

8. Worked Example 4: Using base and height directly

In a circle, the area of a sector is \(40\text{ cm}^2\). The triangle inside the sector has base \(10\) cm and height \(6\) cm. Find the area of the segment.

Step 1: Triangle area

$$\text{Area of triangle} = \frac{1}{2} \times 10 \times 6 = 30\text{ cm}^2$$

Step 2: Segment area

$$\text{Area of segment} = 40 - 30 = 10\text{ cm}^2$$

Answer: The area of the segment is \(10\text{ cm}^2\).

9. Common mistakes to avoid

• Mixing up sector and segment: A sector is the whole slice; a segment is only the curved part between the chord and arc.
• Forgetting to subtract the triangle: The sector alone is not the segment.
• Using the wrong angle fraction: For sector area, always use \(\frac{\theta}{360}\).
• Finding the wrong segment: If the question wants the major segment, first find the minor segment, then subtract from the whole circle.
• Rounding too early: Keep exact values like \(\pi\) or \(\sqrt{3}\) until the final step when possible.

10. Quick problem-solving checklist

• What is the radius?
• What is the central angle?
• Am I finding a minor segment or a major segment?
• Have I found the sector area correctly?
• Have I found the triangle area correctly?
• Did I subtract in the correct order?

11. Summary

A segment of a circle is the region between a chord and an arc. To find its area, first find the area of the sector, then find the area of the triangle formed by the two radii and the chord, and subtract.

The main relationship is:

$$\text{Area of segment} = \text{Area of sector} - \text{Area of triangle}$$

If the question asks for the major segment, subtract the minor segment from the area of the whole circle.

Once you can identify the sector and triangle clearly, these problems become much easier.

Areas of Annuli and Concentric Circles

In earlier work with circles, you may have found the area of a whole circle using the formula \(A = \pi r^2\). In this lesson, we will build on that idea to find the area of a ring-shaped region between two circles.

This ring is called an annulus. It is formed when two circles have the same center but different radii. Circles like this are called concentric circles.

To find the area of an annulus, we do not need a brand-new formula. We use what we already know: find the area of the larger circle, find the area of the smaller circle, and subtract.

Key ideas

• Concentric circles are circles with the same center.
• An annulus is the region between two concentric circles.
• The area of an annulus is the area of the outer circle minus the area of the inner circle.

If the outer radius is \(R\) and the inner radius is \(r\), then:

This can be written more simply as:

Both forms mean the same thing. Use whichever one feels clearer to you.

Understanding the radii

The outer radius is the distance from the center to the larger circle. The inner radius is the distance from the center to the smaller circle.

Be careful not to mix up radius and diameter.

• Radius = distance from center to circle
• Diameter = distance across the whole circle through the center
• Diameter is twice the radius, so \(d = 2r\)

If a question gives diameters, convert them to radii before using the formula.

Why subtraction works

Imagine a large circular disk with a smaller circular hole cut out of the middle. The remaining part is the annulus.

So the area left over is:

• area of the whole large circle
• minus the area of the missing smaller circle

That is why subtraction is the main idea in these problems.

Steps for finding the area of an annulus

1. Identify the outer radius \(R\).
2. Identify the inner radius \(r\).
3. Find the area of the outer circle: \(\pi R^2\).
4. Find the area of the inner circle: \(\pi r^2\).
5. Subtract: \(\pi R^2 - \pi r^2\).
6. Include the correct square units, such as \(\text{cm}^2\), \(\text{m}^2\), or \(\text{in}^2\).

Worked Example 1: Basic annulus area

The outer radius of an annulus is \(8\) cm and the inner radius is \(5\) cm. Find the area.

Step 1: Write the formula

Step 2: Substitute the values

Step 3: Give the final answer

The exact area is \(39\pi\ \text{cm}^2\).

If you need a decimal approximation:

Worked Example 2: Using diameters

A circular path surrounds a fountain. The outer diameter is \(20\) m and the inner diameter is \(12\) m. Find the area of the path.

Step 1: Convert diameters to radii

• Outer radius: \(20 \div 2 = 10\) m
• Inner radius: \(12 \div 2 = 6\) m

Step 2: Use the annulus formula

Step 3: Final answer

The area of the path is \(64\pi\ \text{m}^2\).

Approximate value:

Worked Example 3: Finding a missing radius from the ring width

A ring-shaped garden has an inner radius of \(7\) m. The garden is \(3\) m wide all the way around. Find the area of the garden.

Step 1: Find the outer radius

If the ring is \(3\) m wide, then the outer radius is:

Step 2: Use the formula

Step 3: Final answer

The area of the garden is \(51\pi\ \text{m}^2\).

Approximate value:

Worked Example 4: Solving from areas

The area of a large circle is \(81\pi\ \text{cm}^2\), and the area of the inner circle is \(49\pi\ \text{cm}^2\). Find the area of the annulus.

Step 1: Subtract the two circle areas

Step 2: Final answer

The area of the annulus is \(32\pi\ \text{cm}^2\).

This example shows that sometimes you do not need the radii at all if the areas of both circles are already given.

A useful algebra pattern

Sometimes you may notice this expression:

This is called a difference of squares, but you do not need special algebra to solve these problems. Usually, the easiest way is to square each radius first and then subtract.

For example, if \(R = 9\) and \(r = 4\), then:

So the area would be \(65\pi\).

Common mistakes to avoid

• Using diameters instead of radii: Always check whether the question gives radius or diameter.
• Subtracting before squaring: \((R-r)^2\) is not the same as \(R^2-r^2\).
• Forgetting units: Area must be written in square units.
• Subtracting in the wrong order: Outer circle area should be larger, so subtract inner from outer.

Important note about a common error

If \(R=8\) and \(r=5\), do not do this:

This is incorrect.

The correct calculation is:

These answers are very different, so this mistake matters a lot.

When exact answers and approximations are used

In math, an answer in terms of \(\pi\), such as \(39\pi\), is called an exact answer. This is often preferred unless the question asks for a decimal answer.

If a decimal is needed, use \(\pi \approx 3.14\) or the \(\pi\) button on a calculator and round as instructed.

Real-life examples of annuli

• a running track around a circular field
• a ring-shaped garden around a fountain
• a washer or metal ring
• a circular border around a logo or design

In each case, the area of the ring is found by subtracting the inner circle from the outer circle.

Quick check questions

• If the outer radius is \(11\) cm and the inner radius is \(9\) cm, what expression represents the area? Answer: \(\pi(11^2-9^2)\)
• If the outer diameter is \(18\) m, what is the outer radius? Answer: \(9\) m
• If an annulus has outer area \(100\pi\) and inner area \(36\pi\), what is its area? Answer: \(64\pi\)

Summary

An annulus is the ring-shaped region between two concentric circles. To find its area, subtract the area of the smaller circle from the area of the larger circle.

Always make sure you are using radii, not diameters, and remember to write your answer in square units. If you follow the steps carefully, annulus problems are just circle area problems with subtraction.

Calculating Areas of Composite Geometric Figures means finding the area of a shape made from two or more simpler shapes joined together. In 10th Grade, these often include rectangles, triangles, circles, semicircles, and sectors.

Many shaded regions in geometry are composite figures. They may look complicated at first, but the main idea is simple: break the figure into parts whose areas you already know how to find, then add or subtract those parts carefully.

This skill is especially important in topics involving circles, because curved shapes such as semicircles, quarter circles, and sectors are often attached to or removed from polygons like squares and rectangles.

Goal of this lesson: learn how to identify the smaller shapes inside a composite figure, choose the correct area formulas, and combine the results correctly.

1. Review of basic area formulas

Before finding the area of a composite figure, you need to know the area formulas for common shapes.

• Rectangle: \(A = lw\)
• Square: \(A = s^2\)
• Triangle: \(A = \frac{1}{2}bh\)
• Circle: \(A = \pi r^2\)
• Semicircle: \(A = \frac{1}{2}\pi r^2\)
• Quarter circle: \(A = \frac{1}{4}\pi r^2\)
• Sector: if the central angle is \(\theta^\circ\), then $$A = \frac{\theta}{360}\pi r^2$$

2. Main strategy for composite area problems

Whenever you see a complicated figure, follow these steps.

1. Look for familiar shapes. Can the figure be split into rectangles, triangles, semicircles, or sectors?
2. Decide whether to add or subtract. If shapes are joined together, add their areas. If a part is cut out, subtract its area.
3. Find any missing measurements. You may need to use side lengths, diameters, or simple facts such as: radius = half of diameter.
4. Write the area expression clearly. This helps avoid mistakes.
5. Calculate carefully and include units. Area is always in square units, such as \(\text{cm}^2\), \(\text{m}^2\), or \(\text{in}^2\).

3. Additive and subtractive composite figures

There are two common kinds of composite area problems.

Additive figures: These are shapes made by joining smaller shapes together. For example, a rectangle with a semicircle attached to one side. In this case, find each area and add.

Subtractive figures: These are shapes with a piece removed. For example, a square with a semicircle cut out. In this case, find the whole area first, then subtract the removed part.

4. Important circle ideas in composite figures

Circle pieces often cause the most confusion, so pay close attention to these facts.

• The diameter goes all the way across the circle through the center.
• The radius is half the diameter, so \(r = \frac{d}{2}\).
• If a semicircle sits on a side of a square or rectangle, that side is often the diameter, not the radius.
• A quarter circle is one-fourth of a full circle.
• A sector is a “slice” of a circle.

A very common mistake is using the diameter instead of the radius in \(\pi r^2\). Always check which measurement you were given.

5. Worked Example 1: Rectangle with a semicircle attached

A rectangle is \(10\,\text{cm}\) long and \(6\,\text{cm}\) wide. A semicircle is attached along the width of \(6\,\text{cm}\). Find the total area.

Step 1: Find the area of the rectangle.

$$A_{\text{rectangle}} = lw = 10 \times 6 = 60\,\text{cm}^2$$

Step 2: Find the radius of the semicircle.

The width \(6\,\text{cm}\) is the diameter, so the radius is

$$r = \frac{6}{2} = 3\,\text{cm}$$

Step 3: Find the area of the semicircle.

$$A_{\text{semicircle}} = \frac{1}{2}\pi r^2 = \frac{1}{2}\pi(3^2) = \frac{1}{2}\pi(9) = 4.5\pi$$

Using \(\pi \approx 3.14\),

$$A_{\text{semicircle}} \approx 4.5 \times 3.14 = 14.13\,\text{cm}^2$$

Step 4: Add the areas.

$$A_{\text{total}} = 60 + 14.13 = 74.13\,\text{cm}^2$$

Answer: The total area is approximately \(74.13\,\text{cm}^2\).

6. Worked Example 2: Square with a semicircle cut out

A square has side length \(8\,\text{m}\). A semicircle with diameter \(8\,\text{m}\) is cut out from one side of the square. Find the shaded area remaining.

Step 1: Find the area of the square.

$$A_{\text{square}} = s^2 = 8^2 = 64\,\text{m}^2$$

Step 2: Find the radius of the semicircle.

$$r = \frac{8}{2} = 4\,\text{m}$$

Step 3: Find the area of the semicircle.

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

Using \(\pi \approx 3.14\),

$$A_{\text{semicircle}} \approx 8 \times 3.14 = 25.12\,\text{m}^2$$

Step 4: Subtract.

$$A_{\text{remaining}} = 64 - 25.12 = 38.88\,\text{m}^2$$

Answer: The shaded area is approximately \(38.88\,\text{m}^2\).

7. Worked Example 3: Square and quarter circle combination

A square has side length \(12\,\text{cm}\). Inside the square, a quarter circle of radius \(12\,\text{cm}\) is drawn from one corner to the two nearby corners. Find the area of the part of the square outside the quarter circle.

Step 1: Find the area of the square.

$$A_{\text{square}} = 12^2 = 144\,\text{cm}^2$$

Step 2: Find the area of the quarter circle.

$$A_{\text{quarter circle}} = \frac{1}{4}\pi r^2 = \frac{1}{4}\pi(12^2) = \frac{1}{4}\pi(144) = 36\pi$$

Using \(\pi \approx 3.14\),

$$A_{\text{quarter circle}} \approx 36 \times 3.14 = 113.04\,\text{cm}^2$$

Step 3: Subtract to find the outside region.

$$A_{\text{outside}} = 144 - 113.04 = 30.96\,\text{cm}^2$$

Answer: The area outside the quarter circle is approximately \(30.96\,\text{cm}^2\).

8. Worked Example 4: Composite figure with a triangle and a semicircle

A figure is made of a right triangle and a semicircle attached along one side. The triangle has base \(12\,\text{cm}\) and height \(8\,\text{cm}\). The semicircle has diameter \(12\,\text{cm}\). Find the total area.

Step 1: Find the area of the triangle.

$$A_{\text{triangle}} = \frac{1}{2}bh = \frac{1}{2}(12)(8) = 48\,\text{cm}^2$$

Step 2: Find the radius of the semicircle.

$$r = \frac{12}{2} = 6\,\text{cm}$$

Step 3: Find the area of the semicircle.

$$A_{\text{semicircle}} = \frac{1}{2}\pi r^2 = \frac{1}{2}\pi(6^2) = \frac{1}{2}\pi(36) = 18\pi$$

Using \(\pi \approx 3.14\),

$$A_{\text{semicircle}} \approx 18 \times 3.14 = 56.52\,\text{cm}^2$$

Step 4: Add the areas.

$$A_{\text{total}} = 48 + 56.52 = 104.52\,\text{cm}^2$$

Answer: The total area is approximately \(104.52\,\text{cm}^2\).

9. How to handle irregular shaded regions

Some figures may not show the parts clearly. When that happens, imagine drawing lines to split the figure into simpler pieces.

For example, you might:

• draw a line to divide an L-shaped figure into two rectangles,
• separate a shape into a rectangle and a triangle,
• treat a curved piece as a semicircle or quarter circle,
• find a large simple shape first, then subtract the missing part.

There is often more than one correct way to split a composite figure. The best method is the one that makes the measurements easiest to use.

10. Common mistakes to avoid

• Using diameter instead of radius in circle area formulas.
• Adding when you should subtract, or subtracting when you should add.
• Forgetting the fraction for a semicircle, quarter circle, or sector.
• Mixing units, such as centimeters and meters, without converting.
• Forgetting square units in the final answer.

11. Quick check questions to ask yourself

• What simple shapes make up this figure?
• Which lengths are side lengths, diameters, or radii?
• Am I adding parts together or subtracting a cut-out region?
• Did I use the correct formula for each piece?
• Did I label the answer in square units?

12. Summary

To find the area of a composite geometric figure, break it into smaller familiar shapes such as rectangles, triangles, semicircles, or sectors. Then find each area and combine them using addition or subtraction, depending on whether parts are attached or removed.

When circles are involved, be especially careful to identify the radius correctly. With practice, even complex shaded regions become manageable when you work one piece at a time.

Modeling Sweeping Areas and Rotating Mechanics

In many real-life machines, a part rotates around a fixed point. As it turns, it covers or sweeps out a region. To model this motion in maths, we use ideas from circles, especially sectors and sometimes annular sectors (ring-shaped sectors).

Examples include a windshield wiper clearing rain, a sprinkler spraying water, a radar arm rotating, or a clock hand moving. In each case, the moving part turns through an angle and reaches a certain distance from the center. That creates a curved region whose area can often be found using circle formulas.

This lesson will show how to identify the shape being swept, choose the correct formula, and solve practical problems involving rotating mechanics.

1. Review: area of a circle and area of a sector

A full circle with radius \(r\) has area

$$A = \pi r^2$$

A sector is a slice of a circle. If the central angle is \(\theta\) degrees, then the sector is that fraction of the whole circle:

$$A_{\text{sector}} = \frac{\theta}{360} \pi r^2$$

This formula is the main tool for sweeping-area problems. If something rotates through only part of a full turn, we use the angle to find what fraction of the circle is covered.

2. What does “sweeping area” mean?

When a rod, arm, or blade rotates, every point on it moves in a circular path around the pivot. The furthest point traces part of a circle. The whole moving piece can clear a region shaped like a sector or a ring-shaped sector.

For example, if a sprinkler sprays water up to 8 m away and rotates through \(120^\circ\), the watered region is a sector of radius 8 m and angle \(120^\circ\).

If a windshield wiper has length 45 cm and turns through \(80^\circ\), the tip traces a sector. If the wiper does not start at the exact center of the windshield or if the blade has thickness, the cleared region may be better modeled as a ring-shaped sector.

3. Two common models

Model A: A simple sector

Use this when the rotating object starts at the center and covers everything from the pivot out to radius \(r\).

Formula:

$$A = \frac{\theta}{360} \pi r^2$$

Model B: An annular sector

This is a sector of a large circle with a smaller sector removed. It looks like a curved strip.

This model is useful when only the part between two radii is covered, such as a wiper blade that clears between an inner radius and an outer radius.

If the outer radius is \(R\), the inner radius is \(r\), and the angle is \(\theta\), then

$$A = \frac{\theta}{360} \pi (R^2 - r^2)$$

4. Steps for solving sweeping-area problems

1. Identify the center of rotation. This is the pivot point.

2. Find the angle turned. Be careful to use the angle actually swept, not the angle left over.

3. Find the radius or radii. Use the distance from the pivot to the edge of the region.

4. Choose the correct formula. Use a sector for one radius, or an annular sector for two radii.

5. Substitute carefully and simplify. Include correct square units such as cm\(^2\) or m\(^2\).

5. Important details to watch for

• Degrees matter. Most 10th Grade problems use degrees, so use \(\frac{\theta}{360}\).

• Square the radius. In area formulas, use \(r^2\), not just \(r\).

• Use the cleared region, not the object length alone. Sometimes there is an inner radius and an outer radius.

• Check units. If the radius is in meters, the area is in square meters.

• Estimate reasonableness. A sector must have area less than the full circle.

Worked Example 1: Sprinkler watering a sector

A sprinkler sprays water 6 m from its center and rotates through \(90^\circ\). Find the area of grass watered.

Step 1: Identify the shape

The watered region is a sector with radius \(r = 6\) m and angle \(\theta = 90^\circ\).

Step 2: Use the sector formula

$$A = \frac{\theta}{360} \pi r^2$$

$$A = \frac{90}{360} \pi (6^2)$$

$$A = \frac{1}{4} \pi (36)$$

$$A = 9\pi$$

So the area watered is

$$9\pi \text{ m}^2 \approx 28.3 \text{ m}^2$$

Worked Example 2: Windshield wiper clearing a sector

A wiper blade is 40 cm long and rotates through \(72^\circ\). Model the cleared region as a sector. Find the area cleared.

Step 1: Write the known values

Radius: \(r = 40\) cm

Angle: \(\theta = 72^\circ\)

Step 2: Apply the formula

$$A = \frac{72}{360} \pi (40^2)$$

$$A = \frac{1}{5} \pi (1600)$$

$$A = 320\pi$$

Therefore, the area cleared is

$$320\pi \text{ cm}^2 \approx 1005.3 \text{ cm}^2$$

Worked Example 3: Wiper blade with inner and outer radius

A windshield wiper clears a curved strip between 10 cm and 45 cm from the pivot. It rotates through \(80^\circ\). Find the area cleared.

Step 1: Identify the model

This is not a full sector from the center. It is an annular sector.

Outer radius: \(R = 45\) cm

Inner radius: \(r = 10\) cm

Angle: \(\theta = 80^\circ\)

Step 2: Use the annular sector formula

$$A = \frac{\theta}{360} \pi (R^2 - r^2)$$

$$A = \frac{80}{360} \pi (45^2 - 10^2)$$

$$A = \frac{2}{9} \pi (2025 - 100)$$

$$A = \frac{2}{9} \pi (1925)$$

$$A = \frac{3850}{9}\pi$$

So the area cleared is

$$\frac{3850}{9}\pi \text{ cm}^2 \approx 1344.5 \text{ cm}^2$$

Worked Example 4: Comparing two rotating machines

Machine A has an arm of length 5 m rotating through \(120^\circ\). Machine B has an arm of length 4 m rotating through \(180^\circ\). Which machine sweeps the greater area?

Machine A

$$A_A = \frac{120}{360} \pi (5^2)$$

$$A_A = \frac{1}{3} \pi (25) = \frac{25\pi}{3}$$

$$A_A \approx 26.2 \text{ m}^2$$

Machine B

$$A_B = \frac{180}{360} \pi (4^2)$$

$$A_B = \frac{1}{2} \pi (16) = 8\pi$$

$$A_B \approx 25.1 \text{ m}^2$$

Since \(26.2 > 25.1\), Machine A sweeps the greater area.

6. How to model real situations

In word problems, the hardest part is often deciding what shape to use. Ask yourself these questions:

• Is the motion a rotation around one fixed point?

• Does the moving part cover everything from the center outward, or only a strip?

• What is the maximum distance reached from the pivot?

• What angle does it turn through?

If the object rotates and covers a pie-slice shape, use a sector. If it covers a band between two circular arcs, use an annular sector.

7. Common mistakes

• Using arc length instead of area. Area needs \(\pi r^2\), while arc length uses a different formula.

• Forgetting to square the radii. In \(R^2 - r^2\), both radii must be squared first.

• Subtracting angles incorrectly. Make sure you use the angle actually swept.

• Mixing units. Convert first if one measurement is in cm and another is in m.

• Using the full-circle area by mistake. A sweep is usually only part of a circle.

8. Quick check idea

If the angle doubles and the radius stays the same, the swept area doubles.

If the radius doubles and the angle stays the same, the area becomes four times as large, because area depends on \(r^2\).

This helps you see whether an answer makes sense.

9. Practice-style questions to think about

• A sprinkler rotates through \(150^\circ\) and sprays 10 m. What area is watered?

• A radar arm of length 7 m sweeps through \(45^\circ\). What area is covered?

• A wiper clears between 8 cm and 30 cm from its pivot through \(100^\circ\). What area is cleared?

Brief Summary

Rotating machines often sweep out regions that can be modeled with sectors or annular sectors. For a sector, use

$$A = \frac{\theta}{360} \pi r^2$$

For a ring-shaped sweep, use

$$A = \frac{\theta}{360} \pi (R^2 - r^2)$$

The key is to identify the angle turned, the radius or radii, and the shape of the cleared region. Once you choose the correct model, these real-life rotating-motion problems become straightforward circle-area questions.