Areas Related to Circles Notes

12

Areas Related to Circles

Class 10 Mathematics • Complete Board Exam Notes

Easy English • Step-by-Step Solutions • Formula Focus • Exam Ready

Join WhatsApp Join Telegram University Scope

Chapter Roadmap

1. Circle Basics & Formulas 2. Circumference of a Circle 3. Area of a Circle 4. Sector of a Circle 5. Length of an Arc 6. Segment of a Circle 7. Areas of Combined Figures 8. Board-Exam Questions 9. Ten Detailed Solved Questions 10. Exam Strategy 11. Quick Revision

✍️ For More Notes Visit UniversityScope.com ✍️

Chapter Goal

This chapter is about finding the lengths and areas connected with circles. The most important skills are using the correct formula, substituting values carefully, handling sectors and segments, and solving questions involving shaded or combined regions.

1. Circle Basics & Important Formulas

1.1 Radius and Diameter

The distance from the centre of a circle to any point on the circle is its radius, denoted by \(r\). The diameter \(d\) passes through the centre and joins two points of the circle.

\[ \boxed{d=2r} \qquad \boxed{r=\frac d2} \]

1.2 Circumference

The circumference is the distance around the circle.

\[ \boxed{C=2\pi r=\pi d} \]

1.3 Area

\[ \boxed{A=\pi r^2} \]

1.4 Useful Values of \(\pi\)

Situation	Common Value
Exact form	\(\pi\)
Common approximation	\(\frac{22}{7}\)
Decimal approximation	\(3.14159\ldots\)

Exam tip: Use the value of \(\pi\) specified in the question. If no value is specified, use the form that makes the calculation simplest and is consistent with the data.

Solved Example 1 — Diameter from Radius

The radius of a circle is \(7\) cm. Find its diameter.

1. Use: \[ d=2r. \]
2. Substitute: \[ d=2(7)=14. \]

Answer: \(14\) cm

Solved Example 2 — Radius from Diameter

The diameter of a circle is \(20\) cm. Find its radius.

1. Use: \[ r=\frac d2. \]
2. Therefore: \[ r=\frac{20}{2}=10. \]

Answer: \(10\) cm

Solved Example 3 — Identify the Formula

Which formula gives the area of a circle of radius \(r\)?

1. Area of a circle is: \[ A=\pi r^2. \]

Answer: \(A=\pi r^2\)

2. Circumference of a Circle

2.1 Formula

\[ \boxed{C=2\pi r} \] \[ \boxed{C=\pi d} \]

Choose the formula according to the information given. If radius is given, \(2\pi r\) is direct. If diameter is given, \(\pi d\) is often quicker.

Solved Example 1 — Radius Given

Find the circumference of a circle of radius \(7\) cm. Take \(\pi=\frac{22}{7}\).

1. Use: \[ C=2\pi r. \]
2. Substitute: \[ C=2\times\frac{22}{7}\times7. \]
3. Cancel \(7\): \[ C=44. \]

Answer: \(44\) cm

Solved Example 2 — Diameter Given

Find the circumference of a circle whose diameter is \(14\) cm. Take \(\pi=\frac{22}{7}\).

1. Use: \[ C=\pi d. \]
2. Substitute: \[ C=\frac{22}{7}\times14. \]
3. Therefore: \[ C=44. \]

Answer: \(44\) cm

Solved Example 3 — Find Radius from Circumference

The circumference of a circle is \(88\) cm. Find its radius. Take \(\pi=\frac{22}{7}\).

1. Use: \[ C=2\pi r. \]
2. Substitute: \[ 88=2\times\frac{22}{7}\times r. \]
3. Thus: \[ 88=\frac{44r}{7}. \]
4. Multiply by \(7\): \[ 616=44r. \]
5. Therefore: \[ r=14. \]

Answer: \(14\) cm

3. Area of a Circle

3.1 Formula

\[ \boxed{A=\pi r^2} \]

3.2 Area from Diameter

If diameter \(d\) is given, first find \(r=\frac d2\), then use \(A=\pi r^2\).

Solved Example 1 — Radius Given

Find the area of a circle of radius \(14\) cm. Take \(\pi=\frac{22}{7}\).

1. Use: \[ A=\pi r^2. \]
2. Substitute: \[ A=\frac{22}{7}\times14^2. \]
3. Calculate: \[ A=\frac{22}{7}\times196=616. \]

Answer: \(616\text{ cm}^2\)

Solved Example 2 — Diameter Given

Find the area of a circle of diameter \(14\) cm. Take \(\pi=\frac{22}{7}\).

1. Find radius: \[ r=\frac{14}{2}=7\text{ cm}. \]
2. Use: \[ A=\pi r^2. \]
3. Therefore: \[ A=\frac{22}{7}\times49=154. \]

Answer: \(154\text{ cm}^2\)

Solved Example 3 — Difference of Areas

Find the area of the region between two concentric circles of radii \(10\) cm and \(6\) cm.

1. Outer circle area: \[ \pi(10)^2=100\pi. \]
2. Inner circle area: \[ \pi(6)^2=36\pi. \]
3. Required area: \[ 100\pi-36\pi=64\pi. \]

Answer: \(64\pi\text{ cm}^2\)

4. Sector of a Circle

4.1 Meaning of Sector

A sector is the region enclosed by two radii and the corresponding arc of a circle.

Minor sector: generally has central angle less than \(180^\circ\).
Major sector: generally has central angle greater than \(180^\circ\).

4.2 Area of a Sector

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

\[ \boxed{\text{Area of sector}=\frac{\theta}{360^\circ}\pi r^2} \]

4.3 Perimeter of a Sector

The perimeter includes the two radii and the arc.

\[ \boxed{\text{Perimeter of sector}=2r+\text{arc length}} \]

Solved Example 1 — Area of \(90^\circ\) Sector

Find the area of a sector of angle \(90^\circ\) and radius \(14\) cm. Take \(\pi=\frac{22}{7}\).

1. Use: \[ A_s=\frac{\theta}{360}\pi r^2. \]
2. Substitute: \[ A_s=\frac{90}{360}\times\frac{22}{7}\times14^2. \]
3. Since \(\frac{90}{360}=\frac14\): \[ A_s=\frac14\times616=154. \]

Answer: \(154\text{ cm}^2\)

Solved Example 2 — Area of \(60^\circ\) Sector

Find the area of a \(60^\circ\) sector of radius \(21\) cm. Take \(\pi=\frac{22}{7}\).

1. Use: \[ A_s=\frac{60}{360}\times\frac{22}{7}\times21^2. \]
2. Simplify: \[ A_s=\frac16\times\frac{22}{7}\times441. \]
3. Therefore: \[ A_s=231. \]

Answer: \(231\text{ cm}^2\)

Solved Example 3 — Find Sector Angle

A sector has area \(\frac14\pi r^2\). Find its central angle.

1. Use: \[ \frac{\theta}{360}\pi r^2=\frac14\pi r^2. \]
2. Cancel \(\pi r^2\): \[ \frac{\theta}{360}=\frac14. \]
3. Thus: \[ \theta=90^\circ. \]

Answer: \(90^\circ\)

5. Length of an Arc

5.1 Formula

An arc is a part of the circumference. If the central angle is \(\theta^\circ\), the arc length is the same fraction of the complete circumference as \(\theta/360\).

\[ \boxed{\text{Arc length}=\frac{\theta}{360^\circ}\times2\pi r} \]

Solved Example 1 — Semicircular Arc

Find the length of a semicircular arc of radius \(7\) cm. Take \(\pi=\frac{22}{7}\).

1. A semicircle has angle \(180^\circ\).
2. Use: \[ L=\frac{180}{360}\times2\pi r. \]
3. Therefore: \[ L=\frac12\times2\times\frac{22}{7}\times7=22. \]

Answer: \(22\) cm

Solved Example 2 — Quarter Arc

Find the length of a \(90^\circ\) arc of radius \(14\) cm. Take \(\pi=\frac{22}{7}\).

1. Use: \[ L=\frac{90}{360}\times2\pi r. \]
2. Thus: \[ L=\frac14\times2\times\frac{22}{7}\times14. \]
3. Therefore: \[ L=22. \]

Answer: \(22\) cm

Solved Example 3 — \(120^\circ\) Arc

Find the length of a \(120^\circ\) arc of a circle of radius \(21\) cm. Take \(\pi=\frac{22}{7}\).

1. Use: \[ L=\frac{120}{360}\times2\pi(21). \]
2. Simplify: \[ L=\frac13\times42\pi=14\pi. \]
3. Using \(\pi=\frac{22}{7}\): \[ L=44. \]

Answer: \(44\) cm

6. Segment of a Circle

6.1 Meaning of Segment

A segment of a circle is the region enclosed between a chord and its corresponding arc.

Minor segment: the smaller region cut off by a chord.
Major segment: the larger region cut off by the chord.

6.2 Area of a Minor Segment

\[ \boxed{\text{Area of segment}=\text{Area of sector}-\text{Area of triangle}} \]

This formula is especially useful when a chord, two radii and a sector are involved.

Solved Example 1 — Segment Formula

A minor segment is formed by a sector and a triangle. If the sector area is \(100\text{ cm}^2\) and the triangle area is \(36\text{ cm}^2\), find the segment area.

1. Use: \[ \text{Segment area}=\text{Sector area}-\text{Triangle area}. \]
2. Therefore: \[ 100-36=64. \]

Answer: \(64\text{ cm}^2\)

Solved Example 2 — Semicircle Segment Idea

If a diameter divides a circle into two equal regions, what is the area of either region?

1. A diameter divides a circle into two semicircles.
2. Each semicircle is half the area of the circle.
3. Thus: \[ A_{\text{semicircle}}=\frac12\pi r^2. \]

Answer: \(\frac12\pi r^2\)

Solved Example 3 — Major Segment

If the complete circle has area \(200\text{ cm}^2\) and the minor segment has area \(35\text{ cm}^2\), find the major segment area.

1. The two segments together form the complete circle.
2. Therefore: \[ \text{Major segment}=200-35=165. \]

Answer: \(165\text{ cm}^2\)

7. Areas of Combined Figures

7.1 Basic Method

Many board questions show a figure made from circles, semicircles, quadrants, rectangles, squares or triangles. Do not try to find the shaded area in one step.

1. Identify the complete shapes present.
2. Write the formula for each shape.
3. Find the area of the required larger region.
4. Subtract unwanted regions or add required regions.
5. Keep units as square units.

\[ \boxed{\text{Required area}=\text{Area added}-\text{Area removed}} \]

Solved Example 1 — Square and Circle

A square has side \(14\) cm. A circle of radius \(7\) cm is drawn inside it. Find the area of the part of the square outside the circle. Take \(\pi=\frac{22}{7}\).

1. Area of square: \[ 14^2=196. \]
2. Area of circle: \[ \frac{22}{7}\times7^2=154. \]
3. Required area: \[ 196-154=42. \]

Answer: \(42\text{ cm}^2\)

Solved Example 2 — Two Semicircles

Two semicircles of radius \(7\) cm together form a circle. Find their total area.

1. Two semicircles make one complete circle.
2. Use: \[ A=\pi r^2. \]
3. Therefore: \[ A=\frac{22}{7}\times49=154. \]

Answer: \(154\text{ cm}^2\)

Solved Example 3 — Ring Region

Two concentric circles have radii \(14\) cm and \(7\) cm. Find the area of the ring between them. Take \(\pi=\frac{22}{7}\).

1. Outer area: \[ \frac{22}{7}\times14^2=616. \]
2. Inner area: \[ \frac{22}{7}\times7^2=154. \]
3. Subtract: \[ 616-154=462. \]

Answer: \(462\text{ cm}^2\)

8. Board-Exam Questions

Board Pattern 1

Question: Find the area and circumference of a circle of radius \(7\) cm.

\[ A=\pi r^2=\frac{22}{7}\times49=154\text{ cm}^2 \] \[ C=2\pi r=2\times\frac{22}{7}\times7=44\text{ cm} \]

Answer: Area \(=154\text{ cm}^2\), circumference \(=44\) cm.

Board Pattern 2

Question: Find the area of a sector of angle \(60^\circ\) and radius \(21\) cm.

\[ A_s=\frac{60}{360}\pi(21)^2 =\frac16\times\frac{22}{7}\times441 =231\text{ cm}^2 \]

Answer: \(231\text{ cm}^2\)

Board Pattern 3

Question: Find the length of an arc of \(90^\circ\) in a circle of radius \(14\) cm.

\[ L=\frac{90}{360}\times2\pi(14)=22\text{ cm} \]

Answer: \(22\) cm

Board Pattern 4

Question: Find the area of a shaded region formed by subtracting a circle from a square.

Solution method: Find the square area first, find the circle area next, and subtract the circle area from the square area.

Always write: Required area = larger area − removed area.

9. Ten Detailed Solved Questions

Question 1

Find the circumference of a circle of radius \(14\) cm. Take \(\pi=\frac{22}{7}\).

1. Use: \[ C=2\pi r. \]
2. Substitute: \[ C=2\times\frac{22}{7}\times14. \]
3. Calculate: \[ C=88. \]

Answer: \(88\) cm

Question 2

Find the area of a circle of diameter \(28\) cm. Take \(\pi=\frac{22}{7}\).

1. Find radius: \[ r=\frac{28}{2}=14\text{ cm}. \]
2. Use: \[ A=\pi r^2. \]
3. Therefore: \[ A=\frac{22}{7}\times196=616. \]

Answer: \(616\text{ cm}^2\)

Question 3

Find the area of a \(90^\circ\) sector of radius \(14\) cm. Take \(\pi=\frac{22}{7}\).

1. Use: \[ A_s=\frac{\theta}{360}\pi r^2. \]
2. Substitute: \[ A_s=\frac{90}{360}\times\frac{22}{7}\times196. \]
3. Thus: \[ A_s=\frac14\times616=154. \]

Answer: \(154\text{ cm}^2\)

Question 4

Find the length of a \(120^\circ\) arc of radius \(21\) cm. Take \(\pi=\frac{22}{7}\).

1. Use: \[ L=\frac{\theta}{360}\times2\pi r. \]
2. Substitute: \[ L=\frac{120}{360}\times2\times\frac{22}{7}\times21. \]
3. Therefore: \[ L=44. \]

Answer: \(44\) cm

Question 5

A circular park has radius \(35\) m. Find its area. Take \(\pi=\frac{22}{7}\).

1. Use: \[ A=\pi r^2. \]
2. Substitute: \[ A=\frac{22}{7}\times35^2. \]
3. Calculate: \[ A=\frac{22}{7}\times1225=3850. \]

Answer: \(3850\text{ m}^2\)

Question 6

A sector has radius \(14\) cm and angle \(180^\circ\). Find its area.

1. A \(180^\circ\) sector is a semicircle.
2. Use: \[ A_s=\frac{180}{360}\pi(14)^2. \]
3. Therefore: \[ A_s=\frac12\times196\pi=98\pi. \]
4. With \(\pi=\frac{22}{7}\): \[ A_s=308. \]

Answer: \(308\text{ cm}^2\)

Question 7

The area of a circle is \(154\text{ cm}^2\). Find its radius using \(\pi=\frac{22}{7}\).

1. Use: \[ A=\pi r^2. \]
2. Substitute: \[ 154=\frac{22}{7}r^2. \]
3. Multiply by \(7\): \[ 1078=22r^2. \]
4. Divide by \(22\): \[ r^2=49. \]
5. Therefore: \[ r=7\text{ cm}. \]

Answer: \(7\) cm

Question 8

A square of side \(14\) cm contains a circle of radius \(7\) cm. Find the area of the region of the square outside the circle. Take \(\pi=\frac{22}{7}\).

1. Square area: \[ 14^2=196. \]
2. Circle area: \[ \frac{22}{7}\times49=154. \]
3. Required area: \[ 196-154=42. \]

Answer: \(42\text{ cm}^2\)

Question 9

Two concentric circles have radii \(14\) cm and \(7\) cm. Find the area of the ring between them. Take \(\pi=\frac{22}{7}\).

1. Outer area: \[ \frac{22}{7}\times196=616. \]
2. Inner area: \[ \frac{22}{7}\times49=154. \]
3. Ring area: \[ 616-154=462. \]

Answer: \(462\text{ cm}^2\)

Question 10

A circular field has radius \(14\) m. A sector of angle \(90^\circ\) is used as a garden. Find the area of the garden and the remaining field. Take \(\pi=\frac{22}{7}\).

1. Total field area: \[ A=\frac{22}{7}\times14^2=616\text{ m}^2. \]
2. Garden area: \[ A_g=\frac{90}{360}\times616=154\text{ m}^2. \]
3. Remaining area: \[ 616-154=462\text{ m}^2. \]

Answer: Garden \(=154\text{ m}^2\); remaining field \(=462\text{ m}^2\).

10. Exam Strategy

1. Write the formula first.

Do not start calculations before identifying whether the question asks for circumference, area, sector area, arc length or segment area.

2. Check radius and diameter.

If diameter is given but the formula needs radius, first write: \[ r=\frac d2. \]

3. Sector questions:

Always remember the fraction: \[ \frac{\theta}{360^\circ}. \] Then multiply it by the full-circle quantity.

4. Shaded-area questions:

Break the diagram into familiar shapes. Usually: \[ \text{Required area}=\text{area of larger region}-\text{area of smaller region}. \]

Common mistakes:

• Using diameter in place of radius in \(A=\pi r^2\).
• Forgetting the factor \(2\) in circumference \(2\pi r\).
• Using \(\theta/180\) instead of \(\theta/360\) for a sector or arc fraction.
• Forgetting that area is measured in square units.
• Adding an unwanted region instead of subtracting it.
• Not converting units before calculation.
• Using the wrong angle for a major sector or major arc.

11. Quick Revision Sheet

Concept	Formula / Key Fact
Diameter	\(d=2r\)
Radius	\(r=\frac d2\)
Circumference	\(C=2\pi r=\pi d\)
Area of Circle	\(A=\pi r^2\)
Sector Area	\(\frac{\theta}{360^\circ}\pi r^2\)
Arc Length	\(\frac{\theta}{360^\circ}2\pi r\)
Sector Perimeter	\(2r+\text{arc length}\)
Minor Segment	\(\text{sector area}-\text{triangle area}\)
Semicircle Area	\(\frac12\pi r^2\)
Semicircle Arc	\(\pi r\)
Full Circle Angle	\(360^\circ\)
Straight/semicircle Angle	\(180^\circ\)
Right Angle Sector	\(90^\circ\)
Area Units	cm², m², etc.

30-Second Final Revision

• Circle: \(C=2\pi r,\quad A=\pi r^2\).
• Diameter: \(d=2r\).
• Sector: use \(\theta/360^\circ\).
• Arc: same fraction of circumference.
• Segment: sector area − triangle area.
• Shaded region: split the figure into simple shapes.
• Always check: radius/diameter, units, \(\pi\), and subtraction/addition.

---

✍️ For More Notes Visit UniversityScope.com ✍️

Class 10 Mathematics • Chapter 12: Areas Related to Circles

Prepared in easy, exam-focused handwritten style by University Scope.

Practise formula-based, sector, arc, segment and shaded-region questions for maximum accuracy.