Class 10 Maths Areas Related to Circles Notes

Class 10 Maths Areas Related to Circles Notes

Class 10 Maths — Chapter 11: Areas Related to Circles

1. Sector of a Circle

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

  • Minor sector: smaller part of the circle.
  • Major sector: larger part.
  • If the minor sector has angle θ\theta, the major sector has angle:

360∘−θ360^\circ-\theta

2. Segment of a Circle

A segment is the region enclosed by a chord and its corresponding arc.

  • Minor segment: smaller region.
  • Major segment: larger region.

3. Must-Know Formulas

For a circle of radius rr, with sector angle θ∘\theta^\circ:

Area of sector

Area of sector=θ360∘πr2\boxed{\text{Area of sector}=\frac{\theta}{360^\circ}\pi r^2}

Length of corresponding arc

Arc length=θ360∘ 2πr\boxed{\text{Arc length}=\frac{\theta}{360^\circ}\,2\pi r}

These are the two most important formulas of the chapter.

Area of a segment

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

So, whenever a question asks for a segment, first identify the corresponding sector and triangle.

4. Major Sector & Major Segment

Major sector

Major sector area=πr2−minor sector area\boxed{\text{Major sector area} =\pi r^2-\text{minor sector area}}

or directly,360∘−θ360∘πr2\boxed{\frac{360^\circ-\theta}{360^\circ}\pi r^2}

Major segment

Major segment area=πr2−minor segment area\boxed{\text{Major segment area} =\pi r^2-\text{minor segment area}}

5. Finding the Triangle Area in Segment Problems

When a chord subtends an angle at the centre, the triangle formed by the two radii and the chord is important.

For example, if ∠AOB=120∘\angle AOB=120^\circ, draw a perpendicular from OO to chord ABAB. Since OA=OBOA=OB, this perpendicular bisects the chord and the central angle.

This converts the problem into right triangles, where trigonometric ratios such as sin⁡\sin and cos⁡\cos can be used.

6. Problem-Solving Pattern

If asked for area of a sector:

  1. Identify rr and θ\theta.
  2. Use

θ360πr2\frac{\theta}{360}\pi r^2

If asked for arc length:

Useθ360(2πr)\frac{\theta}{360}(2\pi r)

If asked for minor segment:

Sector area−Triangle area\boxed{\text{Sector area}-\text{Triangle area}}

If asked for major segment:

πr2−minor segment area\boxed{\pi r^2-\text{minor segment area}}

If asked for a major sector:

πr2−minor sector area\boxed{\pi r^2-\text{minor sector area}}

7. Special Cases to Remember

Quadrant

A quadrant is a sector of:θ=90∘\theta=90^\circ

Therefore,Area of quadrant=14πr2\text{Area of quadrant}=\frac14\pi r^2

Semicircle

A semicircle corresponds to:θ=180∘\theta=180^\circ

Therefore,Area of semicircle=12πr2\text{Area of semicircle}=\frac12\pi r^2

8. Exam Tips

  • Sector → use the sector formula.
  • Arc → use the arc-length formula.
  • Segment → sector − triangle.
  • Major part → whole circle − corresponding minor part.
  • Always check whether the angle is in degrees.
  • Use the value of π\pi specified in the question.
  • In word problems involving clocks, wipers, umbrellas, grazing areas, lighthouse beams, etc., identify the sector swept out before calculating. The chapter’s exercises specifically apply sectors to these real-life situations.

Revision

Arc=θ360(2πr)Sector=θ360πr2Segment=Sector−Triangle\boxed{\text{Arc}=\frac{\theta}{360}(2\pi r)} \qquad \boxed{\text{Sector}=\frac{\theta}{360}\pi r^2} \qquad \boxed{\text{Segment}=\text{Sector}-\text{Triangle}}