Is it correct to say: Giving the sector area as the answer when the segment is asked.?

Segment = sector − area of ΔOAB. The segment is bounded by the chord, not by the two radii.

Is it correct to say: Mixing the two formulas: using πr² for arc length or 2πr for sector area.?

Arc length is a length: (θ/360°) × 2πr, in cm. Sector area is an area: (θ/360°) × πr², in cm².

Is it correct to say: Taking the area of ΔOAB as ½ r² when θ = 60°.?

½ r² is only for θ = 90°. For θ = 60° the triangle is equilateral: area = (√3/4) r².

Is it correct to say: Giving only the arc length when the perimeter of a sector is asked.?

Perimeter of a sector = arc length + 2r.

Is it correct to say: Using the diameter in place of the radius.?

If the diameter is given, halve it before substituting. If the circumference is given, use r = C/(2π).

Is it correct to say: Using the minor angle θ for the major sector.?

The major sector has angle 360° − θ. Equivalently, subtract the minor sector from πr².

📄 Formula sheet (PDF)

Areas Related to Circles

Length of an arc, area and perimeter of a sector, and areas of minor and major segments of a circle with simple applications — NCERT Class 10 Maths Ch 11

📐 12 formulas✏️ 4 examples🎯 6 practice⚖️ Unit VI (Mensuration): 10 marks (with Ch 12)🏫 CBSE📚 Class 10✓ 2025–26 syllabus

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Board Exam Tips

→Write θ/360° as a fraction and cancel it first (30° → 1/12, 45° → 1/8, 60° → 1/6, 90° → 1/4). The arithmetic becomes much shorter.
→Use π = 22/7 or π = 3.14 exactly as the question says. If √3 is needed, use the value given (often 1.73).
→Segment = sector − triangle. For θ = 60° the triangle is equilateral (area √3/4 r²); for θ = 90° it is right-angled (area ½ r²); for θ = 120° draw the perpendicular from O to the chord (area √3/4 r²).
→For a major sector or major segment, subtract the minor part from the whole circle πr².
→Clock questions: the minute hand turns 6° per minute, so in t minutes it sweeps a sector of angle 6t°.
→Units: cm for arc length and perimeter, cm² for every area.

📐 Formulas(12)

Circumference of a Circle

C = 2 π r = π d

Symbol	Meaning
$C$	Circumference (cm)
$r$	Radius (cm)
$d$	Diameter, d = 2r (cm)

Area of a Circle

A = π r^{2}

Length of an Arc★ Board fav

l = \frac{θ}{36 0 ^{\circ}} \times 2 π r

Symbol	Meaning
$l$	Length of the arc (cm)
$θ$	Central angle of the sector (degrees)
$r$	Radius (cm)

Area of a Sector★ Board fav

A = \frac{θ}{36 0 ^{\circ}} \times π r^{2}

Symbol	Meaning
$A$	Area of the sector (cm²)
$θ$	Central angle (degrees)
$r$	Radius (cm)

Sector Area from Arc Length

A = \frac{1}{2} l r

Symbol	Meaning
$l$	Length of the arc (cm)
$r$	Radius (cm)

Quadrant and Semicircle

A_{quadrant} = \frac{1}{4} π r^{2}, A_{semicircle} = \frac{1}{2} π r^{2}

Area of the Major Sector

A_{major} = π r^{2} - \frac{θ}{36 0 ^{\circ}} π r^{2} = \frac{36 0 ^{\circ} - θ}{36 0 ^{\circ}} π r^{2}

Perimeter of a Sector

P = 2 r + \frac{θ}{36 0 ^{\circ}} \times 2 π r

Symbol	Meaning
$P$	Perimeter of the sector (cm)
$r$	Radius (cm)
$θ$	Central angle (degrees)

Area of the Minor Segment★ Board fav

A_{segment} = \frac{θ}{36 0 ^{\circ}} π r^{2} - ar (△ O A B)

Triangle OAB for θ = 60°, 90° and 120°

θ = 6 0^{\circ} or 12 0^{\circ} : ar (△ O A B) = \frac{3}{4} r^{2}, θ = 9 0^{\circ} : ar (△ O A B) = \frac{1}{2} r^{2}

Area of the Major Segment

A_{major segment} = π r^{2} - A_{minor segment}

Angle Turned by Clock Hands

minute hand: 6^{\circ} per minute, hour hand: 0. 5^{\circ} per minute

✏️ Solved Examples

1Solved Exampleeasy3 steps

Find the length of the arc and the area of the sector of a circle of radius 14 cm whose central angle is 45°. (Take π = 22/7.)

The fraction of the circle is 45°/360° = 1/8.

2Solved Exampleboard4 steps

A chord of a circle of radius 14 cm subtends a right angle at the centre. Find the areas of the minor segment and the major segment. (Take π = 22/7.)

Area of the minor sector with θ = 90°

\frac{1}{4} \times \frac{22}{7} \times 1 4^{2} = \frac{616}{4} = 154 cm^{2}

3Solved Exampleboard3 steps

A chord of a circle of radius 12 cm subtends an angle of 60° at the centre. Find the area of the minor segment. (Use π = 3.14 and √3 = 1.73.)

Area of the sector with θ = 60° (fraction 1/6)

\frac{1}{6} \times 3.14 \times 144 = 75.36 cm^{2}

4Solved ExampleHOTS4 steps

Three circles, each of radius 7 cm, are drawn with their centres at the vertices of an equilateral triangle of side 14 cm, so that each circle touches the other two. Find the area of the region enclosed between the three circles. (Use π = 22/7 and √3 = 1.73.)

The enclosed region = area of the triangle − the three sectors inside it. Each angle of an equilateral triangle is 60°.

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Giving the sector area as the answer when the segment is asked.
✓Segment = sector − area of ΔOAB. The segment is bounded by the chord, not by the two radii.
2
Mixing the two formulas: using πr² for arc length or 2πr for sector area.
✓Arc length is a length: (θ/360°) × 2πr, in cm. Sector area is an area: (θ/360°) × πr², in cm².
3
Taking the area of ΔOAB as ½ r² when θ = 60°.
✓½ r² is only for θ = 90°. For θ = 60° the triangle is equilateral: area = (√3/4) r².
4
Giving only the arc length when the perimeter of a sector is asked.
✓Perimeter of a sector = arc length + 2r.
5
Using the diameter in place of the radius.
✓If the diameter is given, halve it before substituting. If the circumference is given, use r = C/(2π).
6
Using the minor angle θ for the major sector.
✓The major sector has angle 360° − θ. Equivalently, subtract the minor sector from πr².

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
Find the area of a quadrant of a circle whose circumference is 44 cm. (π = 22/7)
Q2
The minute hand of a clock is 21 cm long. Find the area swept by it in 10 minutes. (π = 22/7)
Q3
Find the perimeter of a sector of a circle of radius 7 cm with central angle 90°. (π = 22/7)
Q4
A horse is tied to a peg at one corner of a square field of side 20 m by a rope 14 m long. Find the area of the field the horse can graze and the area it cannot graze. (π = 22/7)
Q5
A chord of a circle of radius 20 cm subtends a right angle at the centre. Find the area of the major segment. (π = 3.14)
Q6
The arc length of a sector is 22 cm and its area is 231 cm². Find the radius and the central angle. (π = 22/7)

📝 Notes

Areas Related to Circles

Every formula in this chapter takes a fraction of a whole circle. Once you know the central angle θ, the fraction is θ/360°.

Arc, sector and segment

Arc length is that fraction of the circumference: $l = \frac{\theta}{360^\circ} \times 2\pi r$ .
Sector area is that fraction of the circle's area: $A = \frac{\theta}{360^\circ} \times \pi r^2$ .
Segment area is the sector minus triangle OAB, because the chord AB cuts the triangle off the sector.

For the major sector or major segment, either use the angle 360° − θ or subtract the minor part from πr². Subtracting is usually quicker.

The triangle inside the sector

The only extra work in a segment question is the area of ΔOAB. The CBSE syllabus limits segment problems to three central angles:

θ = 90°: a right isosceles triangle, area ½ r².
θ = 60°: an equilateral triangle of side r, area (√3/4) r².
θ = 120°: draw OM ⟂ AB, so ∠AOM = 60°, OM = r/2 and AB = √3 r; area (√3/4) r².

Write the triangle's name and the reason ("OA = OB = r and ∠AOB = 60°, so ΔOAB is equilateral") before substituting.

Circles combined with triangles and squares

Applied questions combine a circle with a simple figure: a horse tied at a corner of a square field, sectors drawn at the vertices of a triangle, a clock hand sweeping a sector, wipers or a lighthouse beam covering a sector. In each case:

Identify the radius (rope, hand, wiper or beam length) and the central angle.
Decide whether the region is a sector, a segment, or a figure minus some sectors.
Compute each piece and add or subtract.

At the vertices of any triangle the three angles add to 180°, so equal-radius sectors there always make half a circle.