Take a circle and cut it with two radii: you get a slice that looks like a piece of pizza. Cut it with a single chord instead: you get a cap shaped like a bowl lid. The first is a sector and the second is a segment. They sound similar, they sit in the same chapter, and they are mixed up in almost every Class 10 paper. The mix-up is rarely about the words. It is about the formulas, because the area of a segment is built from the area of a sector, and the step that joins them is the one students forget.
This article separates the two shapes cleanly, gives the formulas with the reason behind each, lists the slips that cost marks, and then works through examples with the arithmetic checked. If you want the chapter theory alongside, the page on areas of sector and segment of a circle and the full chapter with formulas and questions are good companions.
Telling the Two Shapes Apart in Five Seconds
A sector is bounded by two radii and an arc. Its corner is at the centre of the circle. A segment is bounded by a chord and an arc. It has no corner at the centre at all.
A simple test: if the shape has a point at the centre, it is a sector. If it is cut off by a straight line that does not pass through the centre, it is a segment.
Both come in two sizes. The smaller one, with a central angle below 180 degrees, is the minor sector or segment. The larger one is the major sector or segment. A minor sector and the major sector next to it together make the whole circle, and the same is true of the two segments formed by one chord. That fact solves a surprising number of problems, so keep it in your head.
The angle at the centre, usually called theta, is the quantity that connects everything in the chapter. All the formulas below take a fraction of the full circle, and that fraction is theta divided by 360.
The Formulas, and Why They Look the Way They Do
A full circle has area pi r squared and circumference 2 pi r. A sector with central angle theta is the fraction theta over 360 of the circle, so:
- Area of sector = (theta / 360) x pi r squared
- Length of arc = (theta / 360) x 2 pi r
- Perimeter of sector = 2r + arc length
A segment is what is left when you remove the triangle from the sector. That triangle is formed by the two radii and the chord, so:
- Area of minor segment = area of sector − area of triangle
- Area of major segment = area of circle − area of minor segment
- Perimeter of segment = arc length + chord length
The area of the triangle depends on the angle. For theta = 90 degrees, it is a right triangle with both legs equal to r, so its area is half r squared. For theta = 60 degrees, the triangle is equilateral with side r, so its area is root 3 over 4 times r squared. For theta = 120 degrees, you can drop a perpendicular from the centre to the chord and use the trigonometric ratios of 60 and 30 degrees. A general formula is half r squared sin theta, and many boards expect it only through these special cases, so check which approach your textbook uses.
For related practice on the whole chapter, the important questions for Class 10 Maths Chapter 12 are worth running through after you have this section clear.
The Mistakes That Cost the Most Marks
Using the sector formula and calling it a segment
This is the big one. A question asks for the area of a segment, and the student writes the sector formula and stops. The segment is smaller than the sector by exactly the triangle. If your answer for a segment equals the sector area, you have missed the subtraction.
Using half r squared for every triangle
The shortcut half r squared is true only when the central angle is 90 degrees. For 60 degrees the triangle is equilateral and its area is root 3 over 4 times r squared. Students who apply half r squared to a 60 degree segment get an answer that is too large, and the resulting segment area may even come out negative-looking or suspiciously large. Always ask: what is the angle, and what triangle does that give?
Mixing the arc formula with the area formula
Arc length is linear and uses 2 pi r. Area is square and uses pi r squared. A quick check on units helps: arc length is in centimetres, area is in square centimetres. If you write an arc length with r squared in it, something has gone wrong.
Forgetting what the major segment is
The major segment is not the major sector minus a triangle. It is the whole circle minus the minor segment. Equivalent, but easier to get right: circle area minus minor segment area. Some students also compute the major segment by subtracting the triangle from the major sector, which gives a value that is too small, because for the major side the triangle is added, not subtracted.
Wrong perimeters
The perimeter of a sector includes the two radii as well as the arc. The perimeter of a segment includes the chord, not the radii. Writing 2r plus arc for a segment, or arc alone for a sector, is a regular slip. Draw the boundary and trace it with a finger before writing the formula.
Reading the wrong angle
The angle in the formula is the angle at the centre. If a question gives the angle at the circumference, you have to double it. And if a question describes the major sector by its angle, the minor sector angle is 360 minus that value, so be clear which one the question asks about.
Not identifying what is shaded
In combination figures, such as a square with quarter circles at its corners, students start calculating before they have decided what the shaded part actually is. Write the shaded area in words first, as one shape minus another, and only then substitute numbers.
Choosing the wrong value of pi
When the radius is a multiple of 7, use 22/7. When the radius is something like 10 or 5, the question usually gives 3.14. Check the instruction in the question, because using the wrong value changes the final digits and a marking scheme may not accept it.
Worked Examples With the Arithmetic Checked
Use pi = 22/7 unless stated otherwise.
Example 1: sector, arc and perimeter. A sector of radius 7 cm has a central angle of 90 degrees.
Area = (90/360) x (22/7) x 49 = 38.5 square cm. Arc length = (90/360) x 2 x (22/7) x 7 = 11 cm. Perimeter = 7 + 7 + 11 = 25 cm.
Check: working backwards, arc 11 cm over the full circumference 44 cm gives 1/4, which is 90 degrees. This matches.
Example 2: minor segment at 90 degrees. A chord of a circle of radius 14 cm subtends a right angle at the centre. Find the area of the minor segment.
Sector = (90/360) x (22/7) x 196 = 154 square cm. Triangle = half x 14 x 14 = 98 square cm. Minor segment = 154 − 98 = 56 square cm.
Example 3: the major segment for the same chord. Circle area = (22/7) x 196 = 616 square cm. Major segment = 616 − 56 = 560 square cm.
Check: minor plus major = 56 + 560 = 616, the whole circle.
Example 4: segment at 60 degrees. Radius 10 cm, central angle 60 degrees. Use pi = 3.14 and root 3 = 1.73.
Sector = (60/360) x 3.14 x 100 = 52.33 square cm (to two decimals). The triangle with two radii as sides and a 60 degree angle between them is equilateral with side 10 cm, so its area = (root 3 / 4) x 100 = 43.25 square cm. Minor segment = 52.33 − 43.25 = 9.08 square cm.
Check: the segment is much smaller than the sector, which is what you would expect for a narrow cap.
Example 5: clock hand. The minute hand of a clock is 14 cm long. Find the area swept in 5 minutes.
The minute hand turns 6 degrees per minute, so in 5 minutes it turns 30 degrees. Area = (30/360) x (22/7) x 196 = 616/12 = 51.33 square cm (approximately).
Example 6: finding the angle. A sector of a circle of radius 14 cm has area 154 square cm. Find its central angle.
154 = (theta/360) x 616, so theta/360 = 154/616 = 1/4, and theta = 90 degrees.
Notice how each example is a small variation on one idea: take a fraction of the circle, then adjust. The chapter notes on areas related to circles show how the same method extends to combination figures.
A Quick Way to Decide Which Formula You Need
Ask three questions in order. First, does the shape touch the centre? If yes, it is a sector and you are done after one fraction. If no, it is a segment. Second, is the angle 90, 60 or 120 degrees? That tells you which triangle to subtract. Third, is it the minor or the major piece? If it is the major piece, find the minor and subtract from the circle.
For revision, the MCQs on areas related to circles give quick checks on exactly these distinctions.
Where This Fits in Your Syllabus
Areas Related to Circles is Chapter 12 in the CBSE Class 10 Maths numbering used on the chapter pages linked here. In the Karnataka SSLC textbook the topics may be arranged under a different chapter number, so follow your own book. Boards also differ in whether area-of-combination-figure problems are part of the current syllabus, and in whether the half r squared sin theta formula is expected. Please confirm both points with your board’s current syllabus and sample papers.
The chapter links backwards to tangent properties and the proof mistakes students make, and forwards to surface areas and volumes, where the same circle formulas become the base and the curved surface of solids.
Frequently Asked Questions
Q1. What is the difference between a sector and a segment of a circle?
A sector is bounded by two radii and an arc, so it has a corner at the centre. A segment is bounded by a chord and an arc and has no corner at the centre.
Q2. How do you find the area of a segment?
Find the area of the sector with the same central angle, then subtract the area of the triangle formed by the two radii and the chord. For a major segment, subtract the minor segment from the area of the whole circle.
Q3. What is the perimeter of a sector?
Two radii plus the arc length, that is 2r + (theta/360) x 2 pi r. The chord is not part of a sector’s boundary.
Q4. Is the area of a segment ever equal to the area of a sector?
No. A segment is always smaller than the sector with the same central angle, because the triangle has been removed from it. If your answers come out equal, the subtraction was missed.
Q5. How do I find the area swept by a clock hand?
Find the angle turned. A minute hand turns 6 degrees per minute, so use that angle with the hand length as the radius in the sector formula.
Conclusion: One Fraction, One Subtraction
Every question in this chapter comes down to taking a fraction of a circle and, for a segment, removing a triangle. Once those two moves are automatic, the sector and segment confusion fades.
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