Areas of sectors and segments

Cut a pizza into slices and each slice is a sector; bite off a piece by chomping straight across near the crust and the crescent left on the plate is a segment. Both are pieces of a circle, and once you know the area and circumference of a full circle, finding the area of a slice or a crescent is just a matter of proportion and one triangle. This is the heart of the CBSE Class 10 chapter Areas Related to Circles, in the Mensuration unit. It builds directly on the Class 9 formulas $A = \pi r^2$ and $C = 2\pi r$ , and every calculation uses degrees for the central angle $\theta$ , with $\pi$ taken as $\dfrac{22}{7}$ or $3.14$ as the question specifies. This article shows why a sector's area is a simple fraction of the circle's area, works through a full sector-and-segment problem end to end, and flags the traps — mixing up sector and segment, and using degree fractions correctly — that cost the most marks.

The core insight

A sector's area and its arc length are both just the fraction $\theta/360^\circ$ of the whole circle, and a segment's area is that sector's area minus the triangle cut off by the chord.

The core idea

A sector is the region between two radii and the arc joining them — a "pizza slice." A full circle is a sector with central angle $360^\circ$ , so a sector with angle $\theta$ is exactly that fraction of the whole circle's area and circumference.

A_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2 \qquad \ell = \frac{\theta}{360^\circ} \times 2\pi r

where: $A_{\text{sector}}$ is the sector's area, $\ell$ is the length of its bounding arc, $r$ is the radius, and $\theta$ is the central angle in degrees. A segment is the region between a chord and the arc it cuts off — what is left of a sector once the triangle formed by the two radii and the chord is removed.

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

where: $A_{\triangle}$ is the area of the triangle formed by the two radii and the chord. For the standard angles this syllabus uses, that triangle has a known shape: at $\theta = 60^\circ$ it is equilateral with side $r$ , so $A_{\triangle} = \dfrac{\sqrt{3}}{4}r^2$ ; at $\theta = 90^\circ$ the two radii are perpendicular, so the triangle is right-angled and isosceles with $A_{\triangle} = \dfrac{1}{2}r^2$ . Every chord splits the circle into two segments. The smaller one (central angle less than $180^\circ$ ) is the minor segment; the rest of the circle is the major segment, found as $\pi r^2 - A_{\text{minor segment}}$ — never computed by a separate triangle subtraction.

Figure

A circle split into a shaded sector wedge and a shaded segment crescent side by side with the chord radii and arc labelled

Sector area is a straight fraction of the circle's area

The goal is to show why cutting a circle by a central angle $\theta$ gives a sector whose area is exactly $\dfrac{\theta}{360^\circ}$ of the full circle's area $\pi r^2$ , using nothing beyond the idea that equal central angles always cut off equal areas.

Step 1. Start from the full circle. A full circle has total central angle $360^\circ$ and total area $\pi r^2$ . Every sector of that circle is carved out by choosing some central angle $\theta \le 360^\circ$ between two radii.

Step 2. Use the fact that area scales evenly with angle. Because the circle is perfectly symmetric, any two sectors with equal central angles have equal areas — a $10^\circ$ sliver anywhere on the circle has the same area as any other $10^\circ$ sliver. This means the sector's area is directly proportional to its central angle $\theta$ .

Step 3. Write the proportion and solve for the sector's area. Since angle $360^\circ$ corresponds to area $\pi r^2$ , angle $\theta$ corresponds to area $A_{\text{sector}}$ in the same ratio: $\dfrac{A_{\text{sector}}}{\pi r^2} = \dfrac{\theta}{360^\circ}$ , so $A_{\text{sector}} = \dfrac{\theta}{360^\circ} \times \pi r^2$ .

Step 4. The identical reasoning applies to the arc length, since the arc is likewise a fraction of the full circumference $2\pi r$ : $\ell = \dfrac{\theta}{360^\circ} \times 2\pi r$ . Both formulas rest on the same single idea — that a central angle carves off a fixed, proportional share of the whole circle.

Worked example

A chord of a circle of radius $14\text{ cm}$ subtends a right angle ( $90^\circ$ ) at the centre. Find the area of the sector, the area of the minor segment, and the area of the major segment. Use $\pi = \dfrac{22}{7}$ .

Step 1. Find the area of the sector with $\theta = 90^\circ$ .

A_{\text{sector}} = \frac{90}{360} \times \frac{22}{7} \times 14^2 = \frac{1}{4} \times \frac{22}{7} \times 196 = \frac{1}{4} \times 22 \times 28 = \frac{616}{4} = 154 \text{ cm}^2

Step 2. Find the area of the triangle formed by the two radii and the chord. Since $\theta = 90^\circ$ , the two radii are perpendicular, so the triangle is right-angled with both legs equal to $r = 14\text{ cm}$ .

A_{\triangle} = \frac{1}{2} \times 14 \times 14 = 98 \text{ cm}^2

Step 3. Subtract to get the minor segment's area.

A_{\text{segment}} = A_{\text{sector}} - A_{\triangle} = 154 - 98 = 56 \text{ cm}^2

Step 4. Find the major segment by subtracting the minor segment from the whole circle, never by subtracting a second triangle.

A_{\text{major}} = \pi r^2 - A_{\text{segment}} = \frac{22}{7} \times 196 - 56 = 616 - 56 = 560 \text{ cm}^2

The sector has area $154\text{ cm}^2$ , the minor segment has area $56\text{ cm}^2$ , and the major segment has area $560\text{ cm}^2$ .

Competitive stretch

Check the numbers stay consistent: minor segment plus major segment must equal the whole circle. $56 + 560 = 616\text{ cm}^2$ , which matches $\pi r^2 = \dfrac{22}{7} \times 196 = 616\text{ cm}^2$ exactly — a fast self-check worth doing on any segment question.

Common traps

Forgetting $\theta$ must be in degrees. The fraction $\dfrac{\theta}{360^\circ}$ only works with $\theta$ measured in degrees for this syllabus; do not mix in radian measure.
Confusing sector and segment. A sector is bounded by two radii and an arc; a segment is bounded by a chord and an arc. It is easy to stop at the sector area and forget to subtract the triangle to reach the segment.
Using the wrong triangle for non-standard angles. The neat shortcuts — equilateral at $60^\circ$ , right-angled isosceles at $90^\circ$ — apply only to those specific angles; this chapter sticks to standard angles for that reason.
Mixing up minor and major segments. Always find the minor segment first (sector minus triangle), then get the major segment as $\pi r^2 -$ minor segment; never build a second triangle for the major piece.

Ask the tutor: Want to try a segment problem with the $120^\circ$ standard angle, or see why the equilateral-triangle shortcut works at $60^\circ$ ? Ask the AI tutor to walk through it with you.

Why does the segment formula subtract a triangle instead of just using a different fraction of the circle?

Because a segment is not carved out by a central angle the way a sector is — it is bounded by a straight chord, not two radii meeting at the centre. The only shape whose area you can get by a clean $\theta/360^\circ$ fraction is the sector. So the standard trick is to build the segment out of two pieces you already know how to compute: take the sector (which does scale as a fraction of the circle) and cut away the triangle formed by the two radii and the chord, leaving exactly the curved sliver between chord and arc.

Do I need trigonometry to find the triangle's area in every segment question?

Not in this chapter — CBSE Class 10 keeps segment questions to the standard angles $60^\circ$ , $90^\circ$ and $120^\circ$ , where the triangle formed by the two radii and the chord has a known, simple shape: equilateral at $60^\circ$ , right-angled isosceles at $90^\circ$ . You can find its area with basic geometry formulas rather than trigonometric ratios. Trigonometric area formulas for a general angle belong to a later, more advanced treatment and are not required here.

How do I know whether a question wants the minor or the major segment?

Read carefully: "the segment" or "the minor segment" almost always means the smaller piece cut off directly by the chord, which you get as sector area minus triangle area. "The major segment" or "the region on the other side of the chord" means the larger remaining piece, found as $\pi r^2 -$ (minor segment area). A fast sanity check is that minor segment area plus major segment area must always equal the full circle's area $\pi r^2$ — if your two numbers do not add up to that, one of them is wrong.