Popular Courses
Beginner – Linear Regression
Introduction
Beginner – Decision Trees
Introduction
Beginner – Neural Networks
Introduction

Sign in to Curvingo

or

New Account     Reset Password     

Plans

Details
Areas Related to Circles
Areas of Sector and Segment of a Circle
Practice: Sectors and Segments
Summary
Curvingo · Made in India

Share

Send us Feedback

Please let us know if you have any feedback and suggestions, or if you find any errors and bugs in our content.

Sorry, your message couldn’t be submitted. Please try again!

Thanks for your feedback!

Reset Progress

This will delete your progress and chat data for all chapters in this course, and cannot be undone!

Glossary

Select one of the keywords on the left…

Areas Related to CirclesAreas of Sector and Segment of a Circle

Reading time: ~40 min

It is mela day at school, and Class 10 B has the busiest stall: one giant round cake baked by Riya's mother. Sold whole, it costs ₹720. But nobody wants the whole cake — everybody wants a slice, and everybody wants to pay only for the cake they actually get.

So Riya makes one fair rule: every slice is cut from the centre, and its price depends only on the angle of the slice at the centre.

Three friends are waiting with exact change: Aman has ₹120, Meera has ₹180 and Kabir has ₹240. Each of them wants a slice worth exactly their money — not a rupee more, not a rupee less.

You hold the knife. Drag the handle on the rim to choose how wide the next slice is, then press Cut and sell. The cashier will tell you what that slice costs. Serve all three friends.

Did you crack Riya's rule? The whole cake is 360° and costs ₹720, so every degree of cake is worth ₹720 ÷ 360 = ₹. A 60° slice is \frac{60}{360} = \frac{1}{6} of the cake, so it costs \frac{1}{6} of the price. The 90° still on the plate is worth ₹.

Price is not the only thing that works like this. A slice of angle \theta is the fraction \frac{\theta}{360} of the whole cake — of its price, its weight, the area of its top and the length of its crust. That one idea is the whole of this chapter.

Sectors: Slices from the Centre

The slices you just cut have a proper name. You met the words sector and segment in earlier classes; here they are again, made precise. The part of a circular region enclosed by radii and the arc between them is a sector of the circle — a slice of cake cut from the centre. The part enclosed by a chord and the arc it cuts off is a segment — the piece you get from one straight cut that misses the centre.

Two radii always cut a circle into two sectors. Drag B round the circle. Let go when the larger sector's angle is exactly three times the smaller one's.

In Fig. 11.1, the smaller one, OAPB, is the minor sector, and the larger one, OAQB, is the major sector.

Fig. 11.1 — the minor sector OAPB and the major sector OAQB of a circle with centre O.

The angle \angle AOB is called the angle of the sector, and the major sector gets whatever is left of the full turn: 360^\circ - \angle AOB.

With \angle AOB = 90^\circ, the major sector is 270^\circ = 3 \times 90^\circ. Wherever you put B, the angles of the minor and the major sector always add up to .

Segments: One Straight Cut

Now cut the circle with one straight line instead. In Fig. 11.2, the chord AB splits the disc into two segments: the smaller one, APB, is the minor segment, and the larger one, AQB, is the major segment.

Drag B round the circle and watch the two segments trade size. Let go when the two segments are exactly equal.

Fig. 11.2 — the chord AB cuts off the minor segment APB and the major segment AQB.

The only chord that does it passes through the centre O — it is a , and each segment is a semicircle. Every other chord gives one minor and one major segment.

Remark

From now on, "sector" and "segment" on their own mean the minor sector and the minor segment, unless a question says otherwise.

Speed Round: Name That Region

Every card shows a circle with both radii and the chord drawn in. Look at what bounds the shaded region — two radii or one chord, the short arc or the long one — and name it.

Get five right to move on.

The Area of a Sector

Riya's price rule turns straight into an area formula. Let OAPB be a sector of a circle with centre O and radius r, as in Fig. 11.3, and let the degree measure of \angle AOB be \theta.

You know that the area of a circle — of the whole disc — is \pi r^2. Think of the whole disc as a sector whose angle at the centre O is the full . Then use the unitary method, exactly as the cashier did with rupees.

Angle at the centre 360° → area of the sector = \pi r^2

Angle at the centre 1° → area of the sector =

Angle at the centre \theta → area of the sector = \dfrac{\pi r^2}{360} \times \theta

Fig. 11.3 — the sector OAPB, with radius r and angle \theta.

Area of a sector of angle \theta = \dfrac{\theta}{360} \times \pi r^2, where r is the radius of the circle and \theta is the angle of the sector in degrees.

Same Area, Different Slices

A formula is a machine: feed it r and \theta, and out comes the area. Here it is running live, with \pi = \frac{22}{7}.

Three different sectors on this machine have an area of exactly 154 cm². Find all three.

r = ${r} cm    \theta = ${t}°

Put the three answers side by side: a radius of 7 cm needs 360°, 14 cm needs 90° and 21 cm needs 40°. Doubling the radius makes the whole disc times as big, so the angle has to shrink to a quarter to keep the same area. The area of a sector grows with r^2, not with r.

The Length of an Arc

Riya's cake has a chocolate crust all round its edge. How much crust comes with a slice? That is the length of the arc APB of the sector, and the unitary method answers it too. This time start from the whole of the circle — its circumference 2\pi r, which belongs to the full angle of 360°:

Angle 360° → length of the arc = 2\pi r

Angle 1° → length of the arc = \dfrac{2\pi r}{360}

Angle \theta → length of the arc = \dfrac{2\pi r}{360} \times \theta

Length of an arc of a sector of angle \theta = \dfrac{\theta}{360} \times 2\pi r.

Why the two formulas look alike

A sector of angle \theta is the fraction \frac{\theta}{360} of the whole circle. Take that fraction of the whole area \pi r^2 and you get the sector's area; take it of the whole boundary 2\pi r and you get its arc.

Try both formulas on a wall clock. The minute hand of the clock in the school hall is 21 cm long. It turns 360° in 60 minutes, so it turns 6° every minute. Drag the tip of the minute hand forward and let go — the hand sweeps a sector and its tip draws an arc. Complete all three missions.

Each mission was really a fraction of a full turn. 15 minutes is a quarter of an hour, so the tip travels a quarter of the circumference, 132 \div 4 = 33 cm; 30 minutes is half an hour, so the hand sweeps half the disc, 1386 \div 2 = 693 cm².

The Area of a Segment

A segment needs no new formula — just a subtraction. In Fig. 11.4, the sector OAPB is made of two pieces that meet along the chord AB: the triangle OAB and the segment APB. Take the triangle away from the sector and only the segment is left:

Area of segment APB = area of sector OAPB − area of \triangle OAB

= \dfrac{\theta}{360} \times \pi r^2 - \text{area of } \triangle OAB

The only real work is the triangle. The CBSE syllabus keeps segment questions to central angles of 60°, 90° and 120°, so learn the triangle for those three. Pick an angle, cut the triangle out, and read how its area is found. Do all three.

    Fig. 11.4 — the sector OAPB is the triangle OAB plus the segment APB.

    So at 60° and at 120° the triangle has the same area, \frac{\sqrt{3}}{4}r^2, and at 90° it has area \frac{1}{2}r^2. Only the sector you subtract it from changes.

    Note

    The major pieces are always "whole circle minus the minor piece". From Fig. 11.3 and Fig. 11.4, with Q any point on the major arc:

    Area of the sector OAQB = \pi r^2 − area of the minor sector OAPB

    Area of the major segment AQB = \pi r^2 − area of the segment APB

    Let us now try these results on two examples.

    A Thin Slice and the Rest

    Example 1 : Find the area of the sector of a circle with radius 4 cm and of angle 30°. Also, find the area of the corresponding major sector (Use \pi = 3.14).

    Solution : The sector is the minor sector OAPB in Fig. 11.5; the rest of the disc is the major sector.

    Area of the sector = \dfrac{\theta}{360} \times \pi r^2

    = \dfrac{30}{360} \times 3.14 \times 4 \times 4 \text{ cm}^2

    = \dfrac{12.56}{3} \text{ cm}^2 = cm² (approx.)

    Fig. 11.5 — a 30° sector of a circle of radius 4 cm.

    Area of the corresponding major sector

    = \pi r^2 - \text{area of sector OAPB}

    = (3.14 \times 16 - 4.19) \text{ cm}^2 = cm² \approx 46.1 \text{ cm}^2

    Alternatively, area of the major sector = \dfrac{360 - \theta}{360} \times \pi r^2

    = \dfrac{330}{360} \times 3.14 \times 16 \text{ cm}^2 = 46.05 \text{ cm}^2 \approx 46.1 \text{ cm}^2

    A 120° Segment

    Example 2 : Find the area of the segment AYB shown in Fig. 11.6, if radius of the circle is 21 cm and \angle AOB = 120^\circ. (Use \pi = \frac{22}{7})

    Solution : The segment AYB is the sector OAYB with the triangle OAB taken away:

    Area of segment AYB = area of sector OAYB − area of \triangle OAB    (1)

    Area of sector OAYB = \dfrac{120}{360} \times \dfrac{22}{7} \times 21 \times 21 \text{ cm}^2 = cm²    (2)

    Fig. 11.6 — the segment AYB, with \angle AOB = 120^\circ.

    To find the area of \triangle OAB, draw OM \perp AB, as in Fig. 11.7. Since OA = OB and OM is common, \triangle AMO \cong \triangle BMO by congruence.

    So M is the of AB, and \angle AOM = \angle BOM = \frac{1}{2} \times 120^\circ = 60^\circ.

    Let OM = x cm. In \triangle OMA,

    \dfrac{OM}{OA} = \cos 60^\circ, so \dfrac{x}{21} = \dfrac{1}{2}

    So OM = x = \dfrac{21}{2} cm.

    Also \dfrac{AM}{OA} = \sin 60^\circ = \dfrac{\sqrt{3}}{2}

    So AM = \dfrac{21\sqrt{3}}{2} cm.

    Fig. 11.7 — OM \perp AB splits \triangle OAB into two congruent halves.

    Therefore AB = 2 \times AM = 21\sqrt{3} cm, and

    area of \triangle OAB = \dfrac{1}{2} \times AB \times OM

    = \dfrac{1}{2} \times 21\sqrt{3} \times \dfrac{21}{2} \text{ cm}^2

    = cm²    (3)

    It is the machine's 120° rule again: \frac{\sqrt{3}}{4} r^2 = \frac{\sqrt{3}}{4} \times 441. From (1), (2) and (3),

    area of segment AYB = \left(462 - \dfrac{441\sqrt{3}}{4}\right) \text{ cm}^2 = \dfrac{21}{4}\left(88 - 21\sqrt{3}\right) \text{ cm}^2

    Speed Round: Slice Sprint

    The final round of the chapter. Each card gives a radius and an angle; pick the right area or arc length, with \pi = \frac{22}{7}. The wrong answers are the classic slips: using the other formula, using the diameter instead of the radius, or \frac{\theta}{180} instead of \frac{\theta}{360}.

    Get five right to move on.

    Sina