Areas Related to Circles

1.0Master Perimeters, Sector Areas, and Segment Segregations in Minutes

Unlock the two-dimensional spatial logic governing curved boundaries. Learn how to compute circular tracks using circumference metrics, master the fractional division of spaces using Sectors, and dissect complex chord boundaries using Segment formulas to ace your Class 10 board exams.

Class: 10 Mathematics (CBSE)
Chapter: Areas Related to Circles
Estimated Learning Time: 20–25 Minutes

2.0Learning Outcomes

After completing this lesson, you will be able to:

Recall and apply basic circumference and area parameters for full and semicircular regions.
Compute the exact linear length of a circular Arc using its central subtended angle ($\theta$).
Calculate the area of Minor and Major Sectors using fractional degree tracking.
Set up and evaluate equations for Minor and Major Segments by subtracting triangular footprints from sectors.
Deconstruct complex, multi-shape shaded area designs into simple additions and subtractions.

3.0Introduction to the Areas Related to Circles

Welcome to a highly visual and formula-driven chapter of Class 10 Geometry! In previous chapters, you explored the structural traits of circles—like tangents, secants, and perpendicular radii. In this lesson, we transition from lines to regions, exploring the flat space enclosed by circular arcs and chords.

You will master the mathematics of fractional circular slices, learning how to calculate the exact region covered by a slice of pie (Sector) or the boundary area chopped off by a straight chord (Segment). By connecting angular rotations ($\theta$) to the total area of a circle, you will build a foolproof strategy to tackle word problems based on rotating clock hands, running tracks, car wipers, and complex shaded designs.

Sector: The portion (or part) of the circular region enclosed by two radii and the corresponding arc of a circle is called a sector of the circle.

Here region OAPB is a sector of the circle with centre $0.∠ A OB$ is called the angle of the sector. Here unshaded region OAPB is called minor sector and shaded region OAQB is called major sector. Angle of the major sector $= 36 0^{\circ} - ∠ AOB$ .

Segment: The portion (or part) of the circular region enclosed between a chord and the corresponding arc of a circle is called a segment of the circle. Shaded region APB is a segment of the circle. The unshaded region AQB is another segment of the circle formed by the chord $A B$ .
APB is called the minor segment and AQB is called the major segment.

4.0Method to calculate area of sector and segment

Let OAPB be a sector of a circle with centre 0 and radius $r$ . Let $∠ A OB = θ$ . Now area of the circle (in fact a circular region of disc) $= π r^{2}$ . We can regard this circular region on disc as a sector forming an angle of $36 0^{\circ}$ (i.e. of degree measure 360 ) at the centre 0 . By unitary method, area of the sector OAPB can be calculated as follows When the degree measure of the angle at the centre is 360 , area of the sector $= π r^{2}$

When the degree measure is 1 , area of sector $= \frac{1}{360} \times π r^{2}$ When the degree measure is $θ$ , area of sector $= \frac{θ}{360} \times π r^{2}$ Thus, area of the sector OAPB (i.e., sector of angle $θ$ ) $= \frac{θ}{360} \times π r^{2}$ where $r$ is the radius of the circle and $θ$ is the angle of the sector. Next, we shall find the length of the arc APB corresponding to this sector by unitary method.

If angle subtended at the centre is $36 0^{\circ}$ , then length of the $arc = 2 π r$

∴

If the angle at the centre is

θ

, then length of the arc of sector APB

= \frac{θ}{360} \times 2 π r

Thus, the length of the arc of segment OAPB i.e., length of arc of sector of angle

θ = \frac{θ}{360} \times 2 π r

Now, area of segment APB. $=$ area of sector OAPB - area of $△ OAB$

= \frac{θ}{360} \times π r^{2} -

area of

△ OAB

Cor. 1. Area of major sector

θ

area of the minor sector

OAPB = \frac{360 - θ}{360} \times π r^{2}

Cor. 2. Area of major segment

A QB = π r^{2} -

area of the minor segment

A PB

.

Note: Area of $△ OAB$ with $= \frac{θ}{360} \times 2 π r$ is $\frac{1}{2} r^{2} sin θ$ or $r^{2} sin (\frac{θ}{2}) cos (\frac{θ}{2})$

If $θ = 9 0^{\circ}$ , then $△ A OB$ is right triangle. So, area of $△ A OB = \frac{1}{2} \times A O \times BO = \frac{1}{2} \times r^{2}$
If $θ = 6 0^{\circ}$ , then $△ OAB$ is equilateral triangles. So, of $△ A OB = \frac{3}{4} \times r^{2}$
If $θ = 12 0^{\circ}$ , then $△ A OB$ is isosceles triangle. So, area of $△ A OB = \frac{1}{2} r^{2} sin 12 0^{\circ}$ $= \frac{1}{2} r^{2} \times \frac{3}{2} = \frac{3}{4} r^{2}$

5.0Area of combinations of plane figures

In this section, we shall calculate the areas of some combinations of plane figures. In our daily life, we see figures like flowerbeds, drain covers, window designs, designs on table covers etc. These are combinations of plane figures.

6.0Numerical Ability

The perimeter of a semi-circular protractor is $32.4 cm$ . Calculate: (i) The radius of the protractor in cm (ii) The area of the protractor in $cm^{2}$ . Solution: (i) Let the radius of the protractor be rcm . Then, its perimeter $= (π r + 2 r) c m$ . $∴ π r + 2 r = 32.4$ $\Rightarrow (π + 2) r = 32.4$ $\Rightarrow (\frac{22}{7} + 2) r = 32.4$ $\Rightarrow \frac{36}{7} r = 32.4$ $\Rightarrow r = (32.4 \times \frac{7}{36}) cm$ $= 6.3 cm$ . Radius of the protractor $= 6.3 cm$ . (ii) Area of the protractor $= \frac{1}{2} π r^{2} = (\frac{1}{2} \times \frac{22}{7} \times 6.3 \times 6.3) cm^{2} = 62.37 cm^{2} .$ $∴$ Area of the protractor $= 62.37 cm^{2}$ .
Two circles touch externally. The sum of their areas is $130 π sq$ . cm and the distance between their centres is $14 cm$ . Determine the radii of the circles. Solution: Let the radii of the given circles be Rcm and rcm respectively as shown in figure. As the circles touch externally, distance between their centres $= (R + r) c m$ .
$∴ R + r = 14$ Sum of their areas $= (π R^{2} + π r^{2}) cm^{2} = π (R^{2} + r^{2}) cm^{2}$ . $∴ π (R^{2} + r^{2}) = 130 π$ $\Rightarrow R^{2} + r^{2} = 130$ We have the identity, $(R + r)^{2} + (R - r)^{2} = 2 (R^{2} + r^{2})$ $\Rightarrow (14)^{2} + (R - r)^{2} = 2 \times 130$ [From (i) and (ii)] $\Rightarrow (R - r)^{2} = 64$ $\Rightarrow R - r = 8$ On solving (i) and (iii), we get $R = 11 cm$ and $r = 3 cm$ . Hence, the radii of the given circles are 11 cm and 3 cm .
A chord of a circle of radius 14 cm makes a right angle at the centre. Calculate: (i) The area of the minor segment of the circle. (ii) The area of the major segment of the circle. Solution: Let AB be the chord of a circle with centre 0 and radius 14 cm such that $∠ AOB = 9 0^{\circ}$ . Thus, $r = 14 cm$ and $θ = 9 0^{\circ}$ .
(i) Area of sector $OACB = \frac{π r ^{2} θ}{360} = (\frac{22}{7} \times 14 \times 14 \times \frac{90}{360}) cm^{2} = 154 cm^{2}$ . $Area of △ OAB = \frac{1}{2} r^{2} sin θ = (\frac{1}{2} \times 14 \times 14 \times sin 9 0^{\circ}) cm^{2} = 98 cm^{2} .$ $∴$ Area of minor segment $A CB A = ($ Area of sector $O A CB) - ($ Area of $△ O A B)$ $= (154 - 98) cm^{2} = 56 cm^{2} .$ (ii) Area of major segment $BDAB = ($ Area of the circle $) - ($ Area of minor segment ACBA $)$

$= [(\frac{22}{7} \times 14 \times 14) - 56] cm^{2} = (616 - 56) cm^{2} = 560 cm^{2} .$

The minute hand of a clock is 10.5 cm long. Find the area swept by it in 15 minutes. Solution Angle described by minute hand in 60 minutes $= 36 0^{\circ}$ . Angle described by minute hand in 15 minutes $= (\frac{360}{60} \times 15)^{\circ} = 9 0^{\circ}$ . Thus, the required area is the area of a sector of a circle with central angle, $θ = 9 0^{\circ}$ and radius, $r = 10.5 cm$ . $∴$ Required area $= (\frac{π r ^{2} θ}{360})$ $= (\frac{22}{7} \times 10.5 \times 10.5 \times \frac{90}{360}) cm^{2} = 86.63 cm^{2} .$
The figure shows a running track surrounding a grass enclosure PQRSTU. The enclosure consists of a rectangle $PQST$ with a semi-circular region at each end. Given, $PQ = 200 m$ and $PT = 70 m$ . (i) Calculate the area of the grassed enclosure in $m^{2}$ . (ii) Given that the track is of constant width 7 m , calculate the outer perimeter ABCDEF of the track.
Solution: (i) Diameter of each semi-circular region of grassed enclosure $= PT = 70 m$ , $∴$ Radius of each one of them $= 35 m$ . Area of grassed enclosure $\Rightarrow$ (Area of rect. PQST) +2 (Area of semi-circular region with radius 35 m ) $r = 10.5 cm$ . (ii) Diameter of each outer semi-circle of the track $\Rightarrow AE = (PT + 7 + 7) m = 84 m$ . $∴$ Radius of each one of them $= 42 m$ . Outer perimeter ABCDEF $= (AB + DE +$ semi-circle BCD + semi-circle EFA $)$ $\Rightarrow (2 PQ + 2 \times$ circumference of semi-circle with radius 42 m $)$ $\Rightarrow (2 \times 200 + 2 \times π \times 42) m = [2 \times 200 + 2 \times \frac{22}{7} \times 42] m = 664 m$ .

In an equilateral $Δ$ of side a units, Height $= \frac{3}{2}$ a, Inradius $= \frac{a}{2 3}$ , Circumradius $= \frac{a}{3}$

In an equilateral triangle of side $24 cm$ , a circle is inscribed, touching its sides. Find the area of the remaining portion of the triangle. Take $3 = 1.73$ and $π = 3.14$ . Solution: Let $△ A BC$ be the given equilateral triangle in which a circle is inscribed. Side of the triangle, $a = 24 cm$ .
Height of the triangle, $h = (\frac{3}{2} \times a) cm = (\frac{3}{2} \times 24) cm = 123 cm$ . Radius of the incircle, $r = \frac{1}{3} h = (\frac{1}{3} \times 123) cm = 43 cm$ . $∴$ Required Area $=$ Area of the shaded region $a = 24 cm$ (Area of $△ A BC$ ) - (Area of incircle) $\Rightarrow (\frac{3}{4} \times 24 \times 24 - π \times 43 \times 43) cm^{2}$ $\Rightarrow (1443 - 3.14 \times 48) cm^{2} = (144 \times 1.73 - 3.14 \times 48) cm^{2}$ $\Rightarrow [48 \times (3 \times 1.73 - 3.14)] cm^{2} = (48 \times 2.05) cm^{2} = 98.4 cm^{2}$
In the given figure, ABCPA is a quadrant of a circle of radius 22 cm . With AC as diameter, a semi-circle is drawn. Find the area of the shaded region. Solution: Since, ABCPA is a quadrant of a circle of radius 22 cm .
we have, $AB = BC = 22 cm$ and $∠ ABC = 9 0^{\circ}$ . $∴ AC = AB^{2} + BC^{2}$ [by Pythagoras theorem] $= (22)^{2} + (22)^{2} cm = 2 \times 484 cm = 222 cm$ $∴$ Radius of the semicircle $= \frac{1}{2} AC = 112 cm$ $∴$ Required area $= ($ area of the semicircle with $A C$ as diameter $) + ($ area of $△ A BC) - ($ area of the quadrant with $r = 22 cm$ ) $\Rightarrow [\frac{1}{2} \times \frac{22}{7} \times (112)^{2} + \frac{1}{2} \times 22 \times 22 - \frac{1}{4} \times \frac{22}{7} \times 22 \times 22] cm^{2}$ $\Rightarrow \frac{2662}{7} + 242 - \frac{2662}{7} = 242 cm^{2}$

7.0Important topics in Class 10 Maths: Areas related to Circle

Chords and the angles

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9.0Supporting Study Materials

This study material, containing comprehensive CBSE Notes and NCERT Solutions for Chapter 11 of Class 10 Maths, follows the latest NCERT guidelines. Complete with colorful sector partition charts, step-by-step segment deduction tracks, and angular rotation lookup blocks, this guide ensures thorough preparation for your board examinations.

CBSE Class 10 Maths Notes Chapter 11 Areas Related To Circles	NCERT Solutions for Class 10 Maths Chapter 11: Areas Related to Circles

30-Second Quick Review: Areas related to Circle

Circumference of a circle:= 2πr
Area of a circle= πr²
Area of a sector:= (θ/360°) × πr²
Length of an arc:= (θ/360°) × 2πr
A sector is enclosed by two radii and an arc.
A segment is enclosed by a chord and an arc.
A semicircle has half the area of a circle.
Composite figure questions involve adding or subtracting areas of basic shapes.
Always use consistent units while solving geometry problems.

10.0Previous Year Questions (PYQs) on Areas related to Circle

Question: Find the length of an arc of a circle of radius 14 cm subtending an angle of 90° at the centre.

Answer: Length of arc

= (θ/360°) × 2πr

= (90/360) × 2 × (22/7) × 14

= 22 cm

Question: A sector of a circle has a radius of 10 cm and a central angle of 72°. Find its area.

Answer: Area = (72/360) × π × 10²

= (1/5) × 100π

= 20π cm²

≈ 62.8 cm²

11.0Recommended Next Topics

Coordinate Geometry
Triangles
Introduction to Trigonometry
Applications of Trigonometry
Statistics