Areas Related to Circles: Definitions, Formulas, and Examples

Areas Related to Circles: We are familiar with the methods of finding perimeters and areas of simple plane figures such as rectangles, squares, parallelograms, triangles, and circles. Areas Related to Circles is an important chapter in CBSE Class 10 Mathematics. Concepts taught in this chapter has a wide range of application in higher classes and college. Moreover, having a thorough idea about the concepts discussed in areas related to circles help students to prepare effectively for the Term I exams.

So, at Embibe, we have created a comprehensive summary of the chapter ‘Areas related to Circles’, to help students with their studies and revision. To help the students of Class 10 prepare for their term-wise exams, our experts have formulated a few solved NCERT questions from the chapter based on the latest CBSE syllabus for 2021-22.

Areas Related to Circles

This article gets you detailed information on circle, value or Pi, circle and its terms, and more. Scroll down to know more.

Value of Pi (π)

Mensuration is that branch of mathematics that studies the procedure of measurements. Measurement is a significant human activity. We measure the length of the cloth for stitching. The area of a wall for painting, the perimeter of a plot for fencing. We do many other measurements of similar things in our daily life. All these measurements, we will study in the chapter called mensuration. 

\(\pi \) (pi) occupies the most crucial place in measuring the surface area and volume of various solid and plane figures. The value \(\pi \) of is not precisely known. The story of the accuracy by which the value was estimated is an interesting one. 

\(\pi \) (pi)  is an irrational number. It cannot be expressed as the ratio of whole numbers. However, the ratio \(22:7\) is used as an approximation for it.

Circle and its Terms

A circle is the locus of a point that moves in a plane so that its distance from a given fixed point in the plane is always constant.

The region occupied by a circle in a two-dimensional plane is known as the area of the circle. The area of a circle can be easily determined using the formula \(A = \pi {r^2}\) Where \(r\) is the radius of the circle.

The terms related to a circle are listed below:

1. Radius
2. Chord
3. Diameter
4. Circumference
5. Arc
6. Major arc
7. Minor arc
8. Central Angle
9. Semi-circle
10. Segment
11. Sector

Radius

The distance from the centre to any point on the given circle is called the circle’s radius.

In the above figure, \(O\) is the centre of a circle, and the distance \(OA\) is the radius of the circle.

Chord

A line segment joining any two points on the given circle is called a chord of that circle.

In the above figure, \(AB\) is the chord of the given circle.

Diameter

A chord passing through the centre of the given circle is known as the diameter of the circle.

In the above figure, \(PQ\) is the diameter of the given circle. The diameter of the circle is twice the radius of the circle. If \(r\) is the radius and \(d\) is the diameter of the circle. Then, the relation between them is given by \(d = 2r\) or \(r = \frac{d}{2}\)

Circumference

The perimeter of the circle or the length of the circle’s boundary is called its circumference, i.e., the distance covered by travelling once around a circle is called the perimeter or circumference. The circumference of a circle is given by \(2πr.\)

Arc of a Circle

An uninterrupted section of a circle is called an arc of the circle.

Consider a circle \(C(O,r).\) Let \({A_1},{A_2},{A_3},{A_4},{A_5},{A_6}\) be the points on the circle. Then, the pieces \({A_1}{A_2},{A_3}{A_4},{A_5}{A_6},{A_1}{A_3}\) etc., are all arcs of the circle \(C(O,r).\)

Let \(P \) and \(Q\) be two points on a circle \(C(O,r).\) The circle is divided into two sections, each of which is an arc. We denote the arc from \(P\) to \(Q\) in the counterclockwise direction by \(PQ\) and the arc from \(Q\) to \(P\) in the clockwise direction by \(QP.\) Note that the points \(P\) and \(Q\) lie on both \(PQ\) and \(QP.\)

Length of an Arc

The length of an arc \(PQ\) is the length of the fine thread that just covers the arc completely. We denote the length of an arc by \(l(PQ).\)

From the above discussion, we have, for any two points \(P\) and \(Q\) on a circle either \(l(PQ) < l(QP)\) or \(l(PQ) = l(QP)\) or \(l(PQ) > l(QP)\)

If \(l(PQ) < l(QP),\) then \(PQ\) is called the minor arc, and \(QP\) is known as the major arc. Thus, the arc \(PQ\) will be a minor arc or a major arc according \(l(PQ) < l(QP)\) or \(l(PQ) > l(QP).\)

If the arc subtends an angle of \(\theta \) at the center, then its arc length \( = \frac{\theta }{{180}} \times \pi r\)

Hence the arc length \(l\) of a sector of angle \(\theta \) in a circle of radius \(r\) is given by \(l = \frac{\theta }{{360}} \times 2\pi r\)

Central Angle

Let \(C(O,r)\) be any circle. Then any angle whose vertex is \(O\) is called the central angle.

In the below figure \(∠POQ\) is a central angle of the circle \(C(O,r).\)

If \(l\) is the arc length and \(r\) is the radius of the circle, then the formula to find the central angle \(\theta \) is \(\theta = \frac{l}{r}.\)

Semi Circle

A diameter divides a circle into two equal arcs. Each of these two arcs is called a semicircle.

In the above figure, arc \(APB\) and arc \(BQA\) are semicircles.

The half-circle obtained by cutting the given circle into two halves along a diameter line is known as a semicircle.

The area of the semicircle is given by \(\frac{1}{2}\pi {r^2}\,{\rm{sq}}{\rm{.units}}{\rm{.}}\)

Major Arc

A major arc is one whose length is more than the length of the semi-circle.

In the above figure, arc \(APB\) is the major arc.

Minor Arc

A minor arc is one whose length is less than the length of the semi-circle.

In the above figure, arc \(AQB\) is the minor arc.

Segment of a Circle

Let \(PQ\) be a chord of the circle \(C(O,r).\) Then, \(PQ\) divides the region enclosed by the circle into two parts. Each of the parts is called a segment of the circle.

The segment containing the minor arc is called the minor segment, and the segment containing the major arc is called the major segment of the circle.

The area of a segment of a circle \( = \left\{ {\frac{\pi }{{360}} \times \theta – \sin \frac{\theta }{2}\cos \frac{\theta }{2}} \right\}{r^2}\)

Sector of a Circle

Consider a circle of radius \(r\) having its centre at the point \(O.\) Let \(A, B,\) and \(C\) be three points on the circle, as shown in the below figure. The area enclosed by the circle is divided into two regions, namely, \(OBA\) and \(OBCA.\) These regions are called sectors of the circle.

Each of these two sectors has an arc of the circle as a part of its partition. The sector \(OBA\) has arc \(AB\) as a part of its boundary, whereas \(OBCA\) has arc \(ACB\) as its boundary. These sectors are known as minor and major sectors of the circle, as defined below.

Minor Sector: If the minor arc of a circle is a part of its boundary, then the sector of the circle is called a minor sector.
In the above figure \(OAB\) is the minor sector.

Major Sector: If the major arc of the circle is a part of its dividing line. Then, a sector of a circle is called the major sector. In the above figure, sector \(OACB\) is the major sector.

When the degree measure of the angle at the center is \({360^{\rm{o}}},\) the area of the sector is \(\pi {r^2}.\)
When the degree measure of the angle at the centre is \(1,\) the area of the sector is \(\frac{{\pi {r^2}}}{{360}}\)

Therefore, when the degree measure of the angle at the center is \(\theta ,\) the area of the sector is \(\frac{\theta }{{360}} \times \pi {r^2}\)

Note:

1. The sum of the arcs of a circle’s major and minor sectors is equal to the circle’s circumference.
2. The boundary of a sector is made up of an arc of the circle and the two radii.

Formulas on Areas Related to Circles

For a circle of radius \(r\) units, we have 

1. Circumference of the circle \( = 2\pi r\,{\rm{units}}{\rm{.}}\)

2. Area of the circle \( = \pi {r^2}\,{\rm{sq}}{\rm{.units}}\)

3. Area of the semi-circle \( = \frac{1}{2}\pi {r^2}\,{\rm{sq}}{\rm{.units}}\)

4. Perimeter of the semi-circle \( = (\pi r + 2r) = r(\pi + 2)\)

5. Area of a quadrant of a circle \( = \frac{1}{4}\pi {r^2}\)

6. Area enclosed by the two circles \( = \pi {R^2} – \pi {r^2} = \pi \left( {{R^2} – {r^2}} \right) = \pi (R + r)(R – r)\) Where \(R\) and \(r\) are radii of two concentric circles

7. Area of a segment of a circle \( = \left\{ {\frac{\pi }{{360}} \times \theta – \sin \frac{\theta }{2}\cos \frac{\theta }{2}} \right\}{r^2}\)

Area of Sector

If an arc subtends an angle of \(\theta \) at the centre, then its arc length\( = \frac{\theta }{{180}} \times \pi r\)

Hence the arc length \(l\) of a sector of angle \(\theta \) in a circle of radius \(r\) is given by \(l = \frac{\theta }{{360}} \times 2\pi r\)

Also, the area \(A\) of a sector of angle \(\theta \) in a circle of radius \(r\) is given by \(A = \frac{\theta }{{360}} \times \pi {r^2}\)

Solved Examples – Areas Related to Circles

Q.1. Find the difference of the areas of a sector of an angle \({120^{\rm{o}}}\) and its corresponding major sector of a circle of radius \({\rm{21}}\,{\rm{cm}}{\rm{.}}\)
Ans: Let \({A_1}\) and \({A_2}\) be the areas of the given sector and the corresponding major sector, respectively. We have \(\theta = {120^{\rm{o}}}\) and \(r = 21\;{\rm{cm}}\)
Therefore, \({A_1} = \frac{\theta }{{{{360}^{\rm{o}}}}} \times \pi {r^2} = \frac{{120}}{{360}} \times \pi \times {(21)^2} = 147\pi\, {\rm{c}}{{\rm{m}}^2}\)
Now, \({A_2} = \) area of the circle \( – {A_1}\)
\( \Rightarrow {A_2} = \left\{ {\pi \times {{(21)}^2} – 147\pi } \right\}{\rm{c}}{{\rm{m}}^2}\)
\( = \pi (441 – 147)\,{\rm{c}}{{\rm{m}}^2}\)
\( = 294\pi \,{\rm{c}}{{\rm{m}}^2}\)
Required difference\( = {A_2} – {A_1} = (294\pi – 147\pi )\,{\rm{c}}{{\rm{m}}^2}\)
\( = \left( {147 \times \frac{{22}}{7}} \right){\rm{c}}{{\rm{m}}^2}\)
\( = 462\;{\rm{c}}{{\rm{m}}^2}\)
Therefore, the difference of the areas is \( = 462\;{\rm{c}}{{\rm{m}}^2}.\)

Q.2. Find the area of the segment of a circle, given that the angle of the sector is \({\rm{12}}{{\rm{0}}^{\rm{o}}}\) and the radius of the circle is \({\rm{21}}\,{\rm{cm}}{\rm{.}}\)
Ans: The area \(A\) of a minor segment of a circle of radius \(r\) and the corresponding sector angle is given by
\(A = \left\{ {\frac{\pi }{{360}} \times \theta – \sin \frac{\theta }{2}\cos \frac{\theta }{2}} \right\}{r^2}\)
\( \Rightarrow A = \left\{ {\frac{{22}}{7} \times \frac{{120}}{{360}} – \sin {{60}^{\rm{o}}}\cos {{60}^{\rm{o}}}} \right\}{(21)^2}\;{\rm{c}}{{\rm{m}}^2}\)
\( \Rightarrow A = \left( {462 – \frac{{441}}{4}\sqrt 3 } \right){\rm{c}}{{\rm{m}}^2}\)
\( \Rightarrow A = \frac{{21}}{4}(88 – 21\sqrt 3 ){\rm{c}}{{\rm{m}}^2}\)
Therefore, the area of the segment of a circle is \(\frac{{21}}{4}(88 – 21\sqrt 3 ){\rm{c}}{{\rm{m}}^2}\)

Q.3. In the below figure, two circular flower beds have been shown on two sides of a square lawn \(ABCD\) of side \(56\,{\rm{m}}.\) If the centre of each circular flower bed is the point of intersection \(O\) of the diagonals of the square lawn, find the sum of the areas of the lawn and the flower beds.

Ans: Area of the square lawn \(ABCD = 56 \times 56\;{{\rm{m}}^2}\)
Let \(OA = OB = x\,{\rm{m}}\)
\( \Rightarrow {x^2} + {x^2} = 56 \times 56\)
\( \Rightarrow 2{x^2} = 56 \times 56\)
\( \Rightarrow {x^2} = 28 \times 56\)
Now, total area \(=\)Area of sector \(OAB+\) Area of sector \(ODC+\) Area of \(△OAD+\) Area of \(△OBC\)
\( = \left[ {\frac{{90}}{{360}} \times \frac{{22}}{7} \times 28 \times 56 + \frac{{90}}{{360}} \times \frac{{22}}{7} \times 28 \times 56 + \frac{1}{4} \times 56 \times 56 + \frac{1}{4} \times 56 \times 56} \right]{{\rm{m}}^{\rm{2}}}\)
\( = \frac{1}{4} \times 28 \times 56\left[ {\frac{{22}}{7} + \frac{{22}}{7} + 2 + 2} \right]{{\rm{m}}^2}\)
\( = \frac{{7 \times 56}}{7}(22 + 22 + 14 + 14){{\rm{m}}^{\rm{2}}}\)
\( = 56 \times 72\;{{\rm{m}}^2}\)
\( = 4032\;{{\rm{m}}^2}\)
Therefore, the total area of the lawn and the flower beds is \(4032\;{{\rm{m}}^2}.\)

Q.4. Find the radius of an arc that is \({\rm{156}}\,{\rm{cm}}\) in length and subtends an angle of \(150\) degrees to the circle’s centre.
Ans: We know that, length of an arc \( = \frac{\theta }{{360}} \times 2\pi r\)
\( \Rightarrow 156 = \frac{{150}}{{360}} \times 2 \times 3.14 \times r\)
\( \Rightarrow r = \frac{{360 \times 156}}{{150 \times 2 \times 3.14}}\)
\( \Rightarrow r = 59.62\;{\rm{cm}}\)
Therefore, the radius of the circle is \(59.62\;{\rm{cm}}\)

Q.5. Below figure depicts an archery target marked with its five scoring areas from the centre outwards as Gold, Red, Blue, Black, and white. The diameter of the region representing the Gold score is \({\rm{21}}\,{\rm{cm,}}\) and each of the other bands is \({\rm{10}}{\rm{.5}}\,{\rm{cm}}\) wide. Find the area of each of the five scoring regions.

Ans: We have,
\(r = \) radius of the region representing Gold score \({\rm{10}}{\rm{.5}}\,{\rm{cm}}\)
\({r_1} = \) radius of the region representing Gold and Red scoring areas
\( = (10.5 + 10.5)\,{\rm{cm}} = 21\;{\rm{cm}} = 2{\rm{r}}\,{\rm{cm}}\)
\({r_2} = \) radius of the region representing Gold, Red and Blue scoring areas
\( = (21 + 10.5)\,{\rm{cm}} = 31.5\;{\rm{cm}} = 3{\rm{r}}\,{\rm{cm}}\)
\({r_3} = \) radius of the region representing Gold, Red, Blue and Black scoring areas
\( = (31.5 + 10.5)\,{\rm{cm}} = 42\;{\rm{cm}} = 4{\rm{r}}\,{\rm{cm}}\)
\({r_4} = \) radius of the region representing Gold, Red, Blue, Black, and white scoring areas
\( = (42 + 10.5)\,{\rm{cm}} = 52.5\;{\rm{cm}} = 5{\rm{r}}\,{\rm{cm}}\)
Now,
Area of the region representing gold scoring \({A_1} = \pi {r^2} = \frac{{22}}{7} \times 10.5 \times 10.5 = 346.5\;{\rm{c}}{{\rm{m}}^2}\)
Area of the region representing red scoring \({A_2} = \pi {(2r)^2} – \pi {r^2} = 3\pi {r^2} = 3{A_1} = 3 \times 346.5 = 1030.5\;{\rm{c}}{{\rm{m}}^2}\)
Area of the region representing Blue scoring \({A_3} = \pi {(3r)^2} – \pi {(2r)^2} = 5\pi {r^2} = 5{A_1} = 5 \times 346.5 = 1732.5\;{\rm{c}}{{\rm{m}}^2}\)
Area of the region representing Black scoring \({A_4} = \pi {(4r)^2} – \pi {(3r)^2} = 7\pi {r^2} = 7{A_1} = 7 \times 346.5 = 2425.5\;{\rm{c}}{{\rm{m}}^2}\)
Area of the region representing White scoring \({A_5} = \pi {(5r)^2} – \pi {(4r)^2} = 9\pi {r^2} = 9{A_1} = 9 \times 346.5 = 3118.5\;{\rm{c}}{{\rm{m}}^2}\)

Summary of Areas Related to Circles

In the above article, we have learned the definitions of the terms which is related to circle and the formula for the areas of circle, semi-circle, area of a quadrant of a circle, the area enclosed by the two circles, area of a segment of a circle and area of a sector of a circle. Also, we have solved some example problems based on several areas related to circles and areas of the combination of plane figures.

Frequently Asked Questions – Areas Related to Circles

The most commonly asked questions on areas related to circles are answered here:

We hope you find this detailed article on areas related to circles helpful. If you have any doubts or queries regarding this topic, feel to ask us in the comment section and we will assist you at the earliest. Happy learning!