Ungrouped Data: When a data collection is vast, a frequency distribution table is frequently used to arrange the data. A frequency distribution table provides the...
Ungrouped Data: Know Formulas, Definition, & Applications
December 11, 2024Area of a Circle: The Area of a Circle is the space occupied by the circle in a two-dimensional plane. A circle is a critical geometric figure that is present across many areas such as construction, engineering, and many more. Solving problems related to circles in Geometry requires us to be able to derive the area of a circle.
The formula for the area of a circle is A = πr2, where r is the radius of the circle. The unit of area is the square unit, for example, m2, cm2, in2 etc. In this article, we will learn terms related to circles, like radius, diameter, perimeter, area of the circle, etc. Continue reading to learn more.
Definition of Circle: A circle is the collection of all the points in a plane whose distance from a fixed point is always the same. The fixed point is called the centre of the circle and the boundary of the circle is called the circumference of the circle.
The radius of a circle is a line segment joining the centre and any point on the circumference.
The circle’s diameter is a line segment starting from any point on the circumference of a circle, passing through the centre and ending at the point on the circumference at the opposite side of the circle. The length of the diameter is twice the length of the radius in a circle.
A line segment joining any two points of a circle is called the chord of the circle.
If a chord is the circumference of the circle, it divides the area into two equal parts. However, if the chord is not the circumference of the circle, it divides the area into two unequal parts. A chord is placed at the centre of the circle, which is the Major Segment. A chord placed at the minor region of the circle is called a Minor Segment.
Any part of the circumference of a circle is called an arc of the circle. If the length of an arc is greater than a semicircle, it is called the Major arc, and if the length of an arc is smaller than a semicircle, it is called a Minor arc. The sum of lengths of the major arc and the minor arc will always give the circle’s circumference.
The area enclosed by an arc and the two radii joining the arc’s endpoints with the centre is called the sector of the circle. When the major arc forms the sector, it is called the Major Sector and when the minor arc forms the sector, it is called a Minor Sector.
If two or more circles have the same centre, the circles are called concentric circles.
The area of a circle is defined by the space or region occupied by a circle in a two-dimensional plane.
A constant term \(\pi \) (pi) is used in the formula of area of a circle. \(\pi \) is a constant term, also known as Archimedes’ constant. One way to define \(\pi \) is that it is the ratio of the circumference of a circle to its diameter. It is an irrational number whose value is \(3.141592653589793238.\) For the common use in practice, the value of \(\pi \) is approximately taken as \(3.14\) when used as a decimal number and is taken as \(\frac{{22}}{7}\) when used as a fraction to ease the calculation.
If \(r\) be the radius of a circle, then the area of a circle is given by \(\pi {r^2}.\)
Several techniques are available to derive the formula for the area of a circle. Among them, the most popular techniques used are (a) using the area of a rectangle and (b) using the area of a triangle. Calculus is also used to derive the area of a circle.
Here, the technique of using the area of a rectangle and that of a triangle will be discussed:
In this method, the circle is divided into many small sectors and the arrangement of the sectors has been made in the form of a parallelogram, as shown in the figure below. Clearly, the parallelogram will turn into a rectangle if the circle is divided into more sectors.
As observed in the figure, the circle is divided into \(16\) sectors, \(8\) of them are coloured blue and rest \(8\) are coloured yellow.
The blue and the yellow sectors are arranged alternately as shown in the figure above.
Considering that the circumference of a circle is \(2\,\pi r,\) the total length of the base of the parallelogram (rectangle as the number of sectors becomes more) will be \(\pi r\) (the total length of the bases of \(8\) blue sectors) and height of the parallelogram (or rectangle) will be \(r.\)
So, the area of the rectangle \( = {\rm{length}} \times {\rm{base}} = \pi r \times r = \pi {r^2}\)
The rectangle consists of the area of the sectors of the circle.
Hence, the area of the circle is \(\pi {r^2}.\)
In this method, the entire area of the circle is considered as consisting of infinitely many numbers of concentric circles. If the circle is cut along the radius as shown in the figure and if all those infinitely many lines (circumferences of the concentric circles) are arranged in the form of a right-angled triangle or in the form of an isosceles triangle, then a right-angled triangle or an isosceles triangle is obtained whose base is \(2\,\pi r\) and height is \(r.\)
So, the area of the right-angled triangle or an isosceles triangle \( = \frac{1}{2} \times {\rm{base}} \times {\rm{height}} = \frac{1}{2} \times 2\,\pi r \times r = \pi {r^2}\)
In this case, the area of the circle \( = \) the area of the right-angled triangle or an isosceles triangle \( = \pi {r^2}\)
We will derive a few more formulas based on the basic formula of the area of a circle derived above, and present them below, which are widely used for many purposes, such as:
The length of the diameter is twice the length of the radius. So, if \(r\) be the radius and \(d\) be the diameter of a circle, then \(d = 2r\) or \(r = \frac{d}{2}\)
Hence, the area of a semicircle is \(\frac{1}{2} \times \pi {r^2} = \frac{1}{2} \times \pi \times {(r)^2} = \frac{1}{2} \times \pi \times {\left( {\frac{d}{2}} \right)^2} = \frac{1}{2} \times \pi \times \frac{{{d^2}}}{4} = \frac{{\pi {d^2}}}{8}\)
The length of the diameter is twice the length of the radius. So, if \(r\) be the radius and \(d\) be the diameter of a circle, then \(d = 2r\) or \(r = \frac{d}{2}\)
Hence, the area of a circle is \(\pi {r^2} = \pi \times {(r)^2} = \pi \times {\left( {\frac{d}{2}} \right)^2} = \frac{{\pi {d^2}}}{4}\)
The area of a circle can be calculated directly if the value of the circumference \(C = 2\pi r\) is known.
The area of a circle is \(\pi {r^2} = \frac{{\pi {r^2} \times 4\pi }}{{4\pi }} = \frac{{4{\pi ^2}{r^2}}}{{4\pi }} = \frac{{{{(2\pi r)}^2}}}{{4\pi }} = \frac{{{C^2}}}{{4\pi }}\)
The basic formula for the area of a circle, area \( = \pi {r^2}\) can be applied to find the area enclosed between any two concentric circles.
If \(R\) be the radius of the larger circle and \(r\) be the radius of the smaller circle (both being concentric circles), then the area enclosed between the two concentric circles (shown shaded in the figure) is given by \(\pi {R^2} – \pi {r^2} = \pi \left( {{R^2} – {r^2}} \right).\)
This concept can be extended to find the area enclosed between any two concentric circles, where there are more than two concentric circles with radii \({r_1},\;{r_2},\;{r_3},\;{r_4},\;{r_5},….\) etc.
The basic formula for the area of a circle, area \( = \pi {r^2}\) can be applied to find the area of both the minor and the major segments of the circle.
If \(r\) be the radius of a circle and \(\theta \) (pronounced ‘theta’) be the angle subtended by the chord containing the segment, then the area of segment of the circle is given by, area \( = \frac{{\theta – \sin \left( \theta \right)}}{2} \times {r^2},\) where \(\theta \) is in radians.
And, area \( = \left( {\frac{{\theta \times \pi }}{{{{360}^o}}} – \frac{{\sin \left( \theta \right)}}{2}} \right) \times {r^2},\) where \(\theta \) is in degrees.
These two formulas can be used to find the area of any segment (minor or major) of a circle.
The basic formula for the area of a circle, area \( = \pi {r^2}\) can be applied to find the area of both the minor and the major sectors of the circle.
The full circle has angle \(2\,\pi \) radians around the centre. So, the area of the sector with a central angle \(\theta \) and having radius \(r\) will be proportional to this angle. The larger is the angle \(\theta ,\) the larger will be the area.
So, the area of the sector is given by, area \( = \frac{\theta }{{2\pi }} \times \pi {r^2} = \frac{\theta }{2} \times {r^2},\) where \(\theta \) is in radians.
And, the area of the sector \( = \frac{\theta }{{{{360}^{\rm{o}}}}} \times \pi {r^2} = \frac{{\theta \times \pi }}{{{{360}^{\rm{o}}}}} \times {r^2},\) where \(\theta \) is in the degrees.
The area of a circle is measured in square units. The unit used for area of a circle in CGS system is \({\rm{c}}{{\rm{m}}^2}\) and in SI system the unit used is \({{\rm{m}}^2}.\)
Q.1. Find the area of a circle of diameter \(3.5\,{\rm{cm}}.\) Take \(\pi = \frac{{22}}{7}.\)
Ans: The diameter of the circle \(d = 3.5\,{\rm{cm}} = \frac{7}{2}{\rm{cm}}.\)
So, Area of the circle \( = \rm{Area} = \pi \frac{{{d^2}}}{4} = \frac{1}{4} \times \pi \times {d^2}\)
\( = \frac{1}{4} \times \frac{{22}}{7} \times {\left( {\frac{7}{2}} \right)^2} = \frac{{77}}{8} = 9\frac{5}{8}{\rm{c}}{{\rm{m}}^2}\)
Q.2. Find the area of a circle of radius \(28\,{\rm{cm}}.\) Take \(\pi = 3.14.\)
Ans: The radius of the circle \(r = 14\,{\rm{cm}}\)
So, Area of the circle \( = \rm{Area} = \pi \frac{{{d^2}}}{4} = \frac{1}{4} \times \pi \times {d^2}\)
\( = \frac{1}{4} \times 3.14 \times {14^2} = 153.86\,{\rm{c}}{{\rm{m}}^2}\)
Q.3. Find the area of a circle of circumference \(55\,{\rm{cm}}\) in decimal. Take \(\pi = \frac{{22}}{7}.\)
Ans: The circumference of the circle \(C = 56\,{\rm{cm}}\)
So, Area of the circle \( = {\rm{Area}} = \frac{{{C^2}}}{{4\pi }} = {55^2} \times \frac{1}{4} \times \frac{1}{{\frac{{22}}{7}}} = {55^2} \times \frac{1}{4} \times \frac{7}{{22}} = 240.625\,{\rm{c}}{{\rm{m}}^2}\)
Q.4. Amrita divided a circular disc of radius \(7\,{\rm{cm}}\) into two equal parts. Find the area of each semicircular disc. Take \(\pi = \frac{{22}}{7}.\)
Ans: Area of each semicircular disc \( = \frac{1}{2}\pi {r^2} = \frac{1}{2} \times \frac{{22}}{7} \times {7^2} = 77\,{\rm{c}}{{\rm{m}}^2}\)
Q.5. A circular radar screen has a circumference of \(176\,{\rm{cm}}.\) Only \(90\% \) of its area is effective. Calculate the effective area of the screen.
Ans: Let \(r\,{\rm{cm}}\) be radius of the circular radar screen. Then its circumference \( = 2\pi r\,{\rm{cm}}\)
According to information given, \(2\pi r = 176\)
\(\Rightarrow 2 \times \frac{{22}}{7} \times r = 176\)
\( \Rightarrow r = 176 \times \frac{7}{{22}} \times \frac{1}{2} = 28\,{\rm{cm}}\)
Area of the screen \( = \pi {r^2} = \frac{{22}}{7} \times {28^2} = 2464\,{\rm{c}}{{\rm{m}}^2}.\)
As only \(90\% \) of the area of the radar screen is effective, so the effective area of the screen \( = 90\% \) of \(2464\,{\rm{c}}{{\rm{m}}^2} = \frac{{90}}{{100}} \times 2464\,{\rm{c}}{{\rm{m}}^2} = 2217.6\,{\rm{c}}{{\rm{m}}^2}\)
Q.6. The adjoining figure shows two circles with the same centre. The radius of the larger circle is \(10\,{\rm{cm}}\) and the radius of the smaller circle is \(4\,{\rm{cm}}.\) Find the area of the shaded region. Take \(\pi = \frac{{22}}{7}.\)
Ans: Given the radius of the smaller and the larger circles are \(r\) and \(R\) respectively.
So, \(R = 10\,{\rm{cm}}\) and \(r = 4\,{\rm{cm}}.\)
Hence, the required area of the shaded region \( = \pi \left( {{R^2} – {r^2}} \right) = \frac{{22}}{7} \times \left( {{{10}^2} – {4^2}} \right) = \frac{{22}}{7} \times \left( {100 – 16} \right) = \frac{{22}}{7} \times 84 = 264\,{\rm{c}}{{\rm{m}}^2}.\)
Q.7. The adjoining sketch represents a sports field that consists of a rectangle and two semicircles. Calculate the area of the field.
Ans: The two semicircles put together make a circle of diameter \({\rm{42}}\,{\rm{cm}}\) i.e. of radius \(\frac{{42}}{2} = 21\,{\rm{cm}}.\)
Hence, the area of the field \( = \) area of rectangle \(+ \) area of two semicircles
\( = l \times b + \pi {r^2} = 100 \times 42 + \frac{{22}}{7} \times {21^2} = 4200 + 1386 = 5586\,{{\rm{m}}^2}.\)
Q.8. If two circles have areas in the ratio \(\pi r_1^2:\pi r_2^2,\) then what will be the ratio of their circumference?
Ans: Given \(\frac{{{A_1}}}{{{A_2}}} = \frac{{\pi \times r_1^2}}{{\pi \times r_2^2}} = \frac{{r_1^2}}{{r_2^2}}\)
So, \( \frac{{\frac{{C_1^2}}{{4\pi }}}}{{\frac{{C_2^2}}{{4\pi }}}} = \frac{{r_1^2}}{{r_2^2}}\)
So, \( \frac{{C_1^2}}{{C_2^2}} = \frac{{r_1^2}}{{r_2^2}}\)
Hence, \(\frac{{{C_1}}}{{{C_2}}} = \sqrt {\frac{{{r_1}}}{{{r_2}}}}\)
Q.9. The radius of two circles are in the ratio \(1:x\) What is the ratio of their areas?
Ans: Given \(\frac{{{r_1}}}{{{r_2}}} = \frac{1}{x}\)
So, \(\frac{{{A_1}}}{{{A_2}}} = \frac{{\pi \times r_1^2}}{{\pi \times r_2^2}} = \frac{1}{{{x^2}}}\)
Q.10. The circumferences of two circles are in the ratio \(1:x.\) What is the ratio of their areas?
Ans: Given \(\frac{{{C_1}}}{{{C_2}}} = \frac{{2 \times \pi \times {r_1}}}{{2 \times \pi \times {r_2}}} = \frac{{{r_1}}}{{{r_2}}} = \frac{1}{x}\)
So, \(\frac{{{A_1}}}{{{A_2}}} = \frac{{\pi \times r_1^2}}{{\pi \times r_2^2}} = \frac{1}{{{x^2}}}\)
Below are the frequently asked questions on the area of the circle:
Q.1. How do you find an area with diameter?
Ans: Diameter is twice the radius. So, divide the diameter by 22 and obtain the radius. Then use this radius to find the required area with the formula πr2πr2 Or else, use the formula area =πd24=πd24 directly, where dd is the diameter of the circle.
Q.2. What is the Area of a Circle?
Ans: The Area of a Circle is the space occupied by the circle in a two-dimensional plane. A circle is a critical geometric figure that is present across many areas such as construction, engineering, and many more. Solving problems related to circles in Geometry requires us to be able to derive the area of a circle. The formula for the area of a circle is A = πr2, where r is the radius of the circle. The unit of area is the square unit, for example, m2, cm2, in2 etc.
Q.3. If a square is inscribed in a circle, what will be the ratio of the area of the square to that of the circle?
Ans: Let the radius of the circle be r.r.
Hence, area of the circle =πr2=πr2
Diagonal length of square =2r=2r
So, the length of side of the square =2–√r=2r
Hence, area of the square =2r2=2r2
Hence, the required ratio is 2:π
Q.4. What is the formula of the area of the circle?
Ans: The formula of the area of the circle is \(\pi {r^2}\) where \(r\) is the radius of the circle.
Q.5. How do you find the area and circumference of a circle?
Ans: For finding the area of the circle use area \( = \pi {r^2}\) and for the circumference of a circle use circumference \( = 2\pi r,\) where \(r\) is the radius of the circle.
Q.6. If the radius of a circle is given in metres, what will be the unit of the area of the circle?
Ans: \({\rm{sq}}{\rm{.}}\,{\rm{metre}}{\rm{.}}\)
Q.7. What is the difference between the sector and segment of a circle?
Ans: The sector of a circle is the part of the area of the circle bounded by two radii and the included arc between them of the circle. The segment of a circle is the area of the circle bounded by a chord and the arc of the circle.
Q.8. How does the area of a circle change if the diameter of a circle doubles?
Ans: The area of a circle increases four times if its diameter doubles.
Q.9. What is the difference between the area and perimeter of a circle?
Ans: The area of a circle is the space occupied by the circle in a two-dimensional plane. The perimeter is the length of the boundary of the circle.
Q.10. What is the difference between major and minor segments of a circle?
Ans: If the segment of the circle is smaller than a semicircle, then it is called a minor segment and if it is larger than a semicircle, then it is called a major segment. The sum of the area of a minor segment and major segment is equal to the area of a circle.
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