Areas Related to Circles Revision Notes

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A circle is a simple geometrical figure surrounded by a curved line, which contains points separated by radius from a static point known as a centre. When a line and a circle are lying on the same plane and farther from each other, they do not intersect. But when the line comes in contact with the circle, it is tangent to the circle. That line intersects the two points of the curved line of the circle.

Keyterms: Circle, Tangent, Chords, Diameters, Area, Diameter, Radius, Circumference


What is a Circle?

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A circle is a simple geometrical figure surrounded by a curved line, which contains points separated by radius from a static point known as a centre.

When a line and a circle are lying on the same plane and farther from each other, they do not intersect. But when the line comes in contact with the circle, it is tangent to the circle. That line intersects the two points of the curved line of the circle.

The circle is a figure enclosed within a curved line which divides a plane into two regions : exterior and interior.

A circle can also be understood as an ellipse where the two foci are concurrent and eccentricity is zero.


Tangent

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A line is understood to be tangent to a circle when it touches the circle at a certain point. The point where the tangent and the circle meet refers to be the point of tangency. A tangent has its own mathematical equation.

Basic equation is: x2+ y= a2 at (x1, y1) is xx1+yy1= a2

The line is always perpendicular to the radius of the circle.

There are three criteria which determine the condition of tangency, they are:

  • The point of intersection is in the circle
  • The point of intersection is exterior to the circle.
  • The point of intersection is on top of the circle.

Circumference of a Circle

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Circumference of the circle also refers to the perimeter of the circle that estimates the periphery of the circle. When the circle opens to transform it into a straight line, the circumference determines its length.

The units of measurement of the circumference of the circle are in meters or centimeters.

Formula for the circumference of the circle:

Perimeter of a circle = 2πR;

R stands for the radius of the circle; π stands for Pi and it has a value of 3.14, pi is constant mathematically in every equation.

Circumference of a Semicircle

When a circle splits into two equal halves, the circumference or the perimeter of the circle also gets half.

Equation = πr + 2r

Here, π stands for Pi with the value 3.14 and r stands for radius of the circle.

Area of a Circle

Area of the circle refers to be the enclosed section by the circle or space engaged by the circle itself.

Equation for the area of the circle = A = πr2

Area of a Semicircle

Area of a semicircle is the space engaged by the half circle in the two dimensional plane.

Equation = πr2/2

Sectors of a Circle

Sector of a circle refers to the part of a circle which lies among the two radii and the connecting arch. The commonest example which best explains the sector of the circle is Semi- circle.

There are two types of sector of a circle:

  • Major sector
  • Minor sector

Major sector refers to the part estimated greater than the semi circle.

Minor sector refers to the part estimated lesser than the semi circle.

The only thing calculated in the sector of a circle is the area.

Formula for the area of the sector of a circle = (θ/360º) × πr2

Area of the minor sector of the circle = (θ/360º) × πr2 - r2 sin θ /2

Area of the major sector of the circle is the difference between the area of the circle and area of the minor sector of the circle.

Here, the θ stands for theta and the angle formed in the center, r stands for radius and π (pi) has the value 3.14.

Segment of a Circle

Segment of a circle refers to the area formed by the chord and the corresponding arch. The corresponding arch shall lie between the two end points of the chord.

Formula to calculate area of the segment of the circle:

A = (½) × r2 (θ – Sin θ)

Here, A is the considered area, r is radian, θ is the angle and sin θ has the value according to the value of θ.

Important Formulas: Areas Related to a Circle

  1. Area of a ring – π (R+r) (R-r)
  2. Length of the arch - πr2/ 180
  3. When the chord secant 90o, area of the corresponding segment 
  4. When the chord secant 60o, area of the corresponding segment
  5. When the chord secant 120o, area of the corresponding segment
  6. Number of revolutions in a minute Distance covered in 60 seconds/ circumference
  7. Circumference or perimeter of the sector

Revision Notes Benefits

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The benefits of the revision notes on Areas Related to Circle are:

  • The notes provide good ideas to the students about the probable questions examiners will ask.
  •  The question paper divides the solutions into various sections, so that the students understand better.
  • With these revision notes, the students get a better understanding about the chapter in a simplified language.
  • These revision notes are free from errors and well prepared for the students.
  • These revision notes help students to answer all categories of questions long or short, asked in the exams with clear conception and in proper steps.
  • These revision notes are highly beneficial as it covers the chapter to itscore and is convenient to go through right before the exams.

Preparation Guidelines

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  1. The students should read the questions asked carefully. Tough questions can complicate the thoughts and may lead to writing incorrect answers.
  2. The students should memorize the basic formulas of the chapter like the circumference or the area of the circle as the chapter revolves round these two formulas.
  3. The revision notes also include some practice questions and solving those sums will help the students become well prepared for the exams. Their concepts will be clear with the solving of the sums.
  4. The practice sums are set by math experts who help to clear the doubts of the students. The doubts regarding the chapter areas related to a circle are clear under one roof. The students thereafter, solve the questions with understanding.

Sample Questions

Question: PQRS is a diameter of a circle of radius 6 cm. The equal lengths PQ, QR and RS are drawn on PQ and QS as PToo RT diameters, as shown in Figure. Find the perimeter of the shaded region. (2011, Outside Delhi, 2 marks)

Answer: Radius OS = 6 cm

∴ Diameter PS = 12 cm

∴ PQ, QR and RS, three parts of the diameter are equal.

∴ PQ = QR = RS = 4 cm

and QS = 2 × 4 = 8 cm

∴ Required perimeter

Question: In Figure, a circle is inscribed in an equilateral triangle ABC of side 12 cm. Find the radius of inscribed circle and the area of the shaded region. (Use π = 3.14 and 3 = 1.73) (2014, Delhi, 3 marks)

Answer:. Area of an Equilateral (side)2

Construction: Draw OD ⊥ AB, OE ⊥ BC and OF ⊥ AC. Join OA, OB and OC.

Proof: Area of ΔABC

= ar(ΔAOB) + ar(ΔBOC) + ar(ΔAOC)

Question: In Figure, ABC is a right-angled triangle right angled at A. Semicircles are drawn on AB, AC and BC as diameters. Find the area of the shaded region. (2013, Delhi, 4 marks)

Answer: In rt. ΔBAC,

∴ BC2 = AB2 + AC2 …[Pythagoras’ theorem]

= (3)2 + (4)

= 9 + 16 = 25 cm2

∴ BC = + 5 cm … [Side of Δ can’t be -ve]

Question: In the Figure, the side of square is 28 cm and radius of each circle is half of the length of the side of the square where 0 and Oare centres of the circles. Find the area of shaded region. (2017, Delhi, 4 marks)

Answer: Side = 28 cm, Radius = 28/2 cm = 14 cm

The area of the shade = Area of square + 3/4 (Area of circle) + 3/4 (Area of circle)

= Area of square + 3/2(Area of circle) …[Area of square = (Side)2; Area of circle = πr2

= (28)2 + 3/2 x 22/7 x 14 x 14

= 784 + 924 = 1708 cm2

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Also Check:

CBSE X Related Questions

1.

The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table :

Length (in mm)

Number of leaves

118 - 126

3

127 - 135 

5

136 - 144

9

145 - 153

12

154 - 162

5

163 - 171

4

172 - 180

2

Find the median length of the leaves. 
(Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.)

      2.

      Solve the following pair of linear equations by the substitution method. 
      (i) x + y = 14 
          x – y = 4   

      (ii) s – t = 3 
          \(\frac{s}{3} + \frac{t}{2}\) =6 

      (iii) 3x – y = 3 
            9x – 3y = 9

      (iv) 0.2x + 0.3y = 1.3 
           0.4x + 0.5y = 2.3 

      (v)\(\sqrt2x\) + \(\sqrt3y\)=0
          \(\sqrt3x\) - \(\sqrt8y\) = 0

      (vi) \(\frac{3x}{2} - \frac{5y}{3}\) =-2,
          \(\frac{ x}{3} + \frac{y}{2}\) = \(\frac{ 13}{6}\)

          3.
          An umbrella has 8 ribs which are equally spaced (see Fig. 11.10). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
          An umbrella has 8 ribs which are equally spaced

              4.
              A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

                  5.
                  A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

                      6.

                      Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

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