Areas Related to Circles Formulas CBSE Class 10 Maths

Circumference of a Circle or Perimeter of a Circle

The distance around the circle or the length of a circle is called its circumference or perimeter.
Circumference (perimeter) of a circle = πd or 2πr,
where d is a diameter and r is a radius of the circle and π = \(\frac { 22 }{ 7 }\)
Area of a circle = πr²
Area of a semicircle = \(\frac { 1 }{ 2 }\) πr²
Area of quadrant = \(\frac { 1 }{ 4 }\) πr²

Perimeter of a semicircle:
Perimeter of a semicircle or protractor = πr + 2r
Areas Related To Circles Notes CBSE Class 10 Maths

Area of the ring Formulas :
Area of the ring or an annulus = πR² – πr²
= π(R² – r²)
= π (R + r) (R – r)

Length of the arc AB = \(\frac { 2\pi r\theta }{ { 360 }^{ 0 } }\) = \(\frac { \pi r\theta }{ { 180 }^{ 0 } }\)
Areas Related To Circles Notes CBSE Class 10 Maths

Area of sector formula:

Area of sector OACBO = \(\frac { \pi { r }^{ 2 }\theta }{ { 360 }^{ 0 } }\)
Area of sector OACBO = \(\frac { 1 }{ 2 }\) (r × l).

Perimeter of a sector Formula:

Perimeter of sector OACBO = Length of arc AB + 2r
= \(\frac { \pi r\theta }{ { 180 }^{ 0 } }\) + 2r
Areas Related To Circles Notes CBSE Class 10 Maths

Other important formulae:

Distance moved by a wheel in 1 revolution = Circumference of the wheel.
Number of revolutions in one minute = \(\frac { Distance moved in 1 minute }{ Circumference }\)
Angle described by minute hand in 60 minutes = 360°
Angle described by hour hand in 12 hours = 360°
The mid-point of the hypotenuse of a right triangle is equidistant from the vertices of the triangle.
The angle subtended at the circumference by a diameter is always a right angle.

Area of a segment Formula Class 10 :
Areas Related To Circles Notes CBSE Class 10 Maths

Area of minor segment ACBA = Area of sector OACBO – Area of ΔOAB
= \(\frac { \pi { r }^{ 2 }\theta }{ { 360 }^{ 0 } } -\frac { 1 }{ 2 } { r }^{ 2 }sin\theta\)
Area of major segment BDAB = Area of the circle – Area of minor segment АСВА
= πr² – Area of minor segment ACBA.
If a chord subtends a right angle at the centre, then
Area of the corresponding segment = \(\left( \frac { \pi }{ 4 } -\frac { 1 }{ 2 } \right) { r }^{ 2 }\)
If a chord subtends an angle of 60° at the centre, then
Area of the corresponding segment = \(\left( \frac { \pi }{ 3 } -\frac { \surd 3 }{ 2 } \right) { r }^{ 2 }\)
If a chord subtends an angle of 120° at the centre, then
Area of the corresponding segment = \(\left( \frac { \pi }{ 3 } -\frac { \surd 3 }{ 4 } \right) { r }^{ 2 }\)

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