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
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 = \(\frac { 2\pi r\theta }{ { 360 }^{ 0 } }\) = \(\frac { \pi r\theta }{ { 180 }^{ 0 } }\)
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
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 :
- 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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