## 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 = πr
^{2} - Area of a semicircle = \(\frac { 1 }{ 2 }\) πr
^{2} - Area of quadrant = \(\frac { 1 }{ 4 }\) πr
^{2}

**Perimeter of a semicircle:**

Perimeter of a semicircle or protractor = πr + 2r

**Area of the ring Formulas :**

Area of the ring or an annulus = πR^{2} – πr^{2}

= π(R^{2} – r^{2})

= 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

= πr^{2}– 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 = \(\left( \frac { \pi }{ 3 } -\frac { \surd 3 }{ 4 } \right) { r }^{ 2 }\)

NCERT SolutionsMathsScienceSocialEnglishHindiRD Sharma

- CBSE Board Papers Class 10 Maths Areas Related To Circles
- Areas Related to Circles NCERT Solutions Class 10 Maths VSAQ
- Areas Related to Circles NCERT Solutions Class 10 Maths SAQ 3 Marks
- Areas Related to Circles NCERT Solutions Class 10 Maths SAQ 2 Marks
- Areas Related to Circles NCERT Solutions Class 10 Maths LAQ
- Areas Related to Circles CBSE Class 10 Maths HOTS
- CBSE Class 10 Maths Areas Related to Circles Value Based Questions
- NCERT Exemplar Solutions Class 10 Maths Areas Related to Circles VSAQ
- NCERT Exemplar Solutions Class 10 Maths Areas Related to Circles SAQ
- NCERT Exemplar Solutions Class 10 Maths Areas Related to Circles LAQ
- CBSE Class 10 Maths Areas Related TO Circles Objective Type
- CBSE Class 10 Maths Areas Related to Circles Formative Assessment
- CBSE CCE Summative Assessment Class 10 Maths Areas Related To Circles LAQ
- CBSE CCE Summative Assessment Class 10 Maths Areas Related To Circles SAQ
- CBSE CCE Summative Assessment Class 10 Maths Areas Related To Circles VSAQ