CBSE · Class 10 · Mathematics · Chapter 11

Areas Related to Circles — Formula Sheet

Board Formulas

12 formulas

1.Circumference of a Circle
$C = 2 π r = π d$
$C$ : Circumference (cm) · $r$ : Radius (cm) · $d$ : Diameter, d = 2r (cm)
Recalled from earlier classes. Use it when the circumference is given and you must find the radius first.
2.Area of a Circle
$A = π r^{2}$
Every sector and segment area in this chapter is a part of this. If the diameter is given, halve it first.
3.Length of an Arc★
$l = \frac{θ}{36 0 ^{\circ}} \times 2 π r$
$l$ : Length of the arc (cm) · $θ$ : Central angle of the sector (degrees) · $r$ : Radius (cm)
θ is the angle (in degrees) that the arc subtends at the centre. The arc is that fraction of the whole circumference. Unit: cm.
4.Area of a Sector★
$A = \frac{θ}{36 0 ^{\circ}} \times π r^{2}$
$A$ : Area of the sector (cm²) · $θ$ : Central angle (degrees) · $r$ : Radius (cm)
A sector is the region between two radii and an arc. Its area is the fraction θ/360° of the circle's area. Unit: cm².
5.Sector Area from Arc Length
$A = \frac{1}{2} l r$
$l$ : Length of the arc (cm) · $r$ : Radius (cm)
Follows by dividing the sector-area formula by the arc-length formula: A/l = r/2. Useful when the arc length is given instead of the angle.
6.Quadrant and Semicircle
$A_{quadrant} = \frac{1}{4} π r^{2}, A_{semicircle} = \frac{1}{2} π r^{2}$
Sectors with central angles 90° and 180°.
7.Area of the Major Sector
$A_{major} = π r^{2} - \frac{θ}{36 0 ^{\circ}} π r^{2} = \frac{36 0 ^{\circ} - θ}{36 0 ^{\circ}} π r^{2}$
θ is the angle of the minor sector. The major sector has angle 360° − θ.
8.Perimeter of a Sector
$P = 2 r + \frac{θ}{36 0 ^{\circ}} \times 2 π r$
$P$ : Perimeter of the sector (cm) · $r$ : Radius (cm) · $θ$ : Central angle (degrees)
Two radii plus the arc. A common slip is to give only the arc length. Unit: cm.
9.Area of the Minor Segment★
$A_{segment} = \frac{θ}{36 0 ^{\circ}} π r^{2} - ar (△ O A B)$
The segment is the region between the chord AB and the minor arc. O is the centre and θ = ∠AOB.
10.Triangle OAB for θ = 60°, 90° and 120°
$θ = 6 0^{\circ} or 12 0^{\circ} : ar (△ O A B) = \frac{3}{4} r^{2}, θ = 9 0^{\circ} : ar (△ O A B) = \frac{1}{2} r^{2}$
The CBSE syllabus limits segment problems to these three central angles. For 60°, OA = OB = r and the base angles are also 60°, so the triangle is equilateral with side r. For 90°, it is a right isosceles triangle with legs r. For 120°, draw OM ⟂ AB: ∠AOM = 60°, so OM = r cos 60° = r/2 and AM = r sin 60° = (√3/2) r, giving AB = √3 r and area = ½ × AB × OM = (√3/4) r².
11.Area of the Major Segment
$A_{major segment} = π r^{2} - A_{minor segment}$
Find the minor segment first, then subtract from the area of the whole circle.
12.Angle Turned by Clock Hands
$minute hand: 6^{\circ} per minute, hour hand: 0. 5^{\circ} per minute$
The minute hand turns 360° in 60 minutes; the hour hand turns 360° in 12 hours. The length of the hand is the radius of the sector it sweeps.