Circles

Sector: The part of a circle that is enclosed by an arc and the two radii joining the end points of that arc to the centre of the circle is known as a sector. If the arc of the sector is a minor arc, the sector is known as a minor sector. If the arc of the sector is a major arc, the sector is known as a major sector.

X A

Y

O

E

SQ

ɵ

Segment

Sector

For example, in the above figure, the given sector is denoted as (O - ASQ). Sector (O – ASQ) is a minor sector and sector (O - AYQ) is a major sector.

If r is the radius of a circle and ɵ is the angular measure of the arc, then area of the sector,

Area of the sector, As = ɵ ∏r2

360 If L is the length of the arc of a circle and r is

the radius of the circle, then area of the sector,

Area of the sector, As = 1 Lr 2Perimeter of a sector is given by, Ps = L + 2r

Example:Find the area and the perimeter of a sector whose

corresponding arc subtends an angle of 60o at the centre of a circle of radius 7 cm.

Solution:Area of the sector, As = ɵ/ 360 × ∏r2 = 60 × 22 × 72 360 7 = 77 cm2 3Length of the arc, L = ɵ/ 360 × 2∏r = 60 × 2 × 22 × 7 360 7 = 22 cm2

3

Perimeter of the sector is given by,Ps = length of the arc + 2(radius) = L + 2r = 22 + 2 × 7 3 = 64 cm 3Segment: A chord divides the circular region in two parts,

each part known as a segment. Minor segment: The segment of a circle

corresponding to a minor arc.Major segment: The segment of a circle

corresponding to a major arc.For example, in the given figure, the segment

corresponding to arc XEY is a minor segment and that corresponding to arc XSY is a major segment.

Area of the minor segment (i.e. the segment corresponding to minor arc XEY) = area of sector (O - XEY) – area of triangle XOY

= ɵ × ∏r2 – 1 r2 sinɵ

360 2Area of the major segment = area of the circle –

area of the corresponding minor segmentExample:Find the area of a segment which subtends an

angel equal to 90o at the centre of a circle having radius 1 cm.

Solution:

Area of the segment = ɵ × ∏r2 – 1 r2 sinɵ

360 2

= 90 × 22 × 12 – 1 × 1 × 1 360 7 2 = 11 - 1 = 2 cm2 14 2 7Example:Two identical circles intersect so that their

centers, and the points at which they intersect form a square of side 1 cm. The area in sq. cm of the portion that is common to the two circles is

(1) ∏/ 4 (2) ∏/ 2 – 1(3) ∏/ 5 (4) √2 – 1

Solution:

Let the two circles with centers P and O intersect at M and N

Quadrilateral PMON is a square M angle MPN = m angle MON = 90o The area common to both the circles= 2 (Area of sector P–MN – Area of triangle PMN) = 2 90 × ∏ × 12 - 1 × 12 360 2

1 cm 1 cm

P O

M

N

= ∏ - 1 2 Hence, option 2.

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