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Area of Segment of Circle

Areas Related to Circles - Concepts

Circumference of Circle

Area of Circle

Arc Length of Sector

Area of Sector

Area of Segment of Circle

Area of Shaded Region

Class - 10th CBSE Subjects

Concept Explanation

Area of Segment of Circle

Area of segment of circle

The portion (or part) of the circular region enclosed between a chord and the corresponding arc is called a segment of the circle.

Now let us take the case of the area of the segment AQB of a circle with centre O and radius r

Area of the segment AQB = Area of the sector OAQB - Area of $Delta$ OAB

$large Area; of ;sector; OAQB=frac{theta }{360}times pi r^{2}$

$large Area; of Delta ; OAB=frac{1}{2} times base times height = frac{1}{2}times AB times OP$

As angle of the sector and radius is given to us, we will find the value of AB and OP in terms of angle and radius

Now $large Delta OPB$ is a right angled at P

$large therefore sinleft (frac{theta}{2} right )= frac{PB}{OB}=frac{PB}{r}$ $large Rightarrow PB = r;sinleft (frac{theta}{2} right )$

As we that perpendicular to the chord from the centre bisects the chord

$large AB = 2times PB = 2;r;sinleft (frac{theta}{2} right )$

$large and; cosleft (frac{theta}{2} right )= frac{OP}{OB}=frac{OP}{r}$ $large Rightarrow OP = r;cosleft (frac{theta}{2} right )$

$large Area; of Delta ; OAB=frac{1}{2} times 2r; sinleft ( frac{theta}{2} right ) times r ;cosleft ( frac{theta}{2} right )=r^2sinleft ( frac{theta}{2} right ) cosleft ( frac{theta}{2} right )$ $large large Area; of Delta ; OAB=r^2;frac{sin;theta}{2}=frac{1}{2};r^2;sin;theta$

$large Area; of; segment ; OAQB=frac{theta}{360} times pi ;r^2-frac{1}{2}r^2sin;theta$

ILLUSTRATION: O is the centre of the circle, and OA = 28 cm. Find the area of minor segment AQB. $[Take sqrt{2}=1.4.]$

Solution:

$large Area; of; sector; OAQBO=pi times r^{2}times frac{theta }{360}$

$=frac{22}{7}times 28times 28times frac{45}{360}$

$=308;cm^{2}$

$large Area ;of;Delta OAB=frac{1}{2}r^{2};;sin;theta$

$large =frac{1}{2}times 28times 28times sin;45^0$

$large =frac{1}{2}times 28times 28times frac{1}{sqrt{2}}$

$large =196 ;;sqrt{2};;cm^{2}$

Area of segment AQBA = area of sector OAQBO - area of $Delta OAB$

$=(308-196sqrt{2})cm^{2}$

$=(308-274.4)cm^{2}$

$=33.6 ;;cm^{2}$

Sample Questions

(More Questions for each concept available in Login)

Question : 1

If O is the centre of a circle of radius r and AB is a chord of the circle at a distance r/2 from O, then $angle BAO=$
A. $60^o$	B. $45^o$	C. $30^o$	D. $15^o$
Right Option : C
View Explanation

Question : 2

The area of segment of a circle of radius 21 cm if the arc of the segment has a measure of 60 degrees is:
A. 45.27 sq cm	B. 40.27 sq cm	C. 40.8 sq cm	D. None of these
Right Option : B
View Explanation

Question : 3

The area of a segment of circle of radius 14 cm if the arc of the segment has a measure of 30 degree is:
A. 3.2 sq cm	B. 4.2 sq cm	C. 2.3 sq cm	D. 5.6 sq cm
Right Option : C
View Explanation