Area of Segment of Circle


 
 
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

The area of a segment of circle of radius 14  cm if the arc of the segment has a measure of 30 degree is:

Right Option : C
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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:

Right Option : B
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Explanation
Question : 3

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=

Right Option : C
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Explanation
 
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