Q. 235.0( 1 Vote )

Find the difference of the areas of two segments of a circle formed by a chord of length 5 cm subtending an angle of 90° at the center.


Answer :

Given: Length of chord = 5 cm


Angle of sector = 90°


To find: Difference of areas of two segments.


Explanation:


Length of the chord = 5 cm (Given)


Let r be the radius of the circle.


Then, OA = OB = r cm


Now, angle subtended at the center of the sector OABO = 90°


Angle subtended at the center of the sector OABO (in radians) = θ = π/2


Triangle AOB is a right angled triangle.


So, by Pythagoras theorem, (AB)2 = (OA)2 + (OB)2


25 = 2r2



Also, AOB is an isosceles triangle.


Since, line segment OD is perpendicular on AB, therefore it divides Ab into two equal parts. Thus, AD = DB = 5/2 = 2.5 cm


Let AD = h cm


So, in right angled triangle AOD, by Pythagoras theorem,


(AO)2 = (AD)2 + (OD)2



= 25/4


h = 5/2 = 2.5 cm


Area of the isosceles triangle AOB (1/2) × Base × Height


= (1/2) × 5 × (5/2) = 25/4 cm2


Now, Area of the minor sector = (1/2)r2θ


= = (25π/8) cm2


Area of the minor segment = Area of the minor sector – Area of the isosceles triangle


=


Area of the major segment = Area of the circle – Area of the minor segment


=



=



Difference of the areas of two segments of a circle =


|Area of major segment – Area of minor segment|






Hence, the required difference of the areas of two segments is


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