Find the differen
Given: Length of chord = 5 cm
Angle of sector = 90°
To find: Difference of areas of two segments.
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
⇒ 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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