Find the differnce of the areas of two segments of a circle formed by a chord of length 5cm subtending an angle of `90^(@)` at the centre.

Explanation

Let the radius of the circle be r.
`:.` OA = OB = r cm
Given that, length of chord of a circle , AB = 5cm
and central angle of the rector AOBA `(theta) = 90^(@)`
Now, in `DeltaAOB (AB)^(2) = (OA)^(2) + (OB)^(2)` [by Pythagoras theorem]
`(5)^(2) = r^(2) + r^(2)`
`rArr2r^(2) = 25`
`:. r= (5)/(sqrt(2)cm`
Now, in `DeltaAOB` we drawn a perpendicular line OD, which meets at D on AB and divides chord AB into tow equal parts.
So, `AD=DB= (AB)/(2)=(5)/(2)cm`
[since, the perpendicular drawn from the centre to chord of a circle divides the chord into two equal parts]
By Pythagoas theorem, in `DeltaADO`,
`(OA)^(2) = OD^(2) + AD^(2)`
`rArr OD^(2) = OA^(2) - AD^(2)`
= `((5)/(sqrt(2)))^(2) - ((5)/(2))^(2)= (25)/(2)-(25)/(4)`
= `(50-25)/(4)= (25)/(4)`
`rArrOD= (5)/(2)cm`
`:.` Area of an isosceles `DeltaAOB = (1)/(2)xx"Base"(=AB)xx"Height"(=OD)`
=`(1)/(2)xx55xx(5)/(2)=(25)/(4)cm^(2)`
Now, area of sector AOBA = `((5)/(sqrt(2)))^(2) - ((5)/(2))^(2)= (25)/(2)-(25)/(4)`
= `(50-25)/(4)= (25)/(4)`
`rArrOD= (5)/(2)cm`
`:.` Area of minor segment = Area of sector AOBA - Area of an isosceles `DelatAOB`
= `((25pi)/(8)-(25)/(4)) cm^(2)`
Now, area of the circle = `pir^(2)= pi((5)/(sqrt(2)))=(25pi)/(2)cm^(2)`
`:.` Area of major segment = Area of circle- Area of minor segment
= `(25pi)/(2)- ((25pi)/(8)-(25)/(4))`
= `(25pi)/(8)(4-1)+(25)/(4)`
=`((75pi)/(8)+(25)/(4))cm^(2)`...(i)
`:.` Difference of the areas of two segments of a circla = | Area of major segment - Area of minor segment|
= `|((75pi)/(8)+(25)/(4))-((25pi)/(4)-(25)/(4))|`
=` |((75pi)/(8)-(25pi)/(8))-((25pi)/(8)+(25)/(4))|`
=` |(75pi-25pi)/(8)+(50)/(4)|=|(50pi)/(8)+(50)/(4)|`
=`((25pi)/(4)+(25)/(2))cm^(2)`
Hence, the required difference of the areas of two segments is `((25pi)/(4)+(25)/(2))cm^(2)`.