A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding :
(i) minor segment
(ii) major sector.

Explanation

given, radius of a circle AO = 10 cm
A kperpendicular is drawn from centre of circle to the chord of the circle, which bisects the chord
`:. AD = DC`
Alos, `angleAOD = angleCOD = 45^(@)`
`:. Angle AOC = angle AOD + angleCOD`
`= 45^(@) + 45^(@) = 90^(@)`
In right `Delta AOD`,
`sin 45^(@) = (AD)/(AO) rArr (1)/(sqrt2) = (AD)/(10)`
`rArr AD = 5 sqrt2 cm`
`rArr AD = 5 sqrt2 cm`
and `cost 45^(@) = (OD)/(AO) rArr (1)/(sqrt2) = (OD)/(10)`
`rArr OD = 5 sqrt2 cm`
Now, `AC = 2AD`
`= 2 xx 5 sqrt2 = 10 sqrt2 cm`
Now, area of `Delta AOC = (1)/(2) AC xx OD = (1)/(2) xx 10 sqrt2 xx 5 sqrt2 = 50 cm^(2)`
Now, area of sector `= (theta pi r^(2))/(360^(@)) = (90^(@))/(360^(@)) xx 3.14 xx (10)^(2) = (314)/(4) = 78.5 cm^(2)`
(i) `:.` Area of minor segment `AEC =` Area of sector OAEC - area of `Delta AOC`
(ii) `:.` Area of major sector OAFGCO = Area of circle - area of sector OAEC
`= pi r^(2) - 78.5 = 3.14 xx (10)^(2) - 78.5`
`= 314 - 78.5 = 235.5 cm^(2)`