A chord of a circle of radius 42 cm subtends an angle of measure 60 at the centre. Find the area of the minor segment of the circle. (√3 = 1.73)
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Sorry bro,
I forgot to upload image using gogiya167
please see the attached image
It will form a equilateral triangle with center of circle
As shown in the figure
So, we applied area of sector formula for circle formula
and area of equilateral triangle formula
By applying this formulas we finally got the area of minor segment which is 159.794.
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AB is the chord .
O is the centre of the circle.
OAXB is a minor sector .
OAB is an equilateral triangle.
radius = side of the equilateral triangle
= r = 42 cm
sector angle ( x ) = 60°
i ) Area of the sector( A1 ) = (x/360)×πr²
=> A1 = ( 60/360) × ( 22/7 ) × 42²
=> A1 = 924 cm² ----( 1 )
ii ) Area of an equilateral triangle ( A2)
A2 = ( √3/4 ) r²
= ( 1.73/4 ) × 42²
= 762.93 cm² ----( 2 )
iii ) Area of the minor segment of the circle
= A1 - A2
= 924 cm² - 762.93 cm²
= 161.07 cm²
••••
O is the centre of the circle.
OAXB is a minor sector .
OAB is an equilateral triangle.
radius = side of the equilateral triangle
= r = 42 cm
sector angle ( x ) = 60°
i ) Area of the sector( A1 ) = (x/360)×πr²
=> A1 = ( 60/360) × ( 22/7 ) × 42²
=> A1 = 924 cm² ----( 1 )
ii ) Area of an equilateral triangle ( A2)
A2 = ( √3/4 ) r²
= ( 1.73/4 ) × 42²
= 762.93 cm² ----( 2 )
iii ) Area of the minor segment of the circle
= A1 - A2
= 924 cm² - 762.93 cm²
= 161.07 cm²
••••
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