if a wire is bent into the shape of a square then the area enclosed by the square is 81 cm square when the same wire is bent into a semicircular shape find the area enclosed by the semicircle
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if a wire is bent into the shape of a square then the area enclosed by the square is 81 cm square when the same wire is bent into a semicircular shape find the area enclosed by the semicircle
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area of square is given as 81
area of square = side x side
let side of the square = a
81 = a X a
a= square root of 81
a = 9 cm
since same wire is used perimeter of square and semicircle will be equal
so , perimeter of square = perimeter of semi circle
4 X side = pie X r + 2r
4 X9 = 3.14 X r +2 r
36 = 5.14r
r = 7.003
now area of semicircle is
pie r ^2 / 2
area of semicircle enclosed by the wire is 77.035
Area Of square a^2 = 81 ====> a = 9
The wire when bent in a form of Square i.e. the total perimeter of the square
Therefore,Perimeter of Square = 4a = 4*9 = 36 which is total length of the wire
the wire which is bent in the form of semicircle forms circumference of semicircle
therefore circumference is 22/7r+2r = 36 where 'r' is radius
(22+14)*r/7 = 36
r = 7 cm
Therefore area of semicircle is {(22/7)r^2}/2 = (1/2)*(22/7)*7*7
By simplifying
we get Area of semicircle = 77 cm^2
Hope it helps ☺️
Please mark as brainlist answer
Please mark as brainlist answer