Question

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

Answer 1

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

Answer 2

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