Given: If a wire is bent into the shape of a square, then the area of the square is 81 sq.cm.
That means each side of the square = #sqrt 81 = 9 cm# ---as, area of square =#side^2#
The length #l#, of the wire will be equal to the perimeter of this square:
#l = 4 xx 9 = 36 cm#
Now, when the wire is bent into a semi-circular shape, the perimeter will remain same = #36cm#
And perimeter of semicircle will be given as:
#(1/2 xx 2xxpi xx r) + (2xxr)#------ half the circumference of complete circle + diameter.
SO,
#(1/2 xx 2xxpi xx r) + (2xxr) = 36#
#=> (1/cancel2 xx cancel2xxpi xx r) + (2xxr) = 36#
#=> pi r +2r = 36# -------assume value of #pi =22/7#
#=> 22/7 xx r +2r =36#
#=> 22/7 xx r +2rxx 7/7 =36#
#=> (22r +14r)/7 =36#
#=> 36 r = 36 xx 7 #
# => r = 7 cm#
Area of this semicircle 'a' will be :
#a =1/2 pi r^2#
#=>a = 1/2 xx 22/7 xx 7^2 #
# => a = 1/cancel2^1 xx cancel22^11/cancel7^1 xx cancel49^7 = 11xx7#
#therefore a = 77 sq.cm.# will be the area of the semi-circle.
