Let
ABC------> an equilateral triangle
x-------> length side of the triangle
we know that
if ABC is an equilateral triangle
then
AB=BC=AC------> x
and
angle A=angle B=angle C=60 degrees
Applying the law of sines
Area of triangle=(1/2)*a*b*sin C
in this problem
a=b=x cm
C=60 degrees
Area=64/(√3) cm²-------> 36.95 cm²
A=(1/2)*a*b*sin C------> (1/2)*x*x*sin 60°----> (1/2)*x²*(√3)/2
64/(√3)=(1/2)*x²*(√3)/2-----> 64*4=√3*√3*x²-----> 256=3*x²
x²=256/3------> x=16/√3 cm----> x=(16√3)/3 cm
if each side of the triangle is decreased by 4 cm
the new side of triangle (x1) is equal to
x1=[(16√3)/3]-4-----> x1=[16√3-12]/3 cm
the new area (A1) decreased is equal to
A1=(1/2)*x1²*(√3)/2-----> A1=(√3/4)*[(16√3-12)/3]²----> A1=(√3/36)*[(16√3-12)]²
A1=(√3/36)*[(16√3)²-2*(16√3)*12+12²)]----> A1=(√3/36)*[(768)-(384√3)+144)]
A1=(√3/36)*[(912)-(384√3)]-----> A1=11.879 cm²
therefore
the answer is
the area decreased is 11.879 cm²
