we know that
Based on the FIFA rules for international matches, Â dimensions of fields are: Length: minimum [tex]100[/tex] m, maximum [tex]110[/tex] m. Width: minimum [tex]64[/tex] m, maximum [tex]75[/tex] m.
we know that
The area of a rectangle is
[tex]A=L*W[/tex]
where
L is the length side of the field
W is the width side of the field
In this problem we have
[tex]W=75\ yd\\A=7500\ yd^{2}[/tex]
Step [tex]1[/tex]
Find the value of L
[tex]A=L*W[/tex]
[tex]L=A/W[/tex]
substitute the values
[tex]L=7500/75=100\ yd[/tex]
Step [tex]2[/tex]
Convert yards to meters
we know that
[tex]1 \ yard =0.9144\ meters[/tex]
[tex]L=100\ yard=100*0.9144=91.44\ meters[/tex]
[tex]W=75\ yard=75*0.9144=68.58\ meters[/tex]
Step [tex]3[/tex]
Compare the dimensions of the field with the dimensions of FIFA regulation
Length: minimum [tex]100[/tex] m, maximum [tex]110[/tex] m
Length of the field=[tex]91.44\ meters[/tex]
The length does not satisfy the measures of the FIFA regulation
because
[tex]91.44\ meters < 100\ meters[/tex]
therefore
the answer is
The field does not satisfy the measures of the FIFA regulation
