Answer:
Distance around the stadium is 75.16 monger than the field
Step-by-step explanation:
From the graph attached,
Coordinates of A → (-10, 60)
Coordinates of B → (60, -10)
Coordinates of C → (10, -60)
Coordinates of D → (-60, 10)
Perimeter of field = AB + BC + CD + DA
Since field is in the shape of a rectangle,
AB = CD and BC = DA
Therefore, perimeter of the rectangle = 2(AB + BC)
Length of AB = [tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
= [tex]\sqrt{(60+10)^2+(-10-60)^2}[/tex]
= [tex]\sqrt{2(70)^{2} }[/tex]
= [tex]70\sqrt{2}[/tex] m
Length of BC = [tex]\sqrt{(60-10)^2+(-10+60)^2}[/tex]
= [tex]50\sqrt{2}[/tex] m
Perimeter of ABCD = 2(70√2 + 50√2)
= 240√2 m
= 339.41 m
Perimeter of stadium = 2πr [Here, r = radius of the circle]
= 2π(120)
= 753.98 m
To run around the field twice, distance covered by the players = 2(339.41)
= 678.82 m
Difference in the distance covered to run around the stadium and the rectangular field = 753.98 - 678.82
= 75.16 m
