Answer:
The circular field would require 1232 kg of fertilizer.
Step-by-step explanation:
0.5 [tex]m^{2}[/tex] takes 1 kg of fertilizer.
For a circular field of radius 14 m, its area = area of a circle
area of a circle = [tex]\pi[/tex][tex]r^{2}[/tex]
So that,
Area of the circular field = [tex]\frac{22}{7}[/tex] x 14 x 14
= 22 x 2 x 14
= 616
Area of the field = 616 [tex]m^{2}[/tex]
Since, 0.5 [tex]m^{2}[/tex] takes 1 kg of fertilizer. Let the kilograms of fertilizer required by the field be represented by x.
Thus,
x = [tex]\frac{616*1}{0.5}[/tex]
= 1232
x = 1232 kg
Therefore, the circular field would require 1232 kg of fertilizer.
