Answer:
See step-by-step, as it depends which value of pi you use.
Step-by-step explanation:
Equations
- Diameter = [tex]2r[/tex]
- Area of a circle = [tex]\pi r^2[/tex] (where r is the radius)
Assuming the pond is circular...
Diameter of pond = 50 yd ⇒ r = 50 ÷ 2 = 25 yd
Substitute r = 25 into the equation for the area of a circle:
[tex]\implies A= \pi \times25^2=625\pi[/tex] yd²
Using [tex]\pi =3.14[/tex], area = 1962.5 yd²
Using [tex]\pi=\dfrac{22}{7}[/tex], area = 1964.3 yd² (nearest tenth)
Using [tex]\pi[/tex], area = 1963.5 yd² (nearest tenth)
