Answer:
A) [tex]area=63.585\ m^{2}[/tex], [tex]r=4.5\ m[/tex]
B) [tex]area=28.26\ m^{2}[/tex], [tex]D=6\ m[/tex]
C) [tex]area=120.7016\ m^{2}[/tex], [tex]D=12.4\ m[/tex]
D) [tex]area=12.56\ m^{2}[/tex], [tex]r=2\ m[/tex]
E) [tex]area=482.8064\ m^{2}[/tex], [tex]r=12.4\ m[/tex]
F) [tex]area=113.04\ m^{2}[/tex] , [tex]r=6\ m[/tex]
Step-by-step explanation:
we know that
The area of a circle is equal to
[tex]A=\pi r^{2}[/tex]
solve for r
[tex]r=\sqrt{\frac{A}{\pi}}[/tex]
Verify each case
case A) [tex]area=63.585\ m^{2}[/tex]
substitute in the formula
[tex]r=\sqrt{\frac{63.585}{\pi}}=4.5\ m[/tex]
case B) [tex]area=28.26\ m^{2}[/tex]
substitute in the formula
[tex]r=\sqrt{\frac{28.26}{\pi}}=3\ m[/tex]
the diameter is equal to
[tex]D=2r=2*3=6\ m[/tex]
case C) [tex]area=120.7016\ m^{2}[/tex]
substitute in the formula
[tex]r=\sqrt{\frac{120.7016}{\pi}}=6.2\ m[/tex]
the diameter is equal to
[tex]D=2r=2*6.2=12.4\ m[/tex]
case D) [tex]area=12.56\ m^{2}[/tex]
substitute in the formula
[tex]r=\sqrt{\frac{12.56}{\pi}}=2\ m[/tex]
case E) [tex]area=482.8064\ m^{2}[/tex]
substitute in the formula
[tex]r=\sqrt{\frac{482.8064}{\pi}}=12.4\ m[/tex]
case F) [tex]area=113.04\ m^{2}[/tex]
substitute in the formula
[tex]r=\sqrt{\frac{482.8064}{\pi}}=6\ m[/tex]
