Answer:
[tex]\sf \dfrac{1}{4} \pi \quad or \quad \dfrac{7}{9}[/tex]
Step-by-step explanation:
The width of a square is its side length.
The width of a circle is its diameter.
Therefore, the largest possible circle that can be cut out from a square is a circle whose diameter is equal in length to the side length of the square.
Formulas
[tex]\sf \textsf{Area of a square}=s^2 \quad \textsf{(where s is the side length)}[/tex]
[tex]\sf \textsf{Area of a circle}=\pi r^2 \quad \textsf{(where r is the radius)}[/tex]
[tex]\sf \textsf{Radius of a circle}=\dfrac{1}{2}d \quad \textsf{(where d is the diameter)}[/tex]
If the diameter is equal to the side length of the square, then:
[tex]\implies \sf r=\dfrac{1}{2}s[/tex]
Therefore:
[tex]\begin{aligned}\implies \sf Area\:of\:circle & = \sf \pi \left(\dfrac{s}{2}\right)^2\\& = \sf \pi \left(\dfrac{s^2}{4}\right)\\& = \sf \dfrac{1}{4}\pi s^2 \end{aligned}[/tex]
So the ratio of the area of the circle to the original square is:
[tex]\begin{aligned}\textsf{area of circle} & :\textsf{area of square}\\\sf \dfrac{1}{4}\pi s^2 & : \sf s^2\\\sf \dfrac{1}{4}\pi & : 1\end{aligned}[/tex]
Given:
- side length (s) = 6 in
- radius (r) = 6 ÷ 2 = 3 in
[tex]\implies \sf \textsf{Area of square}=6^2=36\:in^2[/tex]
[tex]\implies \sf \textsf{Area of circle}=\pi \cdot 3^2=28\:in^2\:\:(nearest\:whole\:number)[/tex]
Ratio of circle to square:
[tex]\implies \dfrac{28}{36}=\dfrac{7}{9}[/tex]
