Answer:
![\sf \dfrac{1}{4} \pi \quad or \quad \dfrac{7}{9}](https://tex.z-dn.net/?f=%5Csf%20%5Cdfrac%7B1%7D%7B4%7D%20%5Cpi%20%5Cquad%20or%20%5Cquad%20%5Cdfrac%7B7%7D%7B9%7D)
Step-by-step explanation:
The <u>width</u> of a square is its <u>side length</u>.
The <u>width</u> of a circle is its <u>diameter</u>.
Therefore, the largest possible circle that can be cut out from a square is a circle whose <u>diameter</u> is <u>equal in length</u> to the <u>side length</u> of the square.
<u>Formulas</u>
![\sf \textsf{Area of a square}=s^2 \quad \textsf{(where s is the side length)}](https://tex.z-dn.net/?f=%5Csf%20%5Ctextsf%7BArea%20of%20a%20square%7D%3Ds%5E2%20%5Cquad%20%5Ctextsf%7B%28where%20s%20is%20the%20side%20length%29%7D)
![\sf \textsf{Area of a circle}=\pi r^2 \quad \textsf{(where r is the radius)}](https://tex.z-dn.net/?f=%5Csf%20%5Ctextsf%7BArea%20of%20a%20circle%7D%3D%5Cpi%20r%5E2%20%5Cquad%20%5Ctextsf%7B%28where%20r%20is%20the%20radius%29%7D)
![\sf \textsf{Radius of a circle}=\dfrac{1}{2}d \quad \textsf{(where d is the diameter)}](https://tex.z-dn.net/?f=%5Csf%20%5Ctextsf%7BRadius%20of%20a%20circle%7D%3D%5Cdfrac%7B1%7D%7B2%7Dd%20%5Cquad%20%5Ctextsf%7B%28where%20d%20is%20the%20diameter%29%7D)
If the diameter is equal to the side length of the square, then:
![\implies \sf r=\dfrac{1}{2}s](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%20r%3D%5Cdfrac%7B1%7D%7B2%7Ds)
Therefore:
![\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}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Cimplies%20%5Csf%20Area%5C%3Aof%5C%3Acircle%20%26%20%3D%20%5Csf%20%5Cpi%20%5Cleft%28%5Cdfrac%7Bs%7D%7B2%7D%5Cright%29%5E2%5C%5C%26%20%3D%20%5Csf%20%5Cpi%20%5Cleft%28%5Cdfrac%7Bs%5E2%7D%7B4%7D%5Cright%29%5C%5C%26%20%3D%20%5Csf%20%5Cdfrac%7B1%7D%7B4%7D%5Cpi%20s%5E2%20%5Cend%7Baligned%7D)
So the ratio of the area of the circle to the original square is:
![\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}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Ctextsf%7Barea%20of%20circle%7D%20%26%20%3A%5Ctextsf%7Barea%20of%20square%7D%5C%5C%5Csf%20%5Cdfrac%7B1%7D%7B4%7D%5Cpi%20s%5E2%20%26%20%3A%20%5Csf%20s%5E2%5C%5C%5Csf%20%5Cdfrac%7B1%7D%7B4%7D%5Cpi%20%26%20%3A%201%5Cend%7Baligned%7D)
Given:
- side length (s) = 6 in
- radius (r) = 6 ÷ 2 = 3 in
![\implies \sf \textsf{Area of square}=6^2=36\:in^2](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%20%5Ctextsf%7BArea%20of%20square%7D%3D6%5E2%3D36%5C%3Ain%5E2)
![\implies \sf \textsf{Area of circle}=\pi \cdot 3^2=28\:in^2\:\:(nearest\:whole\:number)](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%20%5Ctextsf%7BArea%20of%20circle%7D%3D%5Cpi%20%5Ccdot%203%5E2%3D28%5C%3Ain%5E2%5C%3A%5C%3A%28nearest%5C%3Awhole%5C%3Anumber%29)
Ratio of circle to square:
![\implies \dfrac{28}{36}=\dfrac{7}{9}](https://tex.z-dn.net/?f=%5Cimplies%20%5Cdfrac%7B28%7D%7B36%7D%3D%5Cdfrac%7B7%7D%7B9%7D)
I believe 25 has to be added as
(x*5)^2 = x^2 + 10x + 25
Answer:
-6.25
Step-by-step explanation:
6.25 - 9= -2.75.
-2.75-3.5= -6.25