The ratio of the area of the inner square to the area of the outer square
[tex]$=\frac{(a-b)^{2}+b^{2}}{a^{2}}$[/tex]
What is the area of the square?
The area of a square exist described as the total number of unit squares in the shape of a square. In other phrases, it is defined as the distance occupied by the square.
Since, By the provided diagram,
The side of the inner square = Distance between the points (0, b) and (a-b, 0)
[tex]&=\sqrt{(a-b-0)^{2}+(0-b)^{2}} \\[/tex]
[tex]&=\sqrt{(a-b)^{2}+b^{2}}[/tex]
Hence, the area of the inner square [tex]$=(\text { side })^{2}$[/tex]
[tex]&=\left(\sqrt{(a-b)^{2}+b^{2}}\right)^{2} \\[/tex]
[tex]&=(a-b)^{2}+b^{2} \text { square } \mathrm{cm}[/tex]
Now, the side of the outer square = Distance between the points (0,0) and (a, 0)
[tex]&=\sqrt{(a-0)^{2}+0^{2}} \\[/tex]
[tex]&=\sqrt{a^{2}}=a[/tex]
Thus, the area of the outer square [tex]$=(\text { side })^{2}$[/tex]
[tex]$=a^{2}$[/tex] square cm
Therefore, the ratio of the area of the inner square to the area of the outer square
[tex]$=\frac{(a-b)^{2}+b^{2}}{a^{2}}$[/tex]
To learn more about the area refer to:
https://brainly.com/question/3948796
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