The ratio of the area of the inner square to the area of the outer square
\($=\frac{(a-b)^{2}+b^{2}}{a^{2}}$\)
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)
\(&=\sqrt{(a-b-0)^{2}+(0-b)^{2}} \\\)
\(&=\sqrt{(a-b)^{2}+b^{2}}\)
Hence, the area of the inner square \($=(\text { side })^{2}$\)
\(&=\left(\sqrt{(a-b)^{2}+b^{2}}\right)^{2} \\\)
\(&=(a-b)^{2}+b^{2} \text { square } \mathrm{cm}\)
Now, the side of the outer square = Distance between the points (0,0) and (a, 0)
\(&=\sqrt{(a-0)^{2}+0^{2}} \\\)
\(&=\sqrt{a^{2}}=a\)
Thus, the area of the outer square \($=(\text { side })^{2}$\)
\($=a^{2}$\) square cm
Therefore, the ratio of the area of the inner square to the area of the outer square
\($=\frac{(a-b)^{2}+b^{2}}{a^{2}}$\)
To learn more about the area refer to:
https://brainly.com/question/3948796
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