Answer:
\(\frac{(a-b)^2+b^2}{a^2}\)
Step-by-step explanation:
Since, By the given diagram,
The side of the inner square = Distance between the points (0,b) and (a-b,0)
\(=\sqrt{(a-b-0)^2+(0-b)^2}\)
\(=\sqrt{(a-b)^2+b^2}\)
Thus the area of the inner square = (side)²
\(=(\sqrt{(a-b)^2+b^2})^2\)
\(=(a-b)^2+b^2\text{ square cm}\)
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 = (side)²
\(=a^2\text{ square cm}\)
Hence, the ratio of the area of the inner square to the area of the outer square
\(=\frac{(a-b)^2+b^2}{a^2}\)
