Answer:
[tex]\frac{(a-b)^2+b^2}{a^2}[/tex]
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)
[tex]=\sqrt{(a-b-0)^2+(0-b)^2}[/tex]
[tex]=\sqrt{(a-b)^2+b^2}[/tex]
Thus the area of the inner square = (side)²
[tex]=(\sqrt{(a-b)^2+b^2})^2[/tex]
[tex]=(a-b)^2+b^2\text{ square 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 = (side)²
[tex]=a^2\text{ square cm}[/tex]
Hence, 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]
