Question

What is the ratio of the area of the inner square to the area of the outer square?

Answer 1

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

$=\frac{(a-b)^2+b^2}{a^2}$

Answer 2

Answer:

I got this fraction:

frac{(a-b)^2+b^2}{a^2}

(I got the answer correct)

Hope this is helpful :)

Step-by-step explanation: