Question

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

Answer 1

Answer:

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

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

The area of the square is

$A=x^{2}$

where

x is the length side of the square

step 1

Find the area of the outer square (A_o)

we have that

$x=a\ units$

substitute in the formula

$A_o=a^{2}\ units^2$

step 2

Find the area of the inner square (A_i)

we know that

The length side of the inner square is equal to the hypotenuse of a right triangle

so

Applying Pythagoras theorem

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

Remember that

$A_i=x^2$

so

$A_i=(a-b)^2+b^2$

step 3

Find the ratio of the area of the inner square to the area of the outer

$ratio=\frac{A_i}{A_o}$

substitute the values

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