Question

what is the ratio of the area of the inner square to the area of the outer square?
a. $\frac{(a - b)^2 + b^2}{a^2}$
b. $\frac{a^2 - b^2}{a^2}$
c. $\frac{(a - b)^2}{(a + b)^2}$
d. $\frac{(ab)^2}{(a + b)^2}$

Explanation:

Step1: Calculate the area of the outer square

The side - length of the outer square is \(a\). Using the formula for the area of a square \(A = s^{2}\) (where \(s\) is the side - length), the area of the outer square \(A_{outer}=a^{2}\).

Step2: Calculate the area of the inner square

We can use the Pythagorean theorem. Consider one of the right - angled triangles formed around the inner square. The two legs of the right - angled triangle: one leg has length \(b\) and the other has length \((a - b)\).
By the Pythagorean theorem, if the side - length of the inner square is \(s\), then \(s^{2}=b^{2}+(a - b)^{2}\) (since the side - length of the inner square is the hypotenuse of the right - angled triangle). So the area of the inner square \(A_{inner}=b^{2}+(a - b)^{2}\).

Step3: Calculate the ratio of the areas

The ratio of the area of the inner square to the area of the outer square is \(\frac{A_{inner}}{A_{outer}}=\frac{b^{2}+(a - b)^{2}}{a^{2}}\).

Answer:

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