91Ó°ÊÓ

The largest square has a side length of \(1\) unit and a square whose side length is \(r%\) as long as the larger square is inscribed with one vertex on each edge of the larger square.

To find the area of the shaded portion let the original starting white square is of side \(a|\). The next black square has a side \(1\) unit.

According to the question,

\(a\times \frac{r}{100}=1\)

\(a=\frac{100}{r}\)

Now, let \(\frac{r}{100}=p\) then \(a=\frac{1}{p}\).

We will assume that the new white square is inscribed in the first black square with sections on the sides \(\left (apx, ap\left ( 1-x \right ) \right)\).

By using Pythagoras' theorem,

\(\sqrt{a^{2}p^{2}x^{2}+a^{2}p^{2}\left ( 1-x \right )^{2}}=ap^{2}\)

\(x^{2}+\left ( 1-x \right )^{2}=p^{2}\)

\(\left ( 1-2 \right )\left ( x-x^{2} \right )=p^{2}\)

\(2\left ( x-x^{2} \right ) =1-p^{2}\)

Now, we will find the area of the dark-shaded regions in the first black square is

\(A_{1}=4\times \frac{1}{2}\left ( apx \right )ap\left ( 1-x \right )\)

\(A_{1}=2a^{2}p^{2}\left ( x-x^{2} \right )\)

\(A_{1} =a^{2}p^{2}\left ( 1-p^{2} \right )\)

Similarly, we get,

\(A_{2}=a^{2}p^{6}\left ( 1-p^{2} \right )\)

\(A_{3}=a^{2}p^{10}\left ( 1-p^{2} \right )\)

\(A_{4}=a^{2}p^{14}\left ( 1-p^{2} \right )\)

By this, we get that the areas \(A_{1},A_{2},A_{3}......A_{n}\) form a geometric series.

Now, the area of the shaded portion is

\(A=A_{1}+A_{2}+A_{3}+...\infty\)

\(A=a^{2}p^{2}\left ( 1-p^{2} \right )+a^{2}p^{6}\left ( 1-p^{2} \right )+a^{2}p^{10}\left ( 1-p^{2} \right )+a^{2}p^{14}\left ( 1-p^{2} \right )+...\)

\(A=a^{2}p^{2}\left ( 1-p^{2} \right )\left [ 1+p^{4}+p^{8}+p^{14}+... \right ]\)

\(A=\frac{a^{2}p^{2}\left ( 1-p^{2} \right )}{1-p^{4}}\)

Put \(a=\frac{1}{p}\), \(p=\frac{r}{100}\)

\(A=\frac{1}{1+\left ( \frac{r}{100} \right )^{2}}\)