Infinity

A square with a side 19 long is an inscribed circle, and the circle is inscribed next square, circle, and so on to infinity. Calculate the sum of the area of all these squares.

Correct answer:

S =  722

Step-by-step explanation:

$$\begin{gathered} a_1=19 \\ {S}_1=a_1^2=19^2=361 \\ {d}_1=a_1=19 \\ {d}_1=\sqrt{2}{a}_2 \\ {a}_2=d_1/\sqrt{2}=19/\sqrt{2}\doteq 13.435 \\ {S}_2=a_2^2=13.435^2=\frac{361}{2}=180\frac{1}{2}=180.5 \\ q=S_2/S_1=\frac{361}{2}/361=\frac{361}{2\cdot 361}=\frac{361}{722}=\frac{1}{2}=0.5 \\ S=S_1\cdot \frac{1}{1-q}=361\cdot \frac{1}{1-\frac{1}{2}}=722 \\ \text{ Verifying Solution: } \\ {d}_2=a_2=13.435\doteq 13.435 \\ {d}_2=\sqrt{2}{a}_3 \\ {a}_3=d_2/\sqrt{2}=13.435/\sqrt{2}=\frac{19}{2}=9\frac{1}{2}=9.5 \\ {S}_3=a_3^2={\frac{19}{2}}^2=\frac{361}{4}=90\frac{1}{4}=90.25 \\ {q}_2=S_3/S_2=\frac{361}{4}/\frac{361}{2}=\frac{361}{4}:\frac{361}{2}=\frac{361}{4}\cdot \frac{2}{361}=\frac{361\cdot 2}{4\cdot 361}=\frac{722}{1444}=\frac{1}{2}=0.5 \end{gathered}$$

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