$ABCD$ is a square with area $1 cm^2$. Draw an isosceles right $\triangle CDE$, right angled at $E$. Draw another square $\square DEFG$, then draw an isosceles right $\triangle FGH$, right angled at $H$. This process is repeated infinitely so that no two figures overlap each other. Find the area of the entire figure.
Area of the first square is $1 cm^2$. Naturally, it's side is $1 cm$ and the side of the isosceles right $\triangle$ is equal to $\frac{1}{\sqrt2}$. And, it's area becomes:
$$\frac12 \times \left(\frac{1}{\sqrt2}\right)^2$$
And, the next square's area: $$\left(\frac{1}{\sqrt2}\right)^2$$
I trust this question has to be answered using mathematical progression. But, I can't quiet keep my hands on simplifying the equation enough. Using the areas from above, I came up with:
$$1+\frac12\times\left(\frac{1}{\sqrt2}\right)^2+\left(\frac{1}{\sqrt2}\right)^2+\frac12\times\left(\frac{1}{\sqrt{2\sqrt2}}\right)^2+\left(\frac{1}{\sqrt{2\sqrt2}}\right)^2+\frac12\times\left(\frac{1}{\sqrt{2\sqrt{2\sqrt2}}}\right)^2...$$ $$\Rightarrow 1+\left(1+\frac12\right)\left(\left(\frac{1}{\sqrt2}\right)^2+\left(\frac{1}{\sqrt{2\sqrt2}}\right)^2+\dotsb\right)$$ Now, squaring all, and taking out $\frac12$ out of the bracket from each, $$1+\left(\frac32\right)\left[\frac12\left\{1+\left(\frac{1}{\sqrt2}\right)+\left(\frac{1}{\sqrt{2\sqrt2}}\right)+\dotsb\right\}\right]$$
The area of the first square and triangle is $\frac54$ units and each succeeding such motif has linear dimensions $\frac1{\sqrt2}$ times that of the last, hence half the area. The area of the entire figure is thus $$\frac54\left(1+\frac12+\frac14+\cdots\right)=\frac52$$