Area of Infinite Shaded Regions I recently came across this WISCONSIN STATE MATHEMATICS MEET problem and solution.
The solution reads that one of the larger triangles has an area of 1/16 th of whole square. How is that? 
 A: Your question has been well-answered. So we concentrate on giving a different solution of the original problem. 
Assume without loss of generality that the big square has area $1$. Let the  area of the shaded region be $A$. 
Then the shaded region consists of our two big shaded triangles, with combined area $\dfrac{1}{8}$, plus the area of the rest of the shaded stuff.
This rest of the area is the original shaded stuff, scaled down by a linear scaling factor $\dfrac{1}{2}$, and therefore by an area scaling factor $\left(\dfrac{1}{2}\right)^2$. Thus we obtain the equation
$$A=\frac{1}{8}+\frac{A}{4}.$$
Solve this linear equation for $A$. We get $A=\dfrac{1}{6}$. No infinite series!
A: Say the area of the largest square is $A$, the area of the next smaller square is $A/4$, the area of half that square (a rectangle, if cut along a median of the square) is $A/8$, so the area of the shaded right-angled triangle is half that, namely $A/16$
Since there are two such triangles for each square size, the total area of the shaded region will be:
$$S=\frac{A}{8}\sum_{n=0}^{\infty} \frac{1}{4^n}=\frac{A}{8}\cdot\frac{4}{3}=\frac{A}{6}$$
A: If we denote the side of the largest square by $x$, it is obvious by symmetry that the base of one of the right triangles is $x/2$, and the height is $x/4$. Since $A=bh/2$ for a triangle, the area is $x^{2}/16$ which is one sixteenth the area of the square.
A: Just take the size of square as 1, and try to find the dimensions of big triangle and then find the area. now try to use this fraction to make a series whose sum gives you the answer.
