# 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?

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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!

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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.

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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:

$$S=\frac{A}{8}\sum_{n=0}^{\infty} \frac{1}{4^n}=\frac{A}{8}\cdot\frac{4}{3}=\frac{A}{6}$$

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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.

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Let us consider the top left square. The triangle in it is 1/4 th of the top left square. How is that? –  Ranjan Yajurvedi Oct 1 '12 at 17:50
That is exactly wht i said .... –  Phani Raj Oct 1 '12 at 18:52