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I was studying for some quizzes when I stumbled across this question. It goes like this:

Two regular quadrilaterals vinyl tiles each of $1$ foot long on each sides overlap each other such that the overlapping region is a regular octagon. What is the area of the overlapping region?

My work:

I imagined the problem like this: enter image description here

The area of the shaded blue region is what I want to get. I don't know if one side of the octagon is one-third of that of the side of square, because my calculated area is slightly larger than that of my book's answer.

How do you get the area of the blue region above?

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Let $a$ be a length-side of the octagon $ABCDE...$, where $BC$ placed on the square side.

Thus, since $\sin45^{\circ}=\frac{1}{\sqrt2}$, we obtain:$$\frac{1}{2}-\frac{a}{2}=a\cdot\frac{1}{\sqrt2},$$ which gives $a=\sqrt2-1$.

Thus, $\frac{1}{2}-\frac{a}{2}=\frac{1}{2}-\frac{\sqrt2-1}{2}=\frac{2-\sqrt2}{2}$ and the needed area is $$1-4\cdot\frac{1}{2}\left(\frac{2-\sqrt2}{2}\right)^2=1-\frac{1}{2}(2-\sqrt2)^2=2\sqrt2-2$$

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  • $\begingroup$ How did you derived the expression above? What is the thought process? $\endgroup$ – Palautot Ka Aug 23 '17 at 15:16
  • $\begingroup$ @Palautot Ka Say me please, are my answer and the answer in your book, same? $\endgroup$ – Michael Rozenberg Aug 23 '17 at 15:35
  • $\begingroup$ The area of a regular polygon is $A = \frac{a^2 n }{4} \cot \frac{180^o}{a}.$ Using your $a = \sqrt{2} -1$ and $n = 8$ sides, I got the area to be 0.8284 square feet. Is your value $4 \sqrt{2} - 5$ the area of the blue octagon? $\endgroup$ – Palautot Ka Aug 23 '17 at 15:35
  • $\begingroup$ @Palautot Ka I fixed my post. See now please. $\endgroup$ – Michael Rozenberg Aug 23 '17 at 15:39
  • $\begingroup$ The book's answer is 119.29 square inchesXD $\endgroup$ – Palautot Ka Aug 23 '17 at 15:40
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The area of the shaded region will be an octagon - one square minus 4 white triangles. Say the legs of the right triangles are l. Then $$2l+{\sqrt{2l^2}} = 1$$ from the Pythagorean theorem. Solving for l we get $$1-{\sqrt2}/2$$ The area will then be $$1-4*l^2/2$$ giving us $$2*{\sqrt2}-2$$

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  • $\begingroup$ I like the insight...... $\endgroup$ – Palautot Ka Aug 23 '17 at 15:45
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The height of a triangle is the half of the diagonal of a square minus the side of a square.

The area of a triangle (isoceles rectangle) is the square of the height.

Hence the blue area

$$1^2-4\frac{\left(\sqrt2-1\right)^2}4=2\sqrt2-2.$$

enter image description here

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