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In the book question is solved by.. Making triangle with 8cm base and 8 cm perpendicular. Then difference of hypotenuse n base is used to draw square.

What's the logic behind this?

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Say the square had side $s$. The diagonal has length $\sqrt 2 s$ so we have $$s+\sqrt 2 s = 8 \implies s(\sqrt 2 +1)=8\implies s=\frac 8{\sqrt 2 +1}=8\times (\sqrt 2 -1)$=8\sqrt 2 - 8$$

Now, in your (isosceles) right triangle, the hypotenuse is $8\sqrt 2$ and the side is $8$ so the desired claim follows at once.

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  1. Draw line segment AX equal to sum of diagonal and side
  2. Draw line at A making an angle of 45 and also draw another line making an angle of 22.5 at X
  3. Mark the intersection point of these two lines to be C
  4. Now complete the square of each side equal to AC

In the image, I'm doing when sum of diagonal and a side is equal to 6.2.

Square when sum of diagonal and a side is given

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