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Is the sum of area of any regular polygon made in the two sides of a right angular triangle equal to the area of the same polygon made at the hypotenuse?

If so how to prove it? The area was equal to the semi circles with diameters of the hypotenuse and the other two sides, does it apply with any other regular shapes? Can it be proved with the original Pythagoras theorem or otherwise?

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    $\begingroup$ Yes. I have seen it in a book which has Euclid's theorems long ago. We can replace squares by 3 of any shapes which has one side lying on the side of the triangle. By shapes I mean they all have to be equal in shape. But different in scale. That's the only help I can give. $\endgroup$ – Isura Manchanayake Dec 6 '15 at 18:02
  • $\begingroup$ yep thankks for the heads up!!!!!!!!! @ isura manchanayake $\endgroup$ – Saha19 Dec 7 '15 at 13:58
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yeah it can be !! Euclid proved it yes!! if we consider the a general term for area, $A = ka^2$ (where $a$ is a side length and $A$ is the Area, area of any polygon is proportional to square of it's side)

$a^2+b^2 = c^2$ (Pythagoras theorem where $c$ is the hypotenuse)

then,

$ka^2 + kb^2 = kc^2$

if we take the area of any regular polygon constructed at sides $a,b$ and $c$ as $A,B$ and $C$, then

$A = ka^2$ $B = kb^2$ $C = kc^2$

then, $A+B = C$

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