MHB Question about a rounded rectangle

[Opalg]

Someone has asked for a fuller explanation of a reply that I gave to this thread on another site eleven years ago.

The question concerns a rectangle (dimensions $l\times w$) whose corners have been replaced by quadrants of a circle of radius $r$. This diagram shows an enlargement of one corner of the rectangle.
[TIKZ]\fill [red!20!white] (0,0) -- (0,-4.5) -- (4,-4.5) -- cycle ;
\draw (0,-6) |- (6,0) ;
\draw (-1,-6) -| (6,1) ;
\draw (0,-6) arc (-90:0:6) ;
\draw (-1,-0.75) -- (6,-6) ;
\draw (5,-5) node{$x$} ;
\draw (3,-6.2) node{$r$} ;
\draw (6.2,-3) node{$r$} ;
\draw (2.3,-2.2) node{$r$} ;
\draw (5.4,-5.8) node{$\theta$} ;
\draw [dashed] (0,-4.5) -- (6,-4.5) ;
\draw [dashed] (4,0) -- (4,-6) ;
\draw [dashed] (0,0) -- (4,-4.5) ;[/TIKZ]
If the diagonal makes an angle $\theta$ with the base of the rectangle then $\tan\theta = \dfrac wl$. So $\cos\theta = \dfrac l{\sqrt{l^2+w^2}}$ and $\sin\theta = \dfrac w{\sqrt{l^2+w^2}}$.
Apply Pythagoras to the coloured triangle to get $$ (r- x\sin\theta)^2 + (r - x\cos\theta)^2 = r^2,$$ $$x^2 - 2rx(\sin\theta + \cos\theta) + r^2 = 0.$$ The solutions to that quadratic equation for $x$ are $$x = r(\sin\theta + \cos\theta) \pm\sqrt{r^2(\sin\theta+\cos\theta)^2 - r^2} = r\bigl(\sin\theta + \cos\theta\pm\sqrt{2\sin\theta\cos\theta}\bigr).$$ We want the smaller solution (the larger one would be where the diagonal meets the other side of the circle), so $$x = r\bigl(\sin\theta + \cos\theta - \sqrt{2\sin\theta\cos\theta}\bigr) = \dfrac{r\bigl(w+l-\sqrt{2lw}\bigr)}{\sqrt{l^2+w^2}}.$$

 

[Svein]

I am not quite sure of what you are looking for, but: Take any corner of the rectangle. Assume that l>2r and w>2r. Place a square with side r in each corner. When you replace the square with a quarter circle, you reduce the area from r² to π/4*r². Thus the total area is reduced from l*w to l*w-πr².