Let triangle *ABC* be a 30-60 right triangle with hypotenuse *AC* and shortest side *AB* \(= 1\) (see Figure 8, below). Let semicircles be constructed on each side of triangle *ABC*.

**Figure 8: **A special case of semicircles. The Pythagorean cut is semicircle *LMK,* shown in red.

The semicircle on *AB* has radius \(\frac{1}{2}\) and so has area \({{\frac{1}{2}}\pi\left({{\frac{1}{2}}}\right)^2,}\) or \(\frac{\pi}{8}.\) Similarly, the area of the semicircle on side *BC* is \(\frac{3\pi}{8}\) and the area of the semicircle on hypotenuse *AC* is \(\frac{\pi}{2}\). Since \(\frac{\pi}{8}+\frac{3\pi}{8}=\frac{\pi}{2},\) the Pythagorean relationship holds:

Area of semicircle on *AB* + Area of semicircle on *BC* = Area of semicircle on *AC*.

Now let *LK* be the segment of length 1 perpendicular to *AC* and let the semicircle be constructed having diameter *LK* as shown. Since *AB* = *LK*, the semicircles on *AB* and on *LK* must have the same area of \(\frac{\pi}{8}.\) Moreover, since quarter circle *AKL* has area one-half the area of semicircle *AKC*, the area of quarter circle *AKL* is \(\frac{\pi}{4}.\) Thus,

Area of region *ALMK* \(=\frac{\pi}{4}-\frac{\pi}{8}=\frac{\pi}{8}\) = Area of semicircle on side *AB*,

and so semicircle *LMK* is a Pythagorean cut for semicircles built on the 30-60 right triangle. A GeoGebra applet showing the Pythagorean cut for semicircles with their diameters along the sides of this special right triangle is shown in Figure 9, below.

**Figure 9.** Semicircles with diameters along the sides of a 30-60 right triangle