By Pythagoras theorem, AP² = AO² + OP² = r² + (r²/4) = (5/4)r²
GP = AP = [(√5)/2] × r
GO = GP - OP = [(√5)-1]/2 × r
GX' = GP + PX' = [(√5)+1]/2 × r
The line GX' has been divided into what Euclid called "Extreme and Mean Ratio at O" (Ø)
Ø = 1.618033989... note: 1 + (1/Ø) = Ø
GX' = Ø × r, GO = (1/Ø) × r
Since AO = OX', again, by Pythagoras theorem,
AG = r√(1 + 1/ز)
AB = BD = DE = CE = AC = AG = r√(1 + 1/ز)
Each triangle has base length a = r√(1 + 1/ز).
The height of triangle:
Again, by Pythagoras theorem, OF² = r² - ¼a²
Now we have both base and height, the area of the triangle is just base × height / 2.
The area of the pentagon is 5 × the area of one triangle.