The golden rectangle is inscribed in a circle with a radius of 10cm. Find the area of that rectangle.

The golden rectangle is inscribed in a circle with a radius of 10cm. Find the area of that rectangle.

The ratio of the sides a and b of the golden rectangle is calculated by the upper formula.

Because the ratio of the sides of this rectangle is constant,  so must be the ratio of the angles that these sides form with rectangle's diagonal!

a = 1,   b = 1/1.61803398874989485 = 0.61803398874989484752

tan(B) = 1/ 0.61803398874989484752

B = ≈58.28^o        A = ≈31.72^o

Circle's radius  r = 10cm,    rectangle's diagonal  d = 20cm      a = ?,   b = ?  and  A = ?

cos(A) = a/d             b = sqrt(d² - a² )                      A = a*b

a = d*cos(A)            b = sqrt(20² - 17.013²)             A = 17.013*10.515

a = 17.013cm           b = 10.515cm                          A =  ≈178.88cm²

Call x one side of the rectangle......then the adjacent side will be (Phi)x

And the diameter of this circle will  = 20cm    ......and by the Pythagorean Theorem, we have.......

x ^2  +  [(Phi)x]^2  = 400

x^2 [ 1 + (Phi)^2]  = 400      and we can note that Phi^2  =  1 + Phi.......so we have

x^2 ( 1 + 1 + Phi)  = 400

x^2 (2 + Phi )  = 400

x^2  =   400 / (2 + Phi)        take the  positive square root of both sides

x = 20 / √(2 + Phi) 

So....the area of this rectangle  = (20/√(2 + Phi))* (Phi * 20/ √(2 + Phi) ) = 400Phi / (2 + Phi)  = about

178.88 cm^2

Why is it a golden rectangle?  What is so special about phi ?

http://en.wikipedia.org/wiki/Golden_rectangle

http://en.wikipedia.org/wiki/Golden_ratio

Thanks anon