A square and an equilateral triangle have the same perimeter. Let Abe the area of the circle circumscribed about the square and B be the area of the circle

A square and an equilateral triangle have the same perimeter. Let Abe the area of the circle circumscribed about the square and B be the area of the circle circumscribed about the triangle. Find A/B

Let the perimeter be 12p

So the sides of the square are 3p each and the sides of the triangle are 4p each

\(A \div B = \frac{4.5\pi p^2}{1}\div \frac{16p^2\pi}{3}\\ A \div B = \frac{4.5\pi p^2}{1}\times \frac{3}{16p^2\pi}\\ \frac{A}{B} = \frac{4.5}{1}\times \frac{3}{16}\\ \frac{A}{B} = \frac{4.5*3}{16}\\ \frac{A}{B} = \frac{13.5}{16}\\ \frac{A}{B} = \frac{27}{32}\\ \)

Very nice, Melody!!!!........great diagrams, too....!!!!!

Here's an alternative method.....

Let the side of the square = (1/4)P.....then the diagonal = [ sqrt(2)/4] P    ....and the radius = [sqrt(2)/8] P

Let the side of the equilateral triangle = (1/3)P.....then by the Law of Sines, the radius of  circle  B can be forund thusly :

(1/3)P / sin(120) = r / sin (30)

r =  [(1/2) * (1/3)P] / [sqrt(3)/2] =  [(1/3)P] / sqrt(3)

And, by Euclid, circles are to each other as the squares of their radiuses....so we have....

A / B   =  ( [sqrt(2)/8] P )^2  /  (   [(1/3)P] / sqrt(3) )^2    =   [ 1/32 ]  /  [ 1/27]  =  27/32