Wednesday

August 31, 2016
Posted by **The Boss** on Tuesday, March 12, 2013 at 9:43am.

- geometry -
**Steve**, Tuesday, March 12, 2013 at 11:30amHmmm. Let's set up a diagram.

Drop a perpendicular from O to AB to intersect at P.

Let OA intersect circle T at Q.

Now, let

b = AE

h = OP

a = AQ

OA is the radius r of T.

EF = 20√(4-π)

let d be the diagonal of the square, s√2

Now, since ABCD and T have the same area,

s^2 = πr^2

s = r√π

c = 1/2 EF = 10√(4-π)

b+c = s/2 = r/2 √π, so

b = r/2 √π - 10√(4-π)

a+r = d/2 = s/√2, so

a = r - r√(π/2)

since we have two secants from A,

b(s-b) = a(d-a)

plug it all in, and you get r=20. - geometry -
**Reiny**, Tuesday, March 12, 2013 at 11:53amI sketched the situation so that, for the square, point A is in quadrant I and B in in quad II

let the radius of the circle be r

area of circle = πr^2

then the area of square is πr^2

and each side of the square is r√π

we can call A(r√π/2 , r√π/2)

EA = r√π/2 - (1/2)EF

= r√π/2 - (1/2)√(1600 - 400π)

= r√π/2 - (1/2)(20)√(4 - π)

= r√π/2 - 10√(4-π)

so now:

r^2 = (r√π/2)^2 + (r√π/2 - 10√(4-π) )^2

arghhhh!!!!!

I then multiplied by 2, simplified and got

r^2(π-2) - 20π√(4-√) + 800-200π = 0

which is a quadratic in r, with

a=π-2

b=-20π√(4-π)

c = 800-200π

I used my calculator and the quadratic formula to get

r = 47.85 or r = 3.14285

checking:

if r = 3.14285..

area of circle = 31.0311...

side of square = r√π = 3.14285√π = 5.5705..

area of square = 31.0311... YEAHHHHH!

if r = 47.85..

area of circle = appr 7193

side of square = 47.85√π = 84.81..

area of square = appr. 7193

both answers are valid,

r = appr 47.85 or r = 3.14285 - geometry -
**Steve**, Tuesday, March 12, 2013 at 12:28pmHmmm. I musta blown it somewhere. Of course, if r=3.14, there's no way that EF could be √(1600-400π) = 18.53

- Go with Steve's - geometry -
**Reiny**, Tuesday, March 12, 2013 at 12:36pm"overthinking" again.

- geometry -
**Reiny**, Tuesday, March 12, 2013 at 12:46pmSteve, my hunch is that you are correct,

your answer came out "too nice" and clearly mine does not work.