Homework Help: geometry

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

Square ABCD and circle T have equal areas and share the same center O.The circle intersects side AB at points E and F.Given that EF=√(1600-400π),what is the radius of T ?

geometry - Steve, Tuesday, March 12, 2013 at 11:30am
Hmmm. 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 11:53am
I 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 12:28pm
Hmmm. 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 12:46pm
Steve, my hunch is that you are correct,
your answer came out "too nice" and clearly mine does not work.

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