Asked by The Boss
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 ?
Answers
Answered by
Steve
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.
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.
Answered by
Reiny
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
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
Answered by
Steve
Hmmm. I musta blown it somewhere. Of course, if r=3.14, there's no way that EF could be √(1600-400π) = 18.53
Answered by
Reiny
"overthinking" again.
Answered by
Reiny
Steve, my hunch is that you are correct,
your answer came out "too nice" and clearly mine does not work.
your answer came out "too nice" and clearly mine does not work.