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 ?
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.
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
Hmmm. I musta blown it somewhere. Of course, if r=3.14, there's no way that EF could be √(1600-400π) = 18.53
"overthinking" again.
Steve, my hunch is that you are correct,
your answer came out "too nice" and clearly mine does not work.
To find the radius of circle T, we first need to find the area of the square ABCD.
The area of a square is given by the formula A = side^2. Let's say the side length of square ABCD is s.
Since the area of the square is equal to the area of circle T, we can set up the equation:
s^2 = πr^2
where r is the radius of the circle T.
Now, we need to find the side length of the square. The problem gives us the length of EF, which is √(1600 - 400π).
EF represents the diagonal of the square. In a square, the diagonal and the sides are related by the Pythagorean theorem:
(2s)^2 = EF^2
4s^2 = (√(1600 - 400π))^2
4s^2 = 1600 - 400π
Dividing both sides by 4, we get:
s^2 = 400 - 100π
Now we can substitute this value of s^2 into our previous equation:
400 - 100π = πr^2
Rearranging the equation and simplifying:
r^2 = (400 - 100π)/π
r^2 = (400/π) - 100
Taking the square root of both sides:
r = √(400/π - 100)
Thus, the radius of circle T is √(400/π - 100).