Circles P and Q are tangent, and each has a radius of 1. Segment PQ extended meets the circles at A and B, and segment AC and segment BTC are tangents to circle P, as shown in the diagram. Compute AC.
Here is what the diagram looks like. http://i58.photobucket.com/albums/g254/poker5495/POWIX.jpg
2007-05-06 14:52:48 · 3 answers · asked by poker5495 4 in Science & Mathematics ➔ Mathematics
BT^2 = 2d^2 = 8
BT = 2√2
AC^2+(2d)^2 = (CT+2√2)^2, CT = AC
Solve for AC,
AC = √2
2007-05-06 15:09:09 · answer #1 · answered by sahsjing 7 · 0⤊ 0⤋
Triangle PTB is a right triangle.
PB = 3, PT = 1
â TB = sqrt(3²-1) = sqrt8.
Triangle CAB is a right triangle, and angles ABC and TBP are one and the same
â Triangles CAB and PTB are similar
â AC : TP = AB : TB
â AC : 1 = 4 : sqrt8
â AC = 4/sqrt8 = sqrt2.
2007-05-06 22:17:26 · answer #2 · answered by Anonymous · 0⤊ 0⤋
4^2+x^2=(x+sqrt(8))^2
16+x^2=x^2+2sqrt(8)x+8
or 4=sqrt(8)x
so x=4/sqrt(x)=sqrt(2)=1.414--this is AC.
I know I did not explain very well but the answer is right.
2007-05-06 22:10:14 · answer #3 · answered by bruinfan 7 · 0⤊ 0⤋