+48 Chords UV,WX andYZ of a circle are parallel. The distance between chords UV and WX is 2 and the distance between chords WX and YZ is also 2. If UV=64 and YZ=48 then find WX.
+129852 Let d be the distance between the center of the circle and UV
We can form a right triangle with legs of d , (UV/2) and a hypotenuse of r
So we have
d^2 + (UV/2)^2 = r^3
d^2 + 32^2 = r^2 (1)
Likewise let the distance between the center of the circle and YZ = d + 2 + 2 = d + 4
And similar to the first case we have
(d + 4)^2 + (YZ / 2)^2 = r^2
(d + 4)^2 + 24^2 = r^2 (2)
Set (1) = (2)
d^2 + 32^2 = (d + 4)^2 + 24^2 simplify
d^2 + 1024 = d^2 + 8d + 16 + 576
1024 - 16 - 576 = 8d
432 = 8d
d =432 / 8 = 54
And
d^2 + 32^2 = r^2
54^2 + 32^2 = r^2
3940 = r^2
So we have another right triangle such that the distance from the center of the circle to WX = d + 2
So
(d + 2)^2 + ( WX / 2)^2 = r^2
(54 + 2)^2 + WX^2 / 4 = 3940
56^2 + WX^2 / 4 =3940
3136 + WX^2 / 4 = 3940
WX^2 = 4 ( 3940 - 3136)
WX^2 = 4 * 804
WX^2 = 3216
WX = sqrt (3216) ≈ 56.7
This makes sense.....WX should be > 48 but < 64
