Given:
In circle O, CD = 56, OM = 20, ON = 16
CD is perpendicular OM and EF is perpendicular to ON
Solution:
The reference image for the answer is attached below.
Part a: Join OC, we get triangle OCM.
OM bisects CD.
CM = MD = [tex]\frac{1}{2} (56)[/tex] = 28
Using Pythagoras theorem:
[tex]OC^2=OM^2+CM^2[/tex]
[tex]OC^2=20^2+28^2[/tex]
[tex]OC^2=1184[/tex]
Taking square root on both sides.
[tex]OC =4\sqrt{74}[/tex]
The radius of the circle is 4√74.
Part b: Join FO, we get triangle FON.
Since radius is [tex]4 \sqrt{74}[/tex], [tex]OF =4 \sqrt{74}[/tex].
Using Pythagoras theorem:
[tex]OF^2=FN^2+ON^2[/tex]
[tex](4\sqrt{74} )^2=FN^2+16^2[/tex]
[tex]1184=FN^2+256[/tex]
Subtract 256 from both sides.
[tex]928=FN^2[/tex]
Taking square root on both sides.
[tex]4\sqrt{58} =FN[/tex]
Therefore, FN = 4√58
Part c: ON bisects EF.
FN = EN = 4√58
EF = FN + EN
[tex]EF = 4\sqrt{58} +4\sqrt{58}[/tex]
[tex]EF = 8\sqrt{58}[/tex]
[tex]EF = 60.9[/tex]
Therefore, EF = 60.9.
