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 =
= 28
Using Pythagoras theorem:
![OC^2=OM^2+CM^2](https://img.qammunity.org/2021/formulas/mathematics/high-school/d94shqkdciwhh1tn2040qxypggvznsr8gs.png)
![OC^2=20^2+28^2](https://img.qammunity.org/2021/formulas/mathematics/high-school/3kj05l7odsplm5nj4x4oavys8kacxn0bme.png)
![OC^2=1184](https://img.qammunity.org/2021/formulas/mathematics/high-school/jlcaqnr7e6ir8hmpji9fmxyqwvwceau95b.png)
Taking square root on both sides.
![OC =4√(74)](https://img.qammunity.org/2021/formulas/mathematics/high-school/ymidkfbl0xhdw19qzxwdeihvrsc0jkhu8z.png)
The radius of the circle is 4√74.
Part b: Join FO, we get triangle FON.
Since radius is
,
.
Using Pythagoras theorem:
![OF^2=FN^2+ON^2](https://img.qammunity.org/2021/formulas/mathematics/high-school/r8zvqkso5o0xl3h3fm4ap99u8kc2qcktb1.png)
![(4√(74) )^2=FN^2+16^2](https://img.qammunity.org/2021/formulas/mathematics/high-school/907r864wng5g8wlhs8sy45hy4ptxqdwjok.png)
![1184=FN^2+256](https://img.qammunity.org/2021/formulas/mathematics/high-school/nd19v8rgb9dp1srx0r4xbu9c49fqn4srxh.png)
Subtract 256 from both sides.
![928=FN^2](https://img.qammunity.org/2021/formulas/mathematics/high-school/lg48lzjphm3bsjl29ve4qwdnlsuf3ljd6x.png)
Taking square root on both sides.
![4√(58) =FN](https://img.qammunity.org/2021/formulas/mathematics/high-school/zeifivcse9kyzea1zeyor9aeumxjwfh7d8.png)
Therefore, FN = 4√58
Part c: ON bisects EF.
FN = EN = 4√58
EF = FN + EN
![EF = 4√(58) +4√(58)](https://img.qammunity.org/2021/formulas/mathematics/high-school/egbbwcl4rtjjoyxbp7pnvh47rw975cmwiw.png)
![EF = 8√(58)](https://img.qammunity.org/2021/formulas/mathematics/high-school/bbaheuyh4cksmcb730ty3zvpmad2pmnz8s.png)
![EF = 60.9](https://img.qammunity.org/2021/formulas/mathematics/high-school/ucltq6frs64ugqwztscckryk5e5i151986.png)
Therefore, EF = 60.9.