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In circle O, CD = 56, OM = 20, ON = 16, CD is perpendicular OM, and EF is perpendicular to ON (The figure is not drawn to scale.)

a. Find the radius. If your answer is not an integer, express it in radical form.
b. Find FN. If your answer is not an integer, express it in radical form.
c. Find EF. Express it as a decimal rounded to the nearest tenth.

In circle O, CD = 56, OM = 20, ON = 16, CD is perpendicular OM, and EF is perpendicular-example-1
User Tattvamasi
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1 Answer

1 vote

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 =
(1)/(2) (56) = 28

Using Pythagoras theorem:


OC^2=OM^2+CM^2


OC^2=20^2+28^2


OC^2=1184

Taking square root on both sides.


OC =4√(74)

The radius of the circle is 4√74.

Part b: Join FO, we get triangle FON.

Since radius is
4 √(74),
OF =4 √(74).

Using Pythagoras theorem:


OF^2=FN^2+ON^2


(4√(74) )^2=FN^2+16^2


1184=FN^2+256

Subtract 256 from both sides.


928=FN^2

Taking square root on both sides.


4√(58) =FN

Therefore, FN = 4√58

Part c: ON bisects EF.

FN = EN = 4√58

EF = FN + EN


EF = 4√(58) +4√(58)


EF = 8√(58)


EF = 60.9

Therefore, EF = 60.9.

In circle O, CD = 56, OM = 20, ON = 16, CD is perpendicular OM, and EF is perpendicular-example-1
User DanielGrech
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3.5k points