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.

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:

Taking square root on both sides.

The radius of the circle is 4√74.

Part b: Join FO, we get triangle FON.

Since radius is , .

Using Pythagoras theorem:

Subtract 256 from both sides.

Taking square root on both sides.

Therefore, FN = 4√58

Part c: ON bisects EF.

FN = EN = 4√58

EF = FN + EN

Therefore, EF = 60.9.

the answer is: d) o, f, and kr.

atomic radius of bromine (br) is 114 pm.

atomic radius of oxygen (o) is 73 pm.

atomic radius of fluorine (f) is 72 pm.

atomic radius of krypron (kr) is 112 pm.

the atomic radius of a chemical element is a measure of the size of its atom.

the atomic radius varies with increasing atomic number, but usually increases because of increasing of number of electrons.

the atomic radius decreases across the periods because an increasing number of protons, because greater attraction between the protons and electrons.