Question 407869
Let {{{MN}}} and {{{PQ}}} be the parallel chords with {{{O}}} as {{{center}}} of the circle

given:

{{{ MN=14cm}}} and {{{PQ=7cm}}}  with distance {{{3cm}}} between the parallel chords {{{MN}}} and {{{PQ}}}. 

Let {{{X}}} and {{{Y}}} be the {{{mid}}}{{{ points}}} of {{{MN}}} and {{{PQ}}}. 

Then {{{OXY}}} is a {{{unique}}} line which is the {{{perpendicular}}}{{{ bisector}}}  to both {{{MN}}} and {{{PQ}}} 

This is the theorem of the line joining the center of the circle and  the mid point of the chord is perpendicular to the chord . 

Or else, a line joining the center of the circle cuts chord perpendicularly, then it bisects the chord.  


So, {{{XN =7cm}}} and {{{YQ = 3.5cm}}}

{{{XY = 3cm}}} ..........given

{{{XN =7cm}}} and {{{YQ = 3.5}}}  

Let {{{OX = x}}}, 

now from the right angled triangle, {{{OXN}}} , by Pythagoras theorem,

{{{OX^2+XN^2 = ON^2 = r^2}}}, where {{{r}}} is the {{{radius}}} of the circle, 

or

{{{x^2 + 7^2 = r^2}}}.........................(1)


Similarly from right triangle {{{OYQ}}},


{{{OY^2+YQ^2=r^2}}} or


{{{(3+x)^2+3.5^2 = r^2}}}...................(2).

From (1) and (2), left sides must be equal because right sides are equal:


{{{x^2+7^2=(x+3)^2+(3.5)^2}}}.solve for {{{x}}} 

{{{x^2+49=x^2+6x+9+12.25}}} 

{{{49-21.22 = 6x}}} 

{{{6x = 27.75}}} 

{{{x=27.5/6 = 4.625 cm}}}. 

Therefore, from(1).

{{{(4.625)^2+7^2 = r^2}}} 

{{{r = sqrt((4.625)^2+7^2) = 8.380012097cm}}}

 {{{r = 8.38cm}}} is the radius of the circle.