Question 631800
<pre>
{{{drawing(200,200,-5,5,-5,5,circle(0,0,5),line(-3,4,3,4),line(-4,-3,4,-3),
locate(-3.45,4.45,A),locate(3.2,4.45,B), locate(-4.2,-3,C), locate(4,-3,D),
red(line(-5,0,5,0)),locate(0,4.75,6), locate(0,-3,8), locate(.2,.8,O),circle(0,0,.1), locate(-.2,0,10)
)}}}
Draw a green diameter perpendicular to the red diameter, which biscects 
chord AB at E into two 3 cm parts, chord CD at F into two 4 cm parts, and the
red diameter into two 5 cm radii 

{{{drawing(200,200,-5,5,-5,5,circle(0,0,5),line(-3,4,3,4),line(-4,-3,4,-3),
locate(-3.45,4.45,A),locate(3.2,4.45,B), locate(-4.2,-3,C), locate(4,-3,D),
red(line(-5,0,5,0)),locate(-1.6,4.75,3), locate(1.4,4.75,3), green(line(0,-5,0,5)),locate(-1.8,-3,4), locate(1.6,-3,4), locate(.2,.8,O),circle(0,0,.1),
locate(-3,0,5),locate(3,0,5),locate(.2,4,E),locate(.2,-3,F)

)}}}

Draw in radii OA and OC, which are 5 cm each.

{{{drawing(200,200,-5,5,-5,5,circle(0,0,5),line(-3,4,3,4),line(-4,-3,4,-3),
locate(-3.45,4.45,A),locate(3.2,4.45,B), locate(-4.2,-3,C), locate(4,-3,D),
red(line(-5,0,5,0)),locate(-1.6,4.75,3), locate(1.4,4.75,3), green(line(0,-5,0,5)),locate(-1.8,-3,4), locate(1.6,-3,4), locate(.2,.8,O),circle(0,0,.1),
locate(-3,0,5),locate(3,0,5), blue(line(0,0,-3,4),line(0,0,-4,-3)), locate(.2,4,E),locate(.2,-3,F), locate(-2,2,5), locate(-2.2,-.7,5)

)}}}
 
We use the Pythagorean theorem on right triangles OAE and OCF to find OE and
OF

   OA² = AE² + OE²           OC² = CF² + OF²
    5² =  3² + OE²            5² =  4² + OF²
    25 =   9 + OE²            25 =  16 + OF²
25 - 9 = OE²             25 - 16 = OF²
    16 = OE²                   9 = OF²  
     4 = OE                    3 = OF

{{{drawing(200,200,-5,5,-5,5,circle(0,0,5),line(-3,4,3,4),line(-4,-3,4,-3),
locate(-3.45,4.45,A),locate(3.2,4.45,B), locate(-4.2,-3,C), locate(4,-3,D),
red(line(-5,0,5,0)),locate(-1.6,4.75,3), locate(1.4,4.75,3), green(line(0,-5,0,5)),locate(-1.8,-3,4), locate(1.6,-3,4), locate(.2,.8,O),circle(0,0,.1),
locate(-3,0,5),locate(3,0,5), blue(line(0,0,-3,4),line(0,0,-4,-3)), locate(.2,4,E),locate(.2,-3,F), locate(-2,2,5), locate(-2.2,-.7,5),
locate(.2,2.5,4), locate(.2,-1.2,3)

)}}}

The distance between the chords = EF = OE + OF = 4 + 3 = 7cm.

Therefore it is none of the choices you gave.  The correct distance
is 7 cm.

Edwin</pre>