Let the chords intersect at right angles at point P. Let the radius of the circle be r. Join the ends of each chord to the centre of the circle O. Call the longer chord AB and the shorter chord CD. AB=50cm, CD=40cm. Let OX=k be the perpendicular from O to AB and OY=h be the perpendicular from O to CD. OA=OB=OC=OD=r.
AP+PB=AB=50cm; CP+PD=CD=40cm (CP and AP are the shorter segments of the chords). OX and OY bisect AB and CD, so AX=XB=25cm, CY=YD=20cm. PB=PX+XB=OY+25=h+25; similarly, PD=k+20. r^2=k^2+25^2=h^2+20^2, so k^2-h^2+25^2-20^2=0, 225=15^2=h^2-k^2. h=sqrt(225+k^2).
AP=50-PB=25-h; CP=40-PD=20-k. AP=25-sqrt(225+k^2).
Example: Let k=8 and h=17; h^2-k^2=289-64=225, r^2=k^2+25^2=64+625=689, so r=sqrt(689). (Also, h^2+20^2=r^2=289+400=689.) AP=25-17=8, CP=20-8=12. AP is the shorter segment of the longer chord.
Now, let k=13, then h=sqrt(394)=19.85 (approx). AP=5.15, CP=7; 394-169=225, r^2=169+625=394+400=794, r=sqrt(794). AP is the shorter segment of the longer chord.
If h=20, k=5sqrt(7)=13.23cm approx. AP=25-20=5cm; CP=20-5sqrt(7)=6.77cm. r=sqrt(800)=20sqrt(2).
If 15<h<25, then 0<k<20, 0<AP<10, 0<CP<20 and 25<r<sqrt(1025).
The example shows that as long as h^2-k^2=225, there are multiple solutions, because we don't know the radius of the circle containing the chords. The question does not specify at what angles the chords intersect (they can't bisect one another, because they would be diameters and have the same length, and there would be no shorter length of the chord, which is what the question is asking for).