the common chord be AB and P \and\ Q be the centers of the two circles.
AP=5cm \and\ AQ=3cm.
PQ=4cm ...given
Now, segPQ⊥chord AB
AR=RB=1/2AB....perpendicular from center to the chord, bisects the chord
Let PR=x cm
so RQ=(4−x)cm
In/_\ARP,AP^2 =AR^2+PR^2
AR^2=5^2−x^2...(1)
In /_\ARQ,AQ^2=AR^2+QR^2
AR^2=3^2−(4−x)^2...(2)
5^2−x^2=3^2−(4−x)^2...from (1) & (2)
25−x^2=9−(16−8x+x^2)
25−x^2=−7+8x−x^2
32=8x
x=4 substituting in equation(1) we get , AR^2=25-16=9
AR=3cm
AB=2xxAR=2xx3=6
AB=6cmSo length of common chord is6 cm.