Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.

Let 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, seg PQ ⊥ 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, AP²=AR²+PR²
AR²=5²-x².. (1)
In △ARQ, AQ²=AR²+QR²
AR²=3²-(4-x)².. (2)
∴5²-x²=3²-(4-x)² from (1) (2)
25-x²=9-(16-8x+x²)
25-x²=9-16+8x-x²
25-x²=-7+8x-x²
32=8x
∴x=4
Substitute in eq(1) we get,
AR²=25-16=9
∴AR=3cm.
∴AB=2×AR=2×3
∴AB=6cm.
So, length of common chord AB is 6cm.