Let ABCD be the parallelogram where AB = 10 m, AD = 12 m, and AC = 8 m (the smaller diagonal). Let P be the intersection of the diagonals AC and BD, such that AP = PC and BP = PD.
We can use the fact that the diagonals of a parallelogram bisect each other to solve for the length of BD (the other diagonal). Since AC is the smaller diagonal, we know that AP = PC = 4 m (half of AC).
Let x be the length of BD. Then, using the Pythagorean theorem in triangles ABP and ADP, we have:
AB^2 - AP^2 = BP^2 (by the Pythagorean theorem in triangle ABP)
AD^2 - AP^2 = DP^2 (by the Pythagorean theorem in triangle ADP)
Substituting the given values, we have:
10^2 - 4^2 = BP^2
12^2 - 4^2 = DP^2
Simplifying, we get:
BP^2 = 84
DP^2 = 128
Since BP = PD (from the definition of P), we have:
2BP^2 = x^2
Substituting the above values, we get:
2(84) = x^2
Solving for x, we get:
x = sqrt(168) = 2sqrt(42) meters
Therefore, the length of the other diagonal BD is 2sqrt(42) meters.