The number of ways of dividing 10 girls into two groups of 5 each so that two shortest girls are in the different groups is:

Excluding two shortest girls, the remaining eight girls can be divided into two groups in  $\frac{1}{2!}\left({ }^8{C}_4\right)\left({ }^4{C}_4\right)=35$ ways. 

The two shortest girls can be put in two different groups in 2 ways. 

Therefore, the required number of ways is 70.