ok... let's say, the pieces are "a", "b", and "c".
"a" being the shortest and "c" being the longest
we know the longest is 38 more than the shortest, so c = a + 38
and the third piece, b, is half the longest, or c/2
we know the three pieces come from the 102in ribbon, thus
a + b + c = 102
[tex]\bf a+b+c=102\quad
\begin{cases}
c=a+38\\
b=\frac{c}{2}\\
\quad =\frac{a+38}{2}
\end{cases}\implies a+(a+38)+\left( \frac{a+38}{2} \right)=102[/tex]
solve for "a".
c = a + 38, and b = c/2
