A circle is inscribed in a triangle touching points A, B and C.
Thus,
Side AV = Side BV;
Side BZ = Side CZ;
Side AW = Side CW.
[tex]\begin{gathered} \text{Given |AV|=14, then |BV|=14;} \\ \text{But |BV| + |BZ|=|VZ|} \\ |BZ|=24-14 \\ |BZ|=10,\text{ then |CZ|=10;} \\ \text{Let |AW|=|CW|=x} \end{gathered}[/tex]
Then, using the perimeter to get the unknown, we have;
[tex]\begin{gathered} P=|AV|+|AW|+|VZ|+|CZ|+|CW| \\ 64=14+x+24+10+x \\ 2x=64-48 \\ 2x=16 \\ x=\frac{16}{2} \\ x=8 \end{gathered}[/tex]
Thus, |AW| is 8
