Check the picture below.
so, we know the perimeter of the base is 17, thus
[tex]s + s + s + s = 17\implies 4s=17\implies s=\cfrac{17}{4} ~\hfill \stackrel{\textit{area of the base}}{s^2\implies \left( \cfrac{17}{4} \right)^2} \\\\[-0.35em] ~\dotfill\\\\ \textit{volume of a pyramid}\\\\ V=\cfrac{Bh}{3}~~ \begin{cases} h=height\\ B=\textit{base's area}\\[-0.5em] \hrulefill\\ h=15\\ B=\frac{17^2}{4^2} \end{cases}\implies \begin{array}{llll} V=\cfrac{1}{3}(15)\left( \cfrac{17^2}{4^2} \right)\implies V=\cfrac{1445}{16} \\\\\\ V \approx 90.31 \end{array}[/tex]
