Check the picture below, so the perimeter of the trapezoid is really 13 + 15 + 13 + x + 15 + x, or just 56 + 2x.
[tex]\textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=\stackrel{hypotenuse}{13}\\ a=\stackrel{adjacent}{12}\\ b=\stackrel{opposite}{x}\\ \end{cases}[/tex]
[tex]\sqrt{13^2 - 12^2}=b\implies \sqrt{169 - 144}=b\implies \sqrt{25}=b\implies 5=b \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{perimeter}{56+2(5)\implies 66}~\hfill ~ \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ cos(\theta)=\cfrac{\stackrel{adjacent}{12}}{\underset{hypotenuse}{13}}\implies \theta =cos^{-1}\left( \cfrac{12}{13} \right)\implies \theta \approx 22.62^o \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{\measuredangle BCD}{\theta +90^o~~\approx 112.62^o}~\hfill[/tex]
