$$x+y+z+15=39$$

The two purple height lines of the trapezoid created by the red line divide the trapezoid into an upper and a lower right-angled triangle and a central rectangle.

Triangles ABC and A₁B₁C are similar to each other therefore, it can be written:

$$\overline{AB} : \overline{A_1{B}_1} =\overline{CB} : \overline{C{B}_1} = \overline{CA} : \overline{C{A}_1}$$

$$15 : x=13 : \left(13-y\right)=14 : \left(14-z\right)$$

$$13x=15\left(13-y\right)$$

$$13x=195-15y$$

$$14x=15\left(14-z\right)$$

$$14x=210-15z$$

$$\begin{array}{ccccccccc}+& x& +& y& +& z& =& +& 24\\ +& 13x& +& 15y& +& 0z& =& +& 195\\ +& 14x& +& 0y& +& 15z& =& +& 210\end{array}$$

$$\begin{array}{ccccccccc}+& x& +& y& +& z& =& +& 24 |·(-13)\\ +& 13x& +& 15y& +& 0z& =& +& 195\\ +& 14x& +& 0y& +& 15z& =& +& 210\end{array}$$

$$\begin{array}{cccccccccc}-& 13x& -& 13y& -& 13z& =& -& 312& \left|\right(I+II)\\ +& 13x& +& 15y& +& 0z& =& +& 195& \\ +& 14x& +& 0y& +& 15z& =& +& 210& \end{array}$$

$$\begin{array}{ccccccccc}+& x& +& y& +& z& =& +& 24\\ & & +& 2y& -& 13z& =& -& 117\\ +& 14x& +& 0y& +& 15z& =& +& 210\end{array}$$

$$\begin{array}{cccccccccc}+& x& +& y& +& z& =& +& 24& |·\left(-14\right)\\ & & -& 2y& -& 13z& =& -& 117& \\ +& 14x& +& 0y& +& 15z& =& +& 210& \end{array}$$

$$\begin{array}{cccccccccc}-& 14x& -& 14y& -& 14z& =& -& 336& |\left(I+III\right)\\ & & -& 2y& -& 13z& =& -& 117& \\ +& 14x& +& 0y& +& 15z& =& +& 210& \end{array}$$

$$\begin{array}{ccccccccc}+& x& +& y& +& z& =& +& 24\\ & & +& 2y& -& 13z& =& -& 117\\ & & -& 14y& +& z& =& -& 126\end{array}$$

$$\begin{array}{cccccccccc}+& x& +& y& +& z& =& +& 24& \\ & & +& 2y& -& 13z& =& -& 117& |·\left(+7\right)\\ & & -& 14y& +& z& =& -& 126& \end{array}$$

$$\begin{array}{cccccccccc}+& x& +& y& +& z& =& +& 24& \\ & & +& 14y& -& 91z& =& -& 819& |·\left(II + III\right)\\ & & -& 14y& +& z& =& -& 126& \end{array}$$

$$\begin{array}{ccccccccc}+& x& +& y& +& z& =& +& 24\\ & & +& 2y& -& 13z& =& +& 195\\ & & & & -& 90z& =& -& 945\end{array}$$

$$z=\frac{-945}{-90}=\frac{21·\bcancel{45}}{2·\bcancel{45}}$$

$$2y-13·\frac{21}{2}=-117 |·2$$

$$2·2y-13·\frac{21}{\bcancel{2}}·\bcancel{2}=-117·2$$

$$4y-273=-234$$

$$4y=273-234$$

$$4y=39$$

$$x+\frac{39}{4}+\frac{21}{2}=24 |·4$$

$$x·4+\frac{39}{\bcancel{4}}·\bcancel{4}+\frac{21}{\bcancel{2}}·\bcancel{2}·2=24·4$$

$$4x+39+42=96$$

$$4x=96-39-42$$

$$4x=15$$

$$p+x+q=15$$

$$p+\frac{15}{4}+q=15$$

$$p+q=15-\frac{15}{4}$$

$$p+q=15·\frac{4}{4}-\frac{15}{4}=\frac{60-15}{4}$$

$${\mathrm{z}}^2=h^2+q^2$$

$$h^2={\mathrm{z}}^2-q^2$$

$$h^2={\left(\frac{21}{2}\right)}^2-q^2$$

$$y^2=h^2+p^2$$

$$h^2=y^2-p^2$$

$$h^2={\left(\frac{39}{4}\right)}^2-p^2$$

$$\frac{441}{4}-q^2=\frac{1521}{16}-p^2$$

$$\frac{441}{4}-q^2=\frac{1521}{16}-p^2 |·16$$

$$\frac{441}{\bcancel{4}}·\bcancel{4}·4-q^2·16=\frac{1521}{\bcancel{16}}·\bcancel{16}-p^2 ·16$$

$$1764-16q^2=1521-16p^2$$

$$16p^2-16q^2=1521-1764$$

$$16p^2-16q^2=-243$$

$$16\left(p^2-q^2\right)=-243$$

$$16\left(p-q\right)\left(p+q\right)=-243$$

$$16\left(p-q\right)·\frac{45}{4}=-243$$

$$4·\bcancel{4}·\left(p-q\right)·\frac{45}{\bcancel{4}}=-243$$

$$180\left(p-q\right)=-243$$

$$p-q=-\frac{243}{180}=-\frac{27·\cancel{9}}{20·\cancel{9}}$$

$$p-q=\frac{-27}{20}$$

$$\begin{array}{ccccccc}p& +& q& =& +& \frac{45}{4}& |\left(I+II\right)\\ p& -& q& =& -& \frac{27}{20}& \end{array}$$

$$2p=\frac{45}{4}-\frac{27}{20}$$

$$2p=\frac{45}{4}-\frac{27}{20}=\frac{45}{4}·\frac{5}{5}-\frac{27}{20}=\frac{225-27}{20}=\frac{198}{20}$$

$$p=\frac{198}{2·20}=\frac{99·\bcancel{2}}{\bcancel{2}·20}$$

$$h^2=y^2-p^2={\left(\frac{39}{4}\right)}^2-{\left(\frac{99}{20}\right)}^2=\frac{1521}{16}-\frac{9801}{400}$$

$$h^2=\frac{1521}{16}·\frac{25}{25}-\frac{981}{400}=\frac{38025-9801}{400}=\frac{28224}{400}$$

$$h=\sqrt{\frac{28224}{400}}=\frac{\sqrt{28224}}{\sqrt{400}}=\frac{168}{20}=\frac{42·\bcancel{4}}{5·\bcancel{4}}$$

$$A=\frac{c+x}{2}·h=\frac{15+{\displaystyle \frac{15}{4}}}{{\displaystyle \frac{2}{1}}}·\frac{42}{5}$$

$$A=\frac{15·{\displaystyle \frac{4}{4}}+{\displaystyle \frac{15}{4}}}{{\displaystyle \frac{2}{1}}}·\frac{42}{5}=\frac{{\displaystyle \frac{60}{4}}+{\displaystyle \frac{15}{4}}}{{\displaystyle \frac{2}{1}}}·\frac{42}{5}=\frac{{\displaystyle \frac{75}{4}}}{{\displaystyle \frac{2}{1}}}·\frac{42}{5}$$

$$A=\frac{75·1}{4·2}·\frac{42}{5}=\frac{15·\cancel{5}}{4·\bcancel{2}}·\frac{21·\bcancel{2}}{\cancel{5}}$$