MR-695 / 14. problem

The sides of the triangle are 13 cm, 14 cm and 15 cm. A line parallel to the largest side of the triangle cuts off a trapezoid with a perimeter of 39 cm. Calculate the area of the trapezoid.

$x+y+z+15=39$

$x+y+z=24$

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:

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

$13x=195-15y$

$13x+15y=195$

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

$14x=210-15z$

$14x+15z=210$

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

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

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

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

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

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

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

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

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

$z=\frac{21}{2}$

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

$4y-273=-234$

$4y=273-234$

$4y=39$

$y=\frac{39}{4}$

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

$4x+39+42=96$

$4x=96-39-42$

$4x=15$

$x=\frac{15}{4}$

$p+x+q=15$

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

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

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

$p+q=\frac{45}{4}$

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

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

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

${h}^{2}=\frac{441}{4}-{q}^{2}$

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

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

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

${h}^{2}=\frac{1521}{16}-{p}^{2}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$p=\frac{99}{20}$

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

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

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

$h=\frac{42}{5}$

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

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

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

$A=\frac{315}{4}=78,75$