MR-692 / 12. problem

The bases of the trapezoid are 142 cm and 89 cm, and the diagonals are 120 cm and 153 cm. Calculate the area of the trapezoid.

The Pythagorean theorem for the triangle AFC:

${h}^{2}+{\left(89+x\right)}^{2}={120}^{2}$

${h}^{2}={120}^{2}-{\left(89+x\right)}^{2}$

The Pythagorean theorem for the triangle EBD:

${h}^{2}+{\left(142-x\right)}^{2}={153}^{2}$

${h}^{2}={153}^{2}-{\left(142-x\right)}^{2}$

${120}^{2}-{\left(89+x\right)}^{2}={153}^{2}-{\left(142-x\right)}^{2}$

$14400-\left(7921+178x+{x}^{2}\right)=23409-\left(20164-284x+{x}^{2}\right)$

$14400-7921-178x\overline{)-{x}^{2}}=23409-20164+284x\overline{)-{x}^{2}}$

$-178x-284x=23409-20164-14400+7921$

$x=7$

$x+89+y=142$

$y=142-x-89$

$y=142-7-89$

$y=46$

${h}^{2}={120}^{2}-{\left(89+x\right)}^{2}={120}^{2}-{\left(89+7\right)}^{2}={120}^{2}-{96}^{2}$

${h}^{2}=14400-6724=5184$

$h=\sqrt{5184}$

$h=72$

$A={A}_{AED}+{A}_{EFCD}+{A}_{FBC}$

${A}_{AED}=\frac{x·h}{2}$

${A}_{AED}=\frac{7·\mathrm{72}}{2}$

${A}_{AED}=\frac{7·\mathrm{72}}{2}=\frac{7·\overline{)2}·36}{\overline{)2}}=7·36$

${A}_{AED}=252$

${A}_{EFCD}=b·h=89·72$

${A}_{EFCD}=6408$

${A}_{FBC}=\frac{y·h}{2}$

${A}_{FBC}=\frac{46·72}{2}=\frac{\overline{)2}·23·72}{\overline{)2}}=23·72$

${A}_{FBC}=1656$

$A={A}_{AED}+{A}_{EFCD}+{A}_{FBC}=252+6408+1656$

$A=8316$