Osnovice trapeza su 142 cm i 89 cm, a dijagonale su 120 cm i 153 cm. Odrediti površinu trapeza.

Pitagorina teorema za trougao AFC:

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

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

Pitagorina teorema za trougao EBD:

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

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

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

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

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

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

$x=7$

$x+89+y=142$

$y=142-x-89$

$y=142-7-89$

$y=46$

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

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

$h=\sqrt{5184}$

$h=72$

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

${P}_{AED}=\frac{x·h}{2}$

${P}_{AED}=\frac{7·\mathrm{72}}{2}$

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

${P}_{AED}=252$

${P}_{EFCD}=b·h=89·72$

${P}_{EFCD}=6408$

${P}_{FBC}=\frac{y·h}{2}$

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

${P}_{FBC}=1656$

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

$P=8316$