The Pythagorean theorem for the triangle AFC:

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

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

The Pythagorean theorem for the triangle EBD:

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

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

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

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

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

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

$$-462x=-3234 |·\left(-1\right)$$

$$462x=3234$$

$$x=\frac{3234 }{462}$$

$$x=7$$

$$x+89+y=142$$

$$y=142-x-89$$

$$y=142-7-89$$

$$y=46$$

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

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

$$h=\sqrt{5184}$$

$$h=72$$

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

$$A_{AED}=\frac{x·h}{2}$$

$$A_{AED}=\frac{7·\mathrm{72}}{2}$$

$$A_{AED}=\frac{7·\mathrm{72}}{2}=\frac{7·\bcancel{2}·36}{\bcancel{2}}=7·36$$

$$A_{EFCD}=b·h=89·72$$

$$A_{FBC}=\frac{y·h}{2}$$

$$A_{FBC}=\frac{46·72}{2}=\frac{\bcancel{2}·23·72}{\bcancel{2}}=23·72$$

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