$$x+10+y=24$$

$$y=24-10-x$$

$$y=14-x$$

Visina h deli trapez na levi i desni desni pravougli trougao. Potrebno je napisati Pitagorinu teoremu, za ova dva pravougla trougla:

$${15}^2=x^2+h^2$$

$${13}^2=h^2+y^2$$

$${15}^2-x^2={13}^2-y^2$$

$${15}^2-x^2={13}^2-{\left(14-x\right)}^2$$

$${15}^2-x^2={13}^2-\left({14}^2-2·14·x+x^2\right)$$

$${15}^2\cancel{-x^2}={13}^2-{14}^2+2·14·x\cancel{-x^2}$$

$${15}^2={13}^2-{14}^2+2·14·x$$

$$225=169-196+28·x$$

$$28x=225-169+196$$

$$28x=252$$

$$x=\frac{252}{28}$$

$$y=14-x=14-9$$

$$y=5$$

$$h^2={15}^2-x^2={15}^2-9^2=225-81=144$$

$$h=\sqrt{144}$$

$$P=m·h=\frac{a+c}{2}·h=\frac{24+10}{2}·12=\frac{34}{2}·12=\frac{\bcancel{2}·17}{\bcancel{2}}·12$$

$$P=204$$