Fie O, intersecția diagonalelor perpendiculare.
[tex]\it \Delta OCD\ \sim\ \Delta OAB \Rightarrow \dfrac{OC}{OA}=\dfrac{CD}{AB} =\dfrac{9}{4}\Rightarrow OC=9k,\ \ OA=4k,\ \ k\in\mathbb{R}_{+}\\ \\ \\ AC=OC+OA=9k+4k=13k[/tex]
Notăm AD = h ( înălțimea trapezului )
Cu teorema catetei în triunghiul ADC, se obține:
[tex]\it h^2=13k\cdot4k=52k^2\ \ \ \ \ (1)\\ \\ \Delta ADC-dreptunghic\ \stackrel{T.Pitagora}{\Longrightarrow}\ AC^2-AD^2=CD^2 \Rightarrow (13k)^2-52k^2=9^2 \Rightarrow \\ \\ \Rightarrow 169k^2-52k^2=81 \Rightarrow 117k^2=81 \Rightarrow k^2=\dfrac{\ 81^{(9}}{117}=\dfrac{9}{13}\ \ \ \ \ (2)\\ \\ \\ (1),\ (2) \Rightarrow h^2=52\cdot\dfrac{9}{13}=4\cdot9=36 \Rightarrow h=6\ cm[/tex]
[tex]\it \mathcal{A}=\dfrac{CD+AB}{2}\cdotAD=\dfrac{9+4}{2}\cdot6=\dfrac{13}{2}\cdot6=13\cdot3=39\ cm^2[/tex]
