ABC triunghi dreptunghic deoarece AB²=BC²+AC² (reciproca t lui Pitagora)
2500=1600+900
2500=2500
fie CM inaltimea trapezului, CM perpendicular pe AB
ABC triunghi dreptunghic sin B=AC/AB=40/50=4/5
BCM triunghi dreptunghic sin B=CM/BC
4/5=CM/30⇒CM=24
Atrap=[tex] \frac{(AB+CD)CM}{2} [/tex]
Δ BCM
MB²=BC²- CM²=900-576=324
MB=18
dar ΔBCM≡ΔADN (Δ drepunghice, CM≡DN,AD≡BC), deci MB=AN=18
DC=AB-(MB+AN))=50-36=14
Atrap=(50+14)24/2=768
ΔDOC≈ΔAOB (latui paralele)
CO/AO=DC/AB
CO/AO=14/50
CO/AO=7/25
25CO=7AO
