Question

In trapezul ABCD (AB//CD AB>CD) notam AC intersectat cu BD in O .Se cunosc AC=25 si AB=15, iar BO/BD=3/5. Calculati AO si CD.
(//=drepte paralele, /=supra)
URGENTTTTTTTTTTTTTT!!!!!!!!!!!!
Dau COROANA!!!!!!!!!

Answer 1

BO/BD=3/5, BO/(BO+OD)=3/5, BO/(BO+OD-BO)=3/(5-3), BO/OD=3/2
deoarece ABII CD, ΔOAB≈ΔODC
AB/CD=BO/ODAO/OC
AB/CD=3/2=AO/OC
AB/CD=3/2, CD=2x15/3=10
AO/OC=3/2
AO/(AO+OC)=3/(3+2)
AO=3x25/5=15

Answer 2

$\displaystyle Deoarece~ \sphericalangle AOB~si~ \sphericalangle COD~sunt~opuse~la~varf,~iar~AB \parallel CD, \\ \\ rezulta~ca~\Delta COD \sim \Delta AOB. \\ \\ Deci~\frac{CO}{AO}= \frac{DO}{BO} \Rightarrow \frac{CO+AO}{AO}= \frac{DO+BO}{BO} \Leftrightarrow \frac{AC}{AO}= \frac{BD}{BO}=\frac{5}{3}. \\ \\ Din ~\frac{AC}{AO}= \frac{5}{3} \Rightarrow AO= \frac{3}{5} \cdot AC= \frac{3}{5} \cdot25=15.$

$\displaystyle Tot ~din~ \Delta COD \sim \Delta AOB~avem:~ \frac{CD}{AB}= \frac{DO}{BO} \Rightarrow \\ \\ \Rightarrow CD= \frac{DO}{BO} \cdot AB= \frac{BD-BO}{BO} \cdot AB=\left( \frac{BD}{BO}-1 \right) \cdot AB= \\ \\ = \left( \frac{5}{3}-1\right) \cdot 15= \frac{2}{3} \cdot 15=10.$