Question

dreptunghiul ABCD cu AB=40 ; BC=30, construim AE_|_BD, si CF_|_BD , F si E apartine lui BD . Calculati EF

Answer 1

In ΔDAB , m(∡A)=90°=>(Teorema lui Pitagora):DB=√(AD²+AB²)=√(40²+30²)=√(1600+900)=√2500=50

AE⊥BD=>AE->inaltime in ΔDAB

In ΔDAB , m(∡A)=90°=>(a doua teorema a inaltimii):AE*DB=AD*AB=>AE=AD*AB/DB=30*40/50=24

InΔDEA , m(∡E)=90°=>(Teorema lui Pitagora):DE=√(AD²-AE²)=√(30²-24²)=√(900-576)=√324=18

In ΔDCB , m(∡C)=90°=>(a doua teorema a inaltimii):FC*DB=DC*BC=>FC=DC*BC/DB=40*30/50=24

In ΔCFB , m(∡F)=90°=>(Teorema lui Pitagora):FB=√(BC²-FC²)=√(30²-24²)=√(900-576)=√324=18

EF=DB-(DE+EF)=50-(18+18)=50-36=14

EF=14