Question

Abcd trapez isoscel AB=25 si cd=9 h=6cm. Aflati aria perimetrul dist de la b la ad.DAU COROANA

Answer 1

Fie DM⊥AB si CN⊥AB => DM si CN->h

DM=CN=6 cm

DC=MN=9 cm

AM=NB=(AB-MN)/2=(25-9)/2=16/2=8 cm

A=(AB+DC)*DM/2=(25+9)*6/2=34*3=102 cm²

In ΔDMA,m(∡M)=90°=>(T.Pitagora):AD²=AM²+DM²=8²+6²=64+36=100=>AD=√100=10 cm

ABCD->trapez isoscel=>AD=BC=10 cm

P=AB+BC+CD+BA=25+10+9+10=54 cm

d(B,AB)=BD

In ΔBDA , m(∡D)=90°=>(T.Pitagora):BD²=AD²+AB²=10²+25²=100+625=725=>BD=√725=5√29 cm