Fie CE⊥AB, E∈AB
mas ∡(B)=45°⇔CEB tr dr isoscel⇔CE=EB
EB=AB-CD (AECDdreptunghi)
EB=DC=14-8=6
CE=6, CE inaltime
Arie trapez = (AB+CD)*CE/2
Arie trap=(14+8)*6/2=22*3=66
EB=6 ⇔CB=6√2
Perimetrul ABCD= 8+6+14+6√2=28+6√2
b) AC²=AD²+DC²=6²+8²=100
AC=10
DB²=AD²+AC²=6²+14²=232
DB=2√58
c)tr MDC asemenea tr MBA cu rap de asemanare =DC/AB=8/14=4/7
deci
si MC=4/7⇒
DM/MB=4/(7+4)=4/11⇒DM/DB=4/11
DM=4/11DB=(4/7) *2 √58=(8√58)/11
MC/MA=4/7⇒MC/AC=4/11
MC=(4/11)*AC=(4/11)*10
Perimetrul
DC +DM+CA=8+(8√58)/11+40/11
