CD si AB prop cu 3 si 5 => exista k nr natural incat AB=5k si CD=3k
construim CE⊥AB si avem
BE=AB-AE= AB-DC= 5k-3k=2k
in ΔADC dr avem AC²=AD²+CD²= 16*6+9k²
in ΔCEB dr avem BC²=CE²+EB²= 16*6+4k²
in ΔACB dr avem AB²= AC²+BC²
⇔ 25k²= 16*6+9k²+16*6+4k²
⇔ 12k²= 16*12
⇒k²= 16
⇒k=4
deci AB= 5k=20
CD= 3k=12
AD=4√6
si BC= √ 9k²+96= √16*9+16*6= 4√15
deci perim=AB+BC+CD+AD (calculezi singur)
O seara buna!