ducem din virful C inaltimea CH=AD=4√2 cm
din ΔCHB-dreptunghic
HB²=CB²-CH²
HB²=(4√6)²-(4√2)²=96-32=64
HB=8cm
dinΔACB-dreptunghic, utilizam teorema inaltimii
CH²=AH*HB
AH=CH²/HB=32/8=4
AB=AH+HB=4+8=12cm
DC=AH=4 cm
deci, bazele au lungimile de 4 si 8 cm
b)d1=AC
AC²=AH²+CH²=16+32=48
AC=4√3cm
d2=BD
BD²=AD²+AB²=32+144=176
BD=4√11 cm
c)fie DM-perpendiculara din D
DM=AC/2=2√3 cm