1.
a)
A=AC*BD/2
ΔDAB, mA=60
AD=AB=10, din cele doua rel. ⇒DB=10 ( triunghiul este ecchilateral)
Notezi O intersectia diagonalelor
Intr-un romb diag sunt perpendiculare, se injumatatesc si sunt si bisectoare
ΔAOB, mO=90, mAOB=60/2=30⇒mB=60⇒th. 30-60-90 AO²=100-25=75
AO=√75=5√5
AC=2*AO=10√5
Aria= 10√5*10/2=50√5
b)
<ABO≡<EBN⇒mEBN=60
2.
ΔABC, construim inaltimea AM
ΔAMB, mM=90, mB=30,mA=60
AB=8⇒th 30-60-90 AM=AB/2
AM=4
A=BC*AM/2
A=5*4/2=10