1) ABCD paralelogram, BD ⊥ AD ⇒ BD= h
AD= 20cm
AB=CD= 25cm
avem ΔBDA dreptunghic in D.
BD²= AB²- AD²
BD²= 25²- 20²
BD= √225⇒ BD= 15cm
A= b·h= BD·AD= 15·20= 300cm²
2) ABCD romb
AC=d₁
BD=d₂
l= 13cm
A= d₁·d₂ :2
d₁∩d₂= O
d₁ ⊥ d₂
avem ΔAOB dreptunghic in O
AB= 13cm
pr AO pe AB= AF= 9cm
AO²= AF·AB
AO²= 13·9
AO=√117
AO=3√13 cm.
d₁= AC= 2AO= 2·3√13= 6√13 cm.
pr BO pe AB= BF
BF= 13-9
BF= 4cm
BO² = 4·13
BO²= 52⇒ BO=√52
BO= 2√13 cm
d₂= BD=2BO=2·2√13= 4√13cm.
A= (4√13· 6·13) : 2
A=( 4·6·13):2
A= 156cm².
uite cele doua figuri in imagine.
