Sa se afle aria unui paralelogram ABCD in care stim ca AC⊥CB, AB=4cm si m(<D)=45 de grade.
zindrag
M(∡B)=m(∡D)=45 grade (∡ op in paralelg) ⇒ΔACB dr isoscel in ΔABC dr isos aplicam teo lui Pitagora si obtinem AC²+BC²=AB², dar AC=BC deci 2AC²=16 AC²=8 ⇒AC=√8=2√2 Aria ΔABC= AC*BC/2= 2√2*2√2/2=4 cm²
⇒ΔACB dr isoscel
in ΔABC dr isos aplicam teo lui Pitagora si obtinem
AC²+BC²=AB²,
dar AC=BC
deci 2AC²=16
AC²=8
⇒AC=√8=2√2
Aria ΔABC= AC*BC/2= 2√2*2√2/2=4 cm²
dar ΔCAD= ΔACB
⇒Aria ABCD=Aria ΔACD + Aria ΔABC=2*Aria ΔABC= 2*4= 8 cm²
O seara buna!