In ΔAOB aplicam sin 60°:
sin 60°=[tex] \frac{c.o}{ip} \ \textless \ =\ \textgreater \ \frac{OB}{AB} [/tex]
stiind valoarea sinusului de 60° voi forma o proportie si voi afla jumatete de diagonala.Diagonalele cum se intersecteaza se injumatatesc,si cealalta jumatate de diagonala va fi congruenta cu cea aflata.
[tex] \frac{ \sqrt{3} }{2}= \frac{OB}{6} =\ \textgreater \ OB= \frac{6 \sqrt{3} }{2} =\ \textgreater \ OB=3 \sqrt{3} [/tex]
Asa cum am scris mai sus,cealalta jumatate de diagonala este congruenta cu asta=>DB=OB+OD=>DB=3√3+3√3=>DB=6√3 cm
Acum,in ΔBOC voi aplica Pitagora pentru a afla jumatate din cea de-a-2-a diagonala,la fel ca la prima,prima jumatate=a doua jumatate
OC²=AC²-OB²
OC²=6²-(3√3)²
OC²=36-27
OC²=9
OC=√9
OC=3
Dupa cum am zis mai sus prima jumatate=a doua jumatate=>AC=3+3=>AC=6 cm
Stiind diagonalele acum putem afla aria:
A=[tex] \frac{d1*d2}{2} \ \textless \ =\ \textgreater \ \frac{AC*DB}{2}=\ \textgreater \ \frac{6*6 \sqrt{3} }{2} =\ \textgreater \ 6*3 \sqrt{3} =\ \textgreater \ 18 \sqrt{3} cm^2[/tex]
