GH = EF = 9cm
Th. cos în ΔFGH :
[tex]\it cosG=\dfrac{FG^2+HG^2-FH^2}{2\cdot FG\cdot HG}=\dfrac{8^2+9^2-7^2}{2\cdot8\cdot9} =\dfrac{2}{3}\\ \\ \\ sinG=\sqrt{1-cos^2G}=\sqrt{1-\dfrac{4}{9}}=\sqrt{\dfrac{5}{9}}=\dfrac{\sqrt5}{3}\\ \\ \\ \mathcal{A}_{EFGH}=FG\cdot GH\cdot sinG=8\cdot9\cdot\dfrac{\sqrt5}{3}=24\sqrt5\ cm^2[/tex]
Răspuns:
Heron ; p=(7+8+9)/2=12 ; A =√p(p-a)(p-b)(p-c) ; (12-7)(12-8)(12-9)=5×4×3; A=√4×15=2√15=A FGH ; A EFGH=2×2√15=4√15
Explicație pas cu pas:
