in this diagram, bac~edf. if the area of bac= 6 in.², what is the area of edf? PLZ HELP PLZ PLZ PLZ

Answer:
2.7 in²
Step-by-step explanation:
Area of ∆BAC : ∆Area of EDF = BC² : EF² (based on the area of similar triangles theorem)
Thus:
[tex] 6 in^2 : x in^2 = (3 in)^2 : (2 in)^2 [/tex]
[tex]\frac{6}{x} = \frac{3^2}{2^2}[/tex]
[tex]\frac{6}{x} = 2.25[/tex]
[tex]\frac{6}{x}*x = 2.25*x[/tex]
[tex]6 = 2.25x[/tex]
[tex]\frac{6}{2.25} = \frac{2.25x}{2.25}[/tex]
[tex]2.67 = x[/tex]
Area of ∆EDF = 2.7 in²