In this diagram, bac~edf. if the area of bac= 6 in.², what is the area of edf? PLZ HELP PLZ PLZ PLZ
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Answer:
2.7 in²
Step-by-step explanation:
Since ∆BAC and ∆EDF are similar, therefore, the ratio of their area = square of the ratio of their corresponding side lengths.
Thus, if area of ∆EDF = x, area of ∆BAC = 6 in², EF = 2 in, BC = 3 in, therefore:
[tex] \frac{6}{x} = (\frac{3}{2})^2 [/tex]
[tex] \frac{6}{x} = (1.5)^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]
[tex] x = 2.7 in^2 [/tex] (nearest tenth)