Answer:
2.7 in²
Step-by-step explanation:
Given:
∆BAC ~ ∆EDF
Area of BAC = 6 in²
EF = 2 in
BC = 3 in
Required:
Area of ∆EDF
SOLUTION:
Let x = area of ∆EDF
[tex] \frac{6 in^2}{x} = (\frac{3 in}{2 in})^2 [/tex] (theorem of area of similar triangles)
[tex] \frac{6 in^2}{x} = (\frac{3 in}{2 in})^2 [/tex]
[tex] \frac{6}{x} = \frac{9}{4} [/tex]
Cross multiply:
[tex] 6*4 = 9*x [/tex]
[tex] 24 = 9x [/tex]
Divide both sides by 9
[tex] \frac{24}{9} = x [/tex]
[tex] 2.67 = x [/tex]
Area of ∆EDF = 2.7 in²
