Area of ΔBAC = 6 in^2
EF = 2 in
BC = 3 in
Both triangles are similar, so:
Area ΔBAC : Area of ΔEDF = BC^2 : EF^2
Replacing:
6 / Area of ΔEDF = 3^2 / 2^2
Cross multiply
6 * 2^2 = 3^2 * Area of ΔEDF
24 = 9 * Area of ΔEDF
24/9 = Area of ΔEDF
Area of ΔEDF = 8/3 in^2 = 2.7 in^2
