Given ¯¯¯¯¯¯¯¯AB∥¯¯¯¯¯¯¯¯¯DE and the area of parallelogram ABFD is 24cm2
ΔAFB,ΔAGB,ΔAEB and parallelogram ABFD are on the same base and between the same parallels
∴ Area of ΔAFB = Area of ΔAGB
= Area of ΔAEB
=1/2× Area of parallelogram ABFD
=1/2×24=12cm2.