三角形面積

BC = 3BD
BD : DC = 1 : 2
so, Area of ΔABD : area of ΔADC = 1 : 2 (same height)
i.e. Area of ΔABD = 10 cm^2 and area of ΔADC = 10 cm^2

AE = ED
so, area of ΔEDC = area of ΔAEC (same height)
= 10 cm^2

Produce CE to meet AB at F. Then join FD and let area of ΔEFD be x cm^2
Then area of ΔAFE is also x cm^2.
Hence, area of ΔFBD is (10 - 2x) cm^2

Area of ΔFBD : area of ΔFDC
= BD : DC (same height)
= 1 : 2

i.e. (10 - 2x) / (10 + x) = 1/2
which gives x = 2

Therefore, area of the shaded region
= (2 + 10) cm^2
= 12 cm^2