In triangle ABC, point D divides side AC so that AD : DC = 1 : 2. Let E be the midpoint of BD and let F be the point of intersection of line BC and line AE. Given that the area of ΔABC is 360 cm2, what is the area of ΔAED?
+128456 Triangles BDA and BDC are under the same height...so....their areas are to each other as their bases
So area of BDA = (1/(1 + 2)) [ABC] = (1/3) [360] = 120
Note that sin angle BEA = sin angle DEA
And BE = ED
And AE= AE
So
area of triangle BEA = (1/2) (BE)(AE) sin (BEA)
area of triangle AED = (1/2) (ED)(AE) sin ( DEA)
So area of triangle BEA = area of trinangle AED
So
[AED] + [ BEA ] = [ BDA ]
2 [ AED ] = 120
[ AED ] = 60
