Let ABCD be a parallelogram. Extend BC past B to F, and let E be the intersection of AB and DF. If the areas of triangles BEF and ADE are 4 and 9, respectively, find the area of parallelogram .
+128408 Note that triangle EAD is similar to triangle EBF
The scale factor from EBF to EAD = sqrt (4/9) = 2/3
Therefore
EB / EA = 2/3
EB = (2/3)EA and
EA = (3/2)EB
Then AB = EA + EB = (3/2)EB + EB = (5/2)EB = DC
And the height of EBF is (2/3) that of the height of EAD...so height of EAD = (3/2) height of EBF
Therefore EBCD is a trapezoid with bases EB and DC and height = height of EAD
So the area of EBCD = (1/2) (height of EAD) (EB + DC) =
(1/2)[ (3/2)height of EBF ] ( EB + (5/2)EB) =
(1/2) (3/2) (height of EBF) [( 7/2)EB ] =
(3/2) (7/2) * (1/2) height of EBF * EB
But (1/2) height of EBF * EB = Area of EBF
So area of trapezoid EBCD = (3/2)(7/2) (4) = 21
So...the area of ABCD = area of EAD + area of EBCD = 9 + 21 = 30
