Question

Look at the given picture and assess given data.

Calculate and explain the area of the blue part (CAED) in the picture attached below.

*** For reference, checkout the attachment ***

Answer 1

The the heights of triangles AEB and BEF are same hence ratio of their areas are also ratio of AE:EF that 2:3. Both triangles AEB and DEF are similar hence ratio of their areas is square of ratio of their sides AE:EF that 2:3 = 4/9 hence area of triangle  DEF=9/4x2=9/2 .Area of DBC=DBF=9/2+3+15/2. Hence area of quadruple CAED= 15/2-2=11/2=5.5

Answer 2

See diagram.
Draw BG ⊥ AF

Ar(ΔAEB) = 2 cm² = 1/2 * AE * BG 
Ar(ΔBEF) = 3 cm² = 1/2 *  EF * BG

Taking their ratio:    EF / AE = 3/2

ΔAEB and ΔDEF ar similar, as the corresponding sides AE||EF, BE||DE, AB||DF.  The ratio of their areas:

Ar(ΔDEF) / Ar(ΔAEB) = AE² / EF² = 9/4
=>  Ar(ΔDEF) = 9/4 * 2 cm² = 4.5 cm²
=>  Ar(ΔBDF) = Ar(ΔDEF) + Ar(ΔBEF)
                       = 4.5 + 3 = 7.5 cm²

As  DCBF is a rectangle and BD is a diagonal, 
 Ar(ΔDCB) = Ar(ΔDBF) = 7.5 cm²

=>  Ar(DCAE) = Ar(ΔDCB) - Ar(ΔAEB)
                       = 7.5 - 2 = 5.5 cm²