Rectangle Divisions

1. We are given the areas of $\Delta ABE$ and $\Delta BEF$
2. Note that $\Delta ABE$ and $\Delta BEF$ share a height, but have different bases, $AE$ and $EF$ respectively.
3. Hence, since their areas are in the ratio $2:3$ and their heights are the same, their bases are in the ratio $2:3 \Longrightarrow AE:EF=2:3$
4. Note that $\Delta ABE$ and $\Delta DEF$ are similar.

$\dfrac{\Delta ABE}{\Delta DEF} = \left( \dfrac{AE}{EF} \right)^{2}$

Since $\Delta ABE = 2 cm^2$ ,

$\Delta DEF = \left( \dfrac{3}{2} \right)^2 \times 2 = 4.5 cm^2$

$\Delta DEF +\Delta BEF = EACD + \Delta ABE$

$\Longrightarrow EACD = \boxed{5.5 cm^2}$

Let EG be the height of △ABE and EH be that of △BEH.

Suppose x is the length of AB. The area of △ABF= 5 = ½ x BF.

So BF is 10/x. Then the area of △BEF= 3 = ½ (10/x) EH; EH= 3x/5.

Likewise, the area of △ABE= 2 = ½ x GE; GE = 4/x.

The rectangle BGEH has one vertex on the diagonal of the big rectangle, so they are similar.

So GE:EH = BF:DF.

Since BF:BH = BF:GE = (10/x)/(4/x) =2.5,

DF = 2.5(EH) = (2.5)(3x/5) = 1.5x.

Thus, the area of rectangle BCDF = 1.5x*10/x = 15.

The area of △DEF = ½ DF HF = ½ 1.5x 6/x =4.5.

As a result, the shaded area = 15-4.5-2-3 = 5.5 cm2.

$\triangle s ~ABE ~ and~ FBE ~ has ~ base ~ AE~ and ~ FE~ but~ same~ height.\\ \therefore ~\dfrac{ AE}{FE}=\dfrac{ 2}{3}.......\triangle s ~ABE ~ and ~FDE ~are~ similar.\\ \therefore ~\dfrac{Area ABE}{Area FDE}=(\dfrac{2}{3})^2=\dfrac{4}{9}~~~~\therefore~Area FDE = 4.5.\\\implies~Area BDF=4.5 + 3= Area BDC=Area AEDC + Area AEB\\\therefore ~Area AEDC =4.5 + 3 - 2 = \boxed{\color{#D61F06}{\Large 5.5} }$

Rotate the rectangle 180 degrees and let B=(0,0), E=(x,y), A=(l,0), D=(b,h). We have $ly=4 \\ hx=6 \\ lh=10 \\ y/x=h/b \\ \implies (lh)*(hx)/(ly)=15 \\ (h^2)(x/y)=15 \\ (h^2)(b/h)=15 \\ bh=15 \\ \triangle BCD = bh/2 - 2=5.5$

(FBAB)/2 = area = 2 cm^2 + 3 cm^2 = 5 cm^2; FBAB = 10 cm^2; AB = 10/FB; AB^2 + FB ^2 = AF^2; (10/FB)^2 + FB^2 = AF^2; 100/FB^2 + FB^2 = AF^2; FB^4 – AF^2FB^2 + 100 = 0; (FB^2 – 10)^2 = 0; FB^2 = 10 FB = sqrt(10); -AF^2FB^2 = -20FB^2, so AF^2 = 20, so AF = sqrt(20) = 2*sqrt(5); Since AB^2 + FB^2 = AF^2, we get AB^2 + 10 = 20. Thus, AB^2 = 10, and AB = sqrt(10) = FB.

(EG * AB)/2 = 2 cm^2 = (EG * sqrt(10))/2. So EG*sqrt(10) = 4. EG = 4/sqrt(10); EH = FB – 4/sqrt(10) = sqrt(10) – 4/sqrt(10) = 10/sqrt(10) – 4/sqrt(10) = 6/sqrt(10);

Triangle AEB is similar to triangle DEF, so EG/EH = AB/DF = 4/sqrt(10) / 6/sqrt(10) = 4/6 = 2/3; AB/DF = 2/3 = sqrt(10)/DF. 2DF = 3sqrt(10). DF = 3sqrt(10)/2; Area of triangle DBF = (EHDF)/2 = (6/sqrt(10) times 3sqrt(10)/2)/2 = (18/2)/2 = 4.5.

So area of triangle BCD = 4.5+3 = 7.5. That means the blue area is 7.5-2 = 5.5.

Triangles \triangle\text ABE and ABF have the same base, so the ratio of their areas (2/(2+3)) = the ratio of their heights. When you go on the diagonal from B to D, by the height you reach E, you have descended 2/5 of the height of the rectangle (BCDF does not need to be a rectangle, it is enough to say it is a parallelogram, we get the same result!), thus BE is the 2/5 of BD. Enlarge triangle ABE keeping B: E goes to D, A goes to A'. Triangle A'CD is congruent to ABF, their area is 5 cm2. The enlarged sides and heights of A'BD are 5/2 times of those of ABE, thus the area of A'BD=5/2 * 5/2 *2 =12.5, of which 5 is A'CD, 2 is ABE, thus 5.5 cm2 is the blue area.

It is clear that,

 [FBA] = [DBA]

hence,

 [DEA] = 3

From the given areas, we have

 [ABE] : [BEF] = AE : EF

also,

 [DEA] : [DEF] = AE : BF

It implies

 [ABE] : [BEF] = [DEA] : [DEF]

So

 3 : [DEF] = 2 : 3

then we get

  [DEF] =  4.5

Finally

  [DEF] + [BEF] = [ACDE] + [AEB]
     4.5   +    3      = [ACDE] +   2

from this,

  [ACDE] = 5.5