Yellow Quadrilateral

Who needs Menelaus? (On the other hand, the following might actually be essentially Menelaus' theorem rediscovered. I'm not good in geometry.)

Label the points as follows:

Denote the area of $DEO$ as $x$ , and the area of $ADE$ as $y$ .

Observe that using $BE$ as the base, $BOD$ and $EOD$ has the same height, and thus the ratio of their areas is the ratio of their bases. That is, $\dfrac{[BOD]}{[EOD]} = \dfrac{BO}{EO}$ , or $\dfrac{3}{x} = \dfrac{BO}{EO}$ .

By observing $BOC$ and $EOC$ , we also get $\dfrac{[BOC]}{[EOC]} = \dfrac{BO}{EO}$ , or $\dfrac{6}{4} = \dfrac{BO}{EO}$ . Equating the two equations gives $\dfrac{3}{x} = \dfrac{6}{4}$ , or $x = 2$ .

Using the same method, now with $AB$ as the base, we have $\dfrac{[ADE]}{[BDE]} = \dfrac{AD}{BD} = \dfrac{[ADC]}{[BDC]}$ , or $\dfrac{y}{3+2} = \dfrac{y+2+4}{3+6}$ . This gives $y = 7.5$ .

The area of the yellow region is thus $x+y = 2+7.5 = \boxed{9.5}$ .

F is the intersection of CD & BE. The calculation of the yellow area is shown on the diagram itself. It is interesting to note that if the triangle is to be formed then b²>ac.

Let the big triangle be denoted by ABC (A at the top , B at left , C at right)

Let the triangle of area 6 be denoted by OBC

Let the triangle of area 4 be denoted by OCD

Let the triangle of area 3 be denoted by OBE

Let the area of triangle ADO be x

Let the area of triangle AEO be y

In triangle AEC

y/(x + 4) = 3/6

Then

2y = x + 4 ........................................... (1)

In triangle ADB

x/(y + 3) = 4/6

Then

2y = 3x - 6 ............................................(2)

From (1), (2)

x= 5, y = 4.5

The area of the yellow region = 5 + 4.5 = 9.5

$\dfrac{4}{x} = \dfrac{10}{3+x+y}$
(This was the crux move in proving Ceva's Theorem. )

Similarly, $\dfrac{3}{y} = \dfrac{9}{4+x+y}$

Now, solve these equations. We get, $x=5$ and $y=4.5$ .

Therefore, area of the yellow quadrilateral= $x+y=9.5$

Let $x = (ABFE).$ By Menelaus' theorem, we have $\frac{AC}{CB} \cdot \frac{BF}{FD} \cdot \frac{FE}{EA} = -1$ in which $\frac{AC}{CB} = \frac{(DCA)}{-(DCB)} = \frac{13 + x}{-9},$ and $\frac{BF}{FD} = \frac{(CFB)}{(CFD)} = \frac{1}{2},$ and $\frac{DE}{EA} = \frac{(CED)}{(CEA)} = \frac{10}{x+3}.$

That is, we have $\frac{x+13}{-9} \cdot \frac{1}{2} \cdot \frac{10}{x+3} = -1$ Solving this equation yields $(ABFE) = x= 9.5$ as required.

Consider the diagram. Recall that the areas of triangles with equal altitudes are proportional to the bases of the triangles. We have, $\dfrac{AD}{DB}=\dfrac{A_{CDA}}{A_{CDB}}=\dfrac{A_{XDA}}{A_{XDB}}$

$\dfrac{a+b+4}{3+6}=\dfrac{a}{3}$ $\implies$ $3(a+b+4)=9a$ $\implies$ $a+b+4=3a$ $\implies$ $b+4=2a$ $\color{#D61F06}(1)$

$\dfrac{AE}{EC}=\dfrac{A_{ABE}}{A_{BEC}}=\dfrac{A_{AXE}}{A_{EXC}}$

$\dfrac{a+b+3}{4+6}=\dfrac{b}{4}$ $\implies$ $4(a+b+3)=10b$ $\implies$ $2a+2b+6=5b$ $\implies$ $2a+6=3b$ $\color{#D61F06}(2)$

Substitute $\color{#D61F06}(1)$ in $\color{#D61F06}(2)$ , we have

$b+4+6=3b$ $\implies$ $10=2b$ $\implies$ $5=b$

It follows that,

$2a=b+4=5+4=9$ $\implies$ $a=4.5$

Finally, the area of the yellow region is $a+b=4.5+5=$ $\boxed{9.5}$

We can skew the triangle without affecting areas. Thus I work in this triangle:

We know the areas of three triangles with based $AB$ . The areas are proportional to the heights, so that $r : s : t : y = 6 : 9 : 10 : A,$ where $A$ is the area of the entire triangle, which is 13 more than the yellow area.

From similar triangles, $\frac{QB}{AB} = \frac{s}{r} = \frac 6 9\ \ \ \ \therefore\ \ \ \ \frac{AQ}{AB} = \frac 1 3;$ also, $\frac{AQ}{AR} = \frac{s}{t} = \frac 6{10}\ \ \ \ \therefore\ \ \ \ \frac{AR}{AB} = \frac{AR}{AQ}\cdot \frac{AQ}{AB} = \frac{10}6\cdot\frac1 3 = \frac 5 9.$ This gives $BR / AB = (AB - AR)/AB = 1 - \frac 59 = \frac 49$ , and a third pair of similar triangles shows $\frac A{10} = \frac y t = \frac{AB}{BR} = \frac 9 4,$ from which follows $A = 22\tfrac12$ and the yellow area is $22\tfrac12 - 13 = \boxed{9\tfrac12}$ .