Insufficient Information On Areas?

As the 3 triangles are similars (because parallels sides), their ratio of areas are proportcional to the square ratio of sides. ${ \left( \frac { PI }{ PF } \right) }^{ 2 }= \frac { [PHI] }{ [PGF]} =\frac { 4 }{ 9 } \quad \therefore \quad \frac { PI }{ PF } =\frac { 2 }{ 3 }$ .

So, we can state $PI = 2k \quad and \quad PF = 3k$

With a similar argument, we can state $DE = 6k$ .

and, because the parallel lines, $BD = PI = 2k \quad and \quad EC = PF = 3k$

and so, $BC = 2k + 6k + 3k$

Therefore, ${ \left( \frac { BC }{ PI } \right) }^{ 2 }= \frac { [ABC] }{ [PHI]} \Rightarrow { \left( \frac { 11k }{ 2k } \right) }^{ 2 }=\frac { [ABC] }{ 4 } \Rightarrow \frac { 121 }{ 4 } =\frac { [ABC]}{ 4 }$

$\boxed {[ABC] = 121}$

Note: Trianges PDE , PFG , PHI , and ABC are all proportional. Futhermore, the sum of the width of the inner triangles equals the width of ABC , and the sum of the height equals the height of ABC .

$\sqrt{36} = 6 \quad \sqrt{9} = 3 \quad \sqrt{4} = 2 \quad \\ (6+3+2)^2 = \boxed{121}$

Labeling of sides, BI = a , IH = b, AH = c, AG = d, GF =e, FC = f, BD = g, DE=h, EC = i, PH = j, PE=k, PD = L PG = m

Gist of strategy : :

We first find ratio of corresponding sides of triangles PDE, HPI and PGF.

Then we move to finding ratio of corresponding sides of triangles PDE and BHE and PDE and GDC and then the rato of their areas and hence the areas of triangles BHE and `GDC (One of them will do).

Then we move from GDC to the triangle ABC ,try to find ratio of the sides and hence their areas and since we know A(GDC), we are done.

We consider similar triangles HPI and PDE..

Using the idea that areas of similar triangles are proportional to the squares of their corresponding sides , essentially A(PDE) / A(HPI) = k^2/j^2 = 36/4 = 9

Therefore, k=3j and k/(j+k) = 3/4

L/(a+b) = 3/4

Using the same principle of ratios of areas of similar triangles being proportional to square of corresponding sides, A(PDE) / A(BHE) = 9/16

A(BHE) = 16*36/9 = 64

Similarly L^2/m^2 = 36/9 = 4

Therefore L = 2m

L / (L + m) = 2/3

A(PDE)/A(GCD) = 4/9

A(GCD) = 9*36/4 = 81

Now, all we have to do is to connect A(GDC) to ABC

We We use the ratio GD/AB = (L+m) / (a+b+m) (l(AB) = a + b +c and c=m since they are opposite sides of a parallelogram). Alternatively, we could have considered triangle FAI and used the fact that DE/FI = PD/(AH+HI) and hence found AH / c. But that would be more work

Now, we know L=2m and L/(a+b) = 3/4

Therefore (a+b) = 4L / 3 = 4*2m / 3 = 8m / 3

GD / AB = (2m+m) / (8m/3 + m)

= 9/11

A(ABC) / A(GDC) = (11/9)^2 = 121 / 81

Therefore A(ABC) = 121/81 * A(GDC) = (121/81)*81

= 121

Since all the four triangles are similar, hence IP : DE : PF = 2 : 6 : 3

But, since IPBD and PFCE are parallelogramms, therefore, IP=BD and PF=EC

Thus, BD : DE : EC = 2 : 6 : 3 and BC= BD+DE+EC=11

Therefore, DE : BC = 6 : 11 and thus area=121*36/36 = 121

$\sqrt{2\times{A}} = \sqrt{2\times36} + \sqrt{2\times9} + \sqrt{2\times4}$

$A = \boxed{121}$

$\sqrt{36}=6$

$\sqrt{9}=3$

$\sqrt{4}=2$

$6+3+2=11$

$11^2=\boxed{121}$