My 200 followers problem!

We can easily figure out that ,

$\Delta GDH \sim \Delta FHE \Rightarrow \frac{GD}{FH} = \frac{GH}{FE} = \frac{DH}{HE} = \sqrt \frac{[GDH]}{[FHE]} = \sqrt{\frac{200}{50}} = \frac{2}{1}$

$\Delta GDH \sim \Delta HIJ \Rightarrow \frac{GD}{HI} = \frac{DH}{IJ} = \frac{GH}{HJ} = \sqrt \frac{[GDH]}{[HIJ]} = \sqrt {\frac{200}{450}} ={\frac{2}{3}}$

$\Delta HIJ \sim \Delta FHE \Rightarrow \frac{HI}{FH} = \frac{IJ}{HE} = \frac{HJ}{FE} = \sqrt \frac{[HIJ]}{[FHE]} = \sqrt{\frac{450}{50}} = \frac{3}{1}$

Hence , Let $FH = x , GD = 2x , IJ = 3x \\ EF = y , GH = 2y , HJ = 3y \\ HE = z , HD = 2z , HI = 3z$

$AGHF , DHIB , HECJ , are , parallelograms \\ AG = x , BI = 2z , CE = 3y$

$\frac{AG}{GD} = \frac{x}{2x} = \frac{1}{2} = \frac{[AGH]}{[GHD]} = \frac{AGH}{200}\Rightarrow [AGH] = [AFH] = 100$

$\frac{BI}{IJ} = \frac{2z}{3z} = \frac{2}{3} = \frac{[BIH]}{[IHJ]} = \frac{BIH}{450}\Rightarrow [BIH] = [BDH] = 300$

$\frac{FE}{CE} = \frac{y}{3y} = \frac{1}{3} = \frac{[HFE]}{[HEC]} = \frac{50}{HEC}\Rightarrow [HEC] = [HCJ] = 150$

$[ABC]=[AGH]+[AFH]+[BIH]+[BDH]+[HEC]+[HCJ]+[DGH]+[HFE]+[HIJ]$

$[ABC] = 100 + 100 + 300 + 300 + 150 + 150 + 200 + 50 + 450$

$\huge \boxed{[ABC] = 1800}$

Simply... first.. take sum of square root of all three similar triangles area. That will be (200^1/2+50^1/2+450^1/2)=30*2^1/2(thirty root two). Now area of triangle ABC will be square of this,i.e. 1800.

Denote the left side of the "450" triangle as 3 ${x}$ such that the sides of the other two triangles are 2 ${x}$ and ${x}$ . Now using parallelograms these can be summed together to form the left side of the largest triangle, which turns out to be 6 ${x}$ , so it is double the scale factor of the "450" triangle, so it's area is 4X larger. $\therefore\ [ABC] = \boxed{1800}$

All the triangles are all similar.

Lets call the triangles with areas $50$ , $200$ , and $450$ $t1$ , $t2$ , and $t3$ , respectively.

The ratio of the side lengths of $t1$ and $ABC$ is $\frac{1}{6}$ , so the ratio of their areas is $\frac{1}{36}$ .

So the area of $ABC$ is $1800$ .

TAKE THE SQUARE ROOTS OF ALL THE THREE GIVEN AREAS SEPERATELY AND ADD THEM. FINALLY SQUARE THE RESULT AND YOU WILL GET IT.

Since area of similar triangles are proportional to the squares of their sides, we have , [ABC]=(√(200)+√(450)+√(50))²=1800 sq. units