What is the area?

First of all it is obvious that all the triangles in this problem are right triangles. Now lets take a look a those two paralellograms in the figure above.Since they are congruent let their area be $S$ . Easily we get $S=30*15=450 cm^{2} \Rightarrow 2S=900$ .Thus the area of the paralelograms together equals the area of the square .Let the area of those four disordered shapes together be $S_1$ .Let the area of the square in the middle be $S_2$ .Let the area of those four shaded triangles each be $S_3$ since they are equal. Then, $S_1+S_2+4S_3=S_1+2S_2 \Rightarrow S_2=4S_3$ .If you look carefully you'll notice four congruent triangles that are similiar to the shaded triangles.The similiarity coeficent is 2 thus the area of each of them is $4S_3$ .By the figure we have : $16S_3+4S_3=900 \Rightarrow 4S_3=\frac{900}{5}=\boxed{180}$ .

Triangle $ADE$ is similar to triangle $ABC$ . It has a right angle at $D$ and the ratio of the legs is $AD:DE=2:1$ .

Pythagorean theorem gives us $4x^2+x^2=15^2$ from which $x^2=45$ .

Area of $\triangle ADE=\frac{1}{2}\times2x\times x=x^2=45$ .

There are four of those triangles, so the area is $4\times45=180$ .