Tricky Treat!

Think about putting three such squares on top of each other.

Then the diagonal will be $90$ cm, and if the shorter side has length $x$ , then, courtesy of our favorite mathematician Pythagoras:

$x^2 + 9x^2 = 90^2$

$10x^2 = 8100$

$x^2 = \boxed{810}$

AB= x

AE= 1/3x

BE= 30

$x^2 + 1/9x^2= 30^2$

$10/9x^2=900$

$x^2= \boxed{810}$

You could simply say that the proportion of the right triangle on the left is 1:3:root 10. Then the edge of the square must be (3*30)/root 10. So the area is 8100/10=810.

Consider $\triangle ABE$ where $\angle ABE = \theta$ .

Let $AE=x$ and thus, by similiar triangles, $AB = 3x$ .

By Pythagoras' Theorem we can say that $BE=x\sqrt{3^2+1^2}=x\sqrt{10}$ .

Since $x\sqrt{10}=30$ we can deduce that $x=3\sqrt{10}$ .

Finally, the total area of the square is equal to $(3x)^2=9x^2=9\times90=810$ .

Not quite as elegant as Geoff's solution but we have effectively done the same thing.

Credit owed to Pythagoras.

This question is NMTC 2016 Kaprekar Contest #1... You should name the title according to this...