A cut above .... and below

Imgur

Let L = Length of the square

The triangles must have base = $\frac{2L}{3}$ to have one third the area of the square.

Hence the side of parallelogram = hypotenuse of the triangles = $\sqrt{L^2+(\frac{2L}{3})^2}$

Area of parallelogram = $6\sqrt{L^2+(\frac{2L}{3})^2} = \frac{L^2}{3}$

Solving for $L^2$ we get $L^2 = 13 \times 36 = 468$

Let the side length of the square be $a$ and distance between the upper left corner or the lower right corner to the cut be $b$ , then we have:

$\dfrac {1}{2} ab = \dfrac {1}{3} a^2 \quad \Rightarrow b = \dfrac {2}{3}a$

Also:

$6\sqrt{a^2+b^2} = 6a \sqrt{1+\frac{4}{9}} = 2a \sqrt{13} = \dfrac {1}{3}a^2\quad \Rightarrow a=6\sqrt{13}$

Therefore, the area of the square $=a^2 = \boxed{468}$

To construct our sought, we will call, big square, we start by drawing a small square which side is distant between parallel we call a. The center of this square it will be the center of the big one.

To determine the corner of the big square we extent one side of the small square, building a segment of length of 2a.

We do counterclockwise the same with the other sides see picture the end of four segments are the corners of big square since we must comply with area ratio imposed for the middle parallelogram being 1/3 of total area of the big square.

Examine closely this central parallelogram we see it contain 4 and 1/3 times the area of central small square. The above means that must be 13 small square contained in the big one.

Being a = 6 Total area of square must be 13 by 36 equal 468

Let $\text{a}$ be the side of the square so that the area of the square is: $\displaystyle\text{A}_{\text{square}} = \text{a}^2$ .

Area of each region are all equal to each other. Therefore: Area of each region = $\displaystyle\frac{\text{a}^2}{3}$

Area of the triangle = $\displaystyle\frac{\text{ab}}{2}$

Both the areas listed above are equal, therefore:

$\displaystyle\frac{\text{a}^2}{3} = \frac{\text{ab}}{2}$

From here, we get $\displaystyle \text{b} = \frac{2\text{a}}{3}$

Observe that the triangle is a right triangle with $\displaystyle \text{a}^2 + \text{b}^2 = \text{x}^2$

Substituting b to the equation, we get this:

$\displaystyle \text{a}^2 + \left(\frac{2\text{a}}{3}\right)^2 = \text{x}^2$

If we take a look back at the areas of each region, we will find the area of the parallelogram is $\displaystyle6\text{x}$

Knowing that area of parallelogram is equal to the area of the triangle, then we can form an equation.

$\displaystyle 6\text{x}=\frac{\text{ab}}{2}$

Solving for x, we get: $\displaystyle\text{x}=\frac{\text{ab}}{12}$

We substitute this x to our equation.

$\displaystyle \text{a}^2 + \left(\frac{2\text{a}}{3}\right)^2 = \left(\frac{\text{ab}}{12}\right)^2$

We wanted to make all variables be $\displaystyle\text{a}$ , so we replace $\displaystyle\text{b}$

$\displaystyle \text{a}^2 + \left(\frac{2\text{a}}{3}\right)^2 = \left(\frac{a\left(\displaystyle\frac{2a}{3}\right)}{12}\right)^2$

Simplifying, we get:

$\displaystyle \text{a}^2 + \frac{4\text{a}^2}{9} = \left(\frac{a^2}{18}\right)^2$

Use $\displaystyle\text{u}=\text{a}^2$

$\displaystyle \text{u} + \frac{4\text{u}}{9} = \left(\frac{u}{18}\right)^2$

Solving u, we get: $\displaystyle\text{u}=468 = \text{a}^2 = \text{area of the square}$

As area of parallelogram is BASE x HEIGHT,
and area of triangle is BASE x 0.5 x HEIGHT.
Also parallelogram has two bases shared with each triangle, so in order to make the areas equal the height of triangle must be twice that of parallelogram.
i.e. HEIGHT of parallelogram is 6cm and that of triangles is 12cm.
And the BASE is 6cm + 12cm + 8cm = 26cm.
Where 8cm comes from the ratio of internal and external triangles (18/12) = (12/T), T = 8cm.
So we get 3 x (26 x 6) = 468 square cm.

Since three regions in Figure 1 have the same area, the vertices of the parallelogram divide the side of the square into two parts, whose ratio is equal to $2:1$ (or $1:2$ ). In Figure 2, the triangles which are marked by the same color have the same area. Therefore, we can cut the original square into parts and merger them to create $13$ squares whose side length is equal to $6 cm.$ The answer: $13x36 = 648$ $cm^2$

Let the side length of the square be $L$ , and let the length of the parallelogram that lies along a side of the square have length $x$ .

Form a right triangle from one end of the parallelogram that has $x$ as its hypotenuse. This triangle is similar to each of the two triangles that have side lengths $L$ and $L - x$ . We thus have that

$\dfrac{6}{x} = \dfrac{L}{\sqrt{L^{2} + (L - x)^{2}}} \Longrightarrow 6\sqrt{L^{2} + (L - x)^{2}} = xL.$

Next, we have from condition (ii) that

$\dfrac{L}{2}(L - x) = 6\sqrt{L^{2} + (L - x)^{2}} \Longrightarrow L(L - x) = 2xL \Longrightarrow L = 3x.$

Thus $6\sqrt{(3x)^{2} + (2x)^{2}} = 3x^{2} \Longrightarrow 6\sqrt{13} = 3x \Longrightarrow x = 2\sqrt{13}.$

So the area of the square is $L^{2} = (3x)^{2} = 36*13 = \boxed{468}$ cm $^{2}$ .

Let a = Side Length of Square

$A_Pink$ = $A_Green$ = $A_Blue$ = $\frac{1}{3}$ $\times a^2$ $\dots$ eq (1)

Lets start with any of the triangles and try to find "\b"" (see in image below)

$\frac{1}{3}$ $\times a^2$ = $\frac{1}{2}$ $\times a$ $\times b$

b = $\frac{2a}{3}$

So now lets calculate hypotenuse \c of that triangle :

c = $\sqrt{a^2 + b^2}$ = $\sqrt{a^2 + (\frac{2a}{3})^2}$

Now lets get to the main part which will remove all variables i.e., Area of Parallelogram

$A_Green$ = $Base Length$ $\times Perpendicular Length$ = $c$ $\times 6$ = $\sqrt{a^2 + (\frac{2a}{3})^2}$ $\times 6$

But from eq 1,

$A_Green$ = $\frac{1}{3}$ $\times a^2$

So, $\frac{1}{3}$ $\times a^2$ = $\sqrt{a^2 + (\frac{2a}{3})^2}$ $\times 6$

$\frac{1}{3}$ $\times a^2$ = $6a$ $\times(\frac{\sqrt{13}}{3})$

On solving, we get

a = 0 [obviously impossible]

and a = $6$ $\times\sqrt{13}$

So Area of Square = $a^2$ = $(6$ $\times\sqrt{13})^2$ = $(36$ $\times13)$ = 468

I have drawn a more realistic figure since the one given is misleading. Let's call a the side of the square. By definition, the area of trangle ADE is equal to parallelogram EBDF or rectangle EBGC. That is, a * AE/2 = a * EB, i.e. AE is 2/3 of a . So the area of the red triangle is a * AE/2 = a * a * (2/3) / 2 = (a^2)/3 . Now, let's draw a vertical from point E to line BF. This will be the given perpendicular distance betwee the two parallel lines. Let's call it h . The area of the rectangle EHDI is the same with that of the parallelogram EBDF, whihch is the same with the area of the read trianngle by definition. That is, (a^2)/3 = h * DE (1). Now in the red triangle, DE^2 = a^2 + (a * 2/3)^2 = a^2 + a^2 * (4/9) = a^2 * (1 + 4/9) (2). Raising both members of the previous equation (1) to power of 2, we get (a^4)/9 = h^2 * DE^2. Replacing DE^2 with the value found in equation (2), we get (a^4)/9 = h^2 * a^2 * (1 + 4/9). Dividing bot members of the equation by a^2, we get (a^2)/9 = h^2 * (1 + 4/9) => a^2 = 9 * h^2 + 9 * h^2 * (4/9) = 9 * h^2 + 4 * h^2 = 13 * h^2. Replacing h with its given value of 6 cm, we finally get the area of the square in cm^2: a^2 = 468 .