1. ## area

Area of black Square?

3. ## Re: area

Hello, parmis!

I will assume that the side of the square is trisected.

Area of black square?

Code:
      A   x   E   x   F   x   B
* - - - * - - - * - - - *
| *       *   *   *   * |
|   *      .*.      *   |
|     *  .*:::*.  *  J  |
3x |       *:::::::*       | 3x
|     *   *:::*   *     |
|   *       *       *   |
| *θ      *θ  *       * |
* - - - * - - - * - - - *
D   x   G   x   H   x   C

Let the side of square $ABCD$ be $3x.$
Let $\angle FDC = \angle BGC = \theta$

Note the right triangle $FJB$ at the upper-right.

Code:
            x
F * * * * * * * B
*      θ *
*     *
*  *
*
J
$FB = x,\;\angle B = \theta$
Hence: . $FJ = x\sin\theta$
This is the side of the shaded square.

Area of shaded square: . $A_1 \:=\:(x\sin\theta)^2 \:=\:x^2\sin^2\!\theta\;\;[1]$

In right triangle $BCG\!:\;\tan\theta \,=\,\tfrac{3x}{2x} \,=\,\tfrac{3}{2}$
Hence: . $\sin\theta \,=\,\tfrac{3}{\sqrt{13}}$

Substitute into [1]: . $A_1 \:=\:x^2\left(\frac{3}{\sqrt{13}}\right)^2 \:=\:\frac{9x^2}{13}$

The area of square $ABCD\!:\;A \,=\,(3x)^2 \,=\,9x^2$

Therefore: . $\frac{A_1}{A} \:=\:\dfrac{\frac{9x^2}{13}}{9x^2} \:=\:\frac{1}{13}$

The area of the shaded square is one-thirteenth that of the original square.

4. ## Re: area

I agree with second poster.
Using Sorobans diagram where the slopes of diagonals are3/2,Iget a rhombus in the middle area = 1/12 of square

5. ## Re: area

Hi all,
I agree with the first response. More information should have been given. Also bjhopper is correct in asserting that the inner quadrilateral is a non-square rhombus when the outer quadrilateral is square. The attachment shows what I think is a reasonable interpretation of the problem and a solution for the area of the inner square.