# area

• Oct 21st 2013, 08:03 AM
parmis
area
Area of black Square?
• Oct 21st 2013, 09:01 AM
phys251
Re: area
• Oct 21st 2013, 06:15 PM
Soroban
Re: area
Hello, parmis!

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

Quote:

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 $\displaystyle ABCD$ be $\displaystyle 3x.$
Let $\displaystyle \angle FDC = \angle BGC = \theta$

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

Code:

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

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

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

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

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

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

The area of the shaded square is one-thirteenth that of the original square.
• Oct 23rd 2013, 06:22 AM
bjhopper
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
• Oct 24th 2013, 04:05 PM
johng
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.

Attachment 29570