Area of black Square?
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 $\displaystyle ABCD$ be $\displaystyle 3x.$
Let $\displaystyle \angle FDC = \angle BGC = \theta$
Note the right triangle $\displaystyle FJB$ at the upper-right.
$\displaystyle FB = x,\;\angle B = \theta$Code:x F * * * * * * * B * θ * * * * * * J
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.
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.