Find The Area

The dimensions of the rectangle are $2a \times a$ ; we must calculate $2a^2$ .

Let $w$ be the width of the green parallelogram. Then $w = \tfrac23a$ to make the three areas equal. The slanted sides of the parallelogram have length $d = \sqrt{w^2 + a^2} = \sqrt{\left(\tfrac 23a\right)^2 + a^2} = \tfrac13\sqrt{13}a.$ Using similar triangles, $\frac a d = \frac p w,$ so that $\tfrac23 a^2 = pd = \tfrac13\sqrt{13}pa.$ Thus we find $a = \tfrac12\sqrt{13}p$ and the area of the rectangle is $2a^2 = \boxed{\dfrac{13}2p^2}.$

Call the width of the rectangle 2L and the height L (we are told that the width is twice the height). The overall area of the rectangle will therefore be $2L^{2}$ .

Calculate the area of one of the trapezoids (the red or blue areas) which has one side and a base of length L. Call the other side y. This trapezoid's area will be 1/3 of the overall area. ( $\frac{2}{3}L^{2}$ ).

Therefore we have $\frac{2}{3}L^{2}=L\frac{(L+y)}{2}$ which gives $y=\frac{L}{3}$ .

Now look at the parallel lines. If you split the rectangle into 2 squares, the area between the parallel lines forms a right-angled triangle and will be 1/6 of the area of the rectangle. We don't know the length of the hypotenuse but we do know the other two sides. One is L and the other is L-y (which is $\frac{2L}{3}$ ).

Using Pythagoras, we can calculate the hypotenuse to be $\frac{\sqrt{13}}{3}L$ .

The area of this triangle will be $\frac{base*height}{2}$ which will be equal to $\frac{2L^{2}}{6}$ . The height of this triangle is p. Therefore we have... $\frac{\sqrt{13}Lp}{6}=\frac{2L^{2}}{6}$ .

Solving this give us $L=\frac{\sqrt{13}}{2}p$ .

And we know the area of the rectangle is $2L^{2}$ . This then gives us...

$A=2\left ( \frac{\sqrt{13}}{2}p \right )^{2}=\frac{13}{2}p^{2}$ .

Is not hard to see that the side of the parallelogram is a third of the longer side. Half of this parallelogram is a rectangle triangle of sides proportional to $2, 3, \sqrt{13}$ Let $p=3x$ The base of the parallelogram is then $\sqrt{13}x$ Then the area of the rectangle is $(3\sqrt{13}x)(3/2\sqrt{13})=\frac{9x^2 13}{2}=\frac{13p^2}{2}$ :)