Ezekiel has a rectangular piece of paper with an area of 40 . The width of the paper is more than twice the height. He folds the bottom left and top right corners at $45^{\circ}$ and creates a parallelogram with an area of 24 . What is the perimeter of the original rectangle?

![](https://cdn.mathpix.com/cropped/2024_04_17_45c89f07dc9351e0476cg-2.jpg?height=198&width=415&top_left_y=324&top_left_x=909)
Suppose that the original rectangle has height $h$ and width $w$.

Since the area of the original rectangle is 40 , then $h w=40$.

Once the corners are folded, the height of the resulting parallelogram is still $h$ and the new base is $w-h$, because the two legs of each folded triangle are equal since these triangles are isosceles.

![](https://cdn.mathpix.com/cropped/2024_04_30_a838ccd738aa6ca69034g-07.jpg?height=276&width=480&top_left_y=1971&top_left_x=882)

Since the area of the parallelogram is 24 , then $h(w-h)=24$ or $h w-h^{2}=24$.

Since $h w=40$, then $h^{2}=40-24=16$.

Since $h>0$, then $h=4$.

Since $h w=40$ and $h=4$, then $w=10$.

The perimeter of the original rectangle is $2 h+2 w$ which equals $2 \cdot 4+2 \cdot 10=28$.