Fold-igami

First, let's find the area of polygon $b$ . Since the corner that is being folded to the opposite corner along the diagonal, the whole rectangle is being folded across the line perpendicular to that. The slope of the diagonal is - $n$ or $n$ , and let's say - $n$ . Therefore, the slope of the line perpendicular is $\frac{1}{n}$ . The area of this polygon can be visualized as 2 congruent triangles in the rectangle, where the base of both triangles is the perpendicular line, and they are both touching corners of the rectangle. These corners are opposite to each other. Therefore, when we fold the rectangle, these 2 triangles are going to lay exactly on top of each other, and then there are 2 other congruent triangles on the outside. The base of these triangles is the length of the perpendicular line. This can be found using the Pythagorean Theorem. Since the base of $R$ is 1, that is one of the legs. The other is found by the slope of the line. The slope of the line is $\frac{1}{n}$ . So, let's call $x$ the other leg. Since slope is rise over run, $\frac{x}{1}$ = $\frac{1}{n}$ , so the other leg is $\frac{1}{n}$ . Using the Pythagorean Theorem: √ $\frac{1}{n^2}$ +1=√ $\frac{n^2+1}{n^2}$ . Multiply this by the height, which is half of the diagonal, or $\frac{√n^2+1}{2}$ , which is √ $\frac{(n^2+1)^2}{(2n)^2}$ = $\frac{n^2+1}{2n}$ . Since this is a triangle, I have to divide it by 2, making $\frac{n^2+1}{4n}$ . I have to subtract this, the overlapping triangle, from the area of the rectangle to get $\frac{4n^2}{4n}$ - $\frac{n^2+1}{4n}$ = $\frac{3n^2-1}{4n}$ . Now, let's deal with polygon $a$ . In this case, the rectangle is being flipped over the line perpendicular to the diagonal of $R$ . We can visualize the area of $a$ as the rectangle minus a triangle with the vertices at 2 opposite corners, and one lying at the end of the line perpendicular to the diagonal (because of 2 triangles lying on top of each other). Wait a second! That sounds like the triangles that we used to solve the area of $b$ ! In this case, these two triangles are the same set of triangles, but they are the triangles cut in half (cut on the diagonal) and put together! These 2 triangles have the same areas as the previous triangles, because half of the base of the original triangles is these new triangles' heights, and double the height of the old triangles is the same as the base of these new triangles. Therefore, the area of $a$ is $\frac{3n^2-1}{4n}$ , which is the same as $b$ . This is because of the same exact process, subtracting an area of a triangle from the area of the rectangle. So, the answer would be: 2(3+(-1)+4)=2(6)= 12 .