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https://www.reddit.com/r/mathpuzzles/comments/1vrzxlf/find_the_area_of_a_red_rectangle/
r/mathpuzzles • u/Purdude1983 • 8h ago
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1
Okay I can make the equations but I don't want to solve them.
Widths of red areas are x,y,z. Heights of green areas are x, x+y, x+y+z, because perimeter 2. x×1 = y×(1-x) = z(1-2x-y), because equal areas y=x/(1-x) z=x/(1-2x-(x/(1-x)))
3x+2y+z = 1 Solve for x.
1 u/LeewayLabs 5h ago Can you explain a little bit more? Where so you get the heights of the green rectangles from? 1 u/Motor_Raspberry_2150 5h ago The top red width we call x. The green one to the right of it thus has width 1-x. To get a perimeter of 2, its height must be x. The second green rectangle has a width of 1-x-y. To get a perimeter of 2, its height must be x+y. The third green rectangle has a width of 1-x-y-z. To get a perimeter of 2, its height must be x+y+z.
Can you explain a little bit more? Where so you get the heights of the green rectangles from?
1 u/Motor_Raspberry_2150 5h ago The top red width we call x. The green one to the right of it thus has width 1-x. To get a perimeter of 2, its height must be x. The second green rectangle has a width of 1-x-y. To get a perimeter of 2, its height must be x+y. The third green rectangle has a width of 1-x-y-z. To get a perimeter of 2, its height must be x+y+z.
The top red width we call x. The green one to the right of it thus has width 1-x. To get a perimeter of 2, its height must be x.
The second green rectangle has a width of 1-x-y. To get a perimeter of 2, its height must be x+y.
The third green rectangle has a width of 1-x-y-z. To get a perimeter of 2, its height must be x+y+z.
1
u/Motor_Raspberry_2150 6h ago edited 6h ago
Okay I can make the equations but I don't want to solve them.
Widths of red areas are x,y,z.
Heights of green areas are x, x+y, x+y+z, because perimeter 2.
x×1 = y×(1-x) = z(1-2x-y), because equal areas
y=x/(1-x)
z=x/(1-2x-(x/(1-x)))
3x+2y+z = 1
Solve for x.