Given two unequal squares, how can you create a third square that combines both their areas? or: Given two unequal squares, how can you find a third square that is equal to the difference between the areas of the larger and smaller squares? Easy!
A rectangle with sides equal to the sides of the two squares will have its diagonal equal to the side of the square which combines them.
Conversely: The point at which the side of the larger square, when swung through an arc, intersects the (extended) side of the smaller square will give the length of the side of the median square whose area is sought. The closer to a square the original rectangle, the greater the distance between the median and large squares.
The greater the difference between the two sides of the rectangle (the small and median squares) the more the median square will depart from the smaller square and approach the larger.
In the special case of the square, the small and median squares are equal. The diagonal of this square produces two right angled isosceles triangles. Formula:
For any rectangle with sides a and b and diagonal c: a²
= 2a(c-b) + 2(b-a)(c-b) + (c-b)² also: a²
= c(c-b)·b(c-b)
If not a sacrifice of 100 lambs, do I at least get a glass of milk?
I like: c2 = 2a2 + e2 + 2ae
and it works well in practice!
not so sure about: a(a + e) = (c2 - e2) / 2 kinda nice to visualize but seems like a tautology?
a(a + e) is the source rectangle, however, (of which c is the diagonal), so worth keeping an eye on? The above is where I decided to switch from illustrations to puzzles!
I switched d and e a few times but I quite like this way round: b - a = d c - b = e It is important to distinguish these from the Montessori "Trinomial Square", where a2, b2 and c2 are the three unit squares. In the system illustrated above b2 and c2 are increasingly compound.
Here is a translation: a = Montessori a b = Montessori a + b c = Montessori a + b + c d = b - a = Montessori b e = c - b = Montessori c
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