4 ms·
You got a bunch of responses already, here is an intuitive reason. In similar triangles all distances are scaled by a factor k, by definition. Then, intuitivel
by aaplok 10mo ago
You got a bunch of responses already, here is an intuitive reason.
In similar triangles all distances are scaled by a factor k, by definition. Then, intuitively the areas are scaled by a factor of k^2, since you obtain an area by multiplying two distances.
So the ratio of the area over the hypothenuse is scaled by a factor of k^2/k=k.
It is not hard to confirm the intuition that the areas are scaled by a factor of k^2, since it is precisely the product of the lengths of the two sides adjacent to the right angle.
- xeonmc 10mo agoI think it would be clearer with an explicit function of a rectangle's area with respect to its diagonals: A = x y x = r cos(θ) y = r sin(θ) => A(r) = r²cos(θ)sin(θ) A(c) = A(a) + A(b) = c²cos(θ)sin(θ) = a²cos(θ)sin(θ) + b²cos(θ)sin(θ) => c² = a² + b²