For a complete proof, you still need to show that the smaller square always fits inside the larger one, which you do by computing its diagonal and show that it is smaller than the half-side of the larger square.

Good solution but it isnt provable because the right angle of the triangle isnt labeled as the middle of the square

Very beautiful solution👍

Good Question

My method is to rotate the triangle around the center until the right-angled sides are parallel to the square, and then the area is equal to the triangle minus the small square.

It's actually not necessary to add the entire square construction to solve the problem. Simply drawing perpendiculars from the center of the square to the right and bottom midpoints will allow you to show via congruent triangles that the overlap area will always be 1/4 the area of the square. As long as the distance from the top triangle vertex to the midpoint of the triangle base is greater than the distance from the square center to its corners. In this case that's no problem, as 16 > 21/√2 (≈14.8). As for determining the area of the triangle, for an isosceles right triangle, that's simply the base squared and divided by 4. That can be proven by either constructing a full square as in the video (where the base length would become the square side length in A = s² ==> s²/4) or by mirroring the triangle about the base, where it becomes the diagonal of a square, whereupon you halve the area formula with respect to the diagonal (A = d²/2 ==> d²/4).

Amazing and beautiful!

Notice that the angle the triangle is rotated to is NOT SPECIFIED in this problem. And yet we are asked for a NUMBER. There is only one way for this to be possible - the shaded area must be INDEPENDENT of that angle. If complete the triangle into a new square with side length 32, it has an area of 32^2. We subtract the inner square area of 21^2. We are interested in a quarter of that area, so the shaded area is (32^2 - 21^2)/4 = 583/4. This will be the result regardless of the orientation of the triangle, provided that the 21-square doesn't extend beyond the 32-square side.

I extended the legs of the isosceles triangle to the edges of the square and then went off the rails. I proved that the four resulting quadrilaterals were congruent based on perpendicular/parallel line - angle theorems. This also means that the area of the quadrilateral is always 1/4 of the area of the square. So I rotated the triangle until the legs passed through the corners of the square. This turned the shaded area into a trapezoid with bases of 21 and 32. I calculated the height of the trapezoid as 5.5, and used the area formula ((b1+b2)*h/2). I sometimes have trouble seeing the easy solution. 😁😁😁

Çok hoş çözüm

Just make three extra rotated copies of the isosceles triangle. Then the shaded is (32^2-24^2)/4. Now I'll watch the video.