**Puzzle:** Find all values of n > 1 for which one can divide a rectangle into n right-angled triangles.

**Answer:**

We can divide any rectangle into **n right-angled triangles, for any n > 1**.

There are two methods to divide:

**Method 1:****Case 1: n = 2**, we can get two right-angled triangles by cutting the rectangle along the diagonal as shown in the figure.**For example:**In the rectangle ABDC, we draw a diagonal BC and get two right-angled triangles, BDC and ABC.**Case 2: n > 2**, we first cut the rectangle along the diagonal and continue with n-2 cuts of any of the available right-angled triangles into two right-angled triangles, by cutting them along the height onto its hypotenuse.**For example:**Consider the case of n = 6, in the rectangle ABDC, we first draw diagonal BC, and then (n-2) i.e 4 cuts are made in the triangle ABC along with the height onto the hypotenuse, to get six right-angled triangles. These are BDC, ACE, AEF, FEG, FGH, HGB.

**Method 2:**We can also solve the puzzle by considering it case-wise.**Case 1: n is even**: divide the rectangle into n/2 small rectangles and then can divide each of the smaller rectangle obtained, along its diagonal.**For example:**For n=6, in the rectangle AGHB, we first form 3 smaller rectangles as ABCD, CEFD, EGHF. Then, we divide each rectangle along its hypotenuse BC, DE, FG respectively, to get 6 right-angled triangles**Case 2: n is odd:**In this case, first, divide the rectangle into n-1 small triangles using the approach mentioned above and then can cut any of the triangles along with the height onto its hypotenuse.**For example:**For n=7, we first form 6 triangles as mentioned in the above point, and then we cut triangle ABC along with the height onto its hypotenuse, to obtain the 7th triangle AIB.