How many isosceles triangles can be formed with no side exceeding 10?excluding equilateral triangles

Answer is 65 ,can anyone explain?

Let x, x, y are integer sides of an isosceles triangle. So for x = 10, y can take values from 1 to 9 satisfying the constraints (So 9 triangles). Now keep decreasing the value of x by 1 and find how many values of y exist. For x = 9, y can take values from 1 to 10 except 11 (So 9 triangles). Similarly, for x = 8, the number of values of y can be 9. This pattern remains the same until x = 6 (9 values of y) For x = 5, y cannot be 5 and 10 (so 8 values). For x = 4, y cannot be 4, 10, 9, 8 (so 6 values). For x = 3, y cannot be 3, 10, 9, 8, 7, 6 (so 4 values). For x = 2, y can be 1, 3 (so 2 values). For x = 1, the only value of y is 1 with the given constraints but this becomes an equilateral triangle so not possible. So if u sum up the values, it'll be 2+4+…+8+(9∗5) = 20+45 =65 So the total number of triangles possible is 65.