We consider two families of Pascal-like triangles that have all ones on the left side and ones separated by m – 1 zeros on the right side. The m = 1 cases are Pascal's triangle and the two families also coincide when m = 2. Members of the first family obey Pascal's recurrence everywhere inside the triangle. We show that the m-th triangle can also be obtained by reversing the elements up to and including the main diagonal in each row of the (1/(1 – x^m), x/(1 – x)) Riordan array. Properties of this family of triangles can be obtained quickly as a result. The (n, k)-th entry in the m-th member of the second family of triangles is the number of tilings of an (n + k) × 1 board that use k (1, m – 1)-fences and n – k unit squares. A (1, g)-fence is composed of two unit square sub-tiles separated by a gap of width g. We show that the entries in the antidiagonals of these triangles are coefficients of products of powers of two consecutive Fibonacci polynomials and give a bijective proof that these coefficients give the number of k-subsets of {1, 2, ... , n − m} such that no two elements of a subset differ by m. Other properties of the second family of triangles are also obtained via a combinatorial approach. Finally, we give necessary and sufficient conditions for any Pascal-like triangle (or its row-reversed version) derived from tiling (n × 1)-boards to be a Riordan array.