Summary: 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 \((n + k) \times 1\) 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 \(\{1, 2, \ldots, n - m\}\) 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 \((n \times 1)\)-boards to be a Riordan array.