We will prove that the desired covering is impossible. Let us assume the opposite, i.e., a square with side length \( a \) can be tiled with \( k \) congruent right-angled triangles, whose sides are of lengths \( b \), \( b\sqrt{3} \), and \( 2b \). Then, the area of such a triangle is \[ \frac{b^2 \sqrt{3}}{2} \] , and the area of the square is \[ S_{\text{sq}} = k \cdot \frac{b^2 \sqrt{3}}{2}. \tag{1} \] Furthermore, the length of the side of the square, \( a \), is obtained by the contribution of an integer number of lengths \( b \), \( 2b \), and \( b\sqrt{3} \), hence \[ a = m b \sqrt{3} + n b, \]where \( m, n \in \mathbb{N} \cup \{0\} \), and at least one of the numbers \( m \) and \( n \) is different from zero. So the area of the square is \[ S_{\text{sq}} = a^2 = \left(m b \sqrt{3} + n b\right)^2 = b^2 \left(3m^2 + n^2 + 2 \sqrt{3} m n\right). \tag{2} \] Now, from (1) and (2), it follows that \[ 3m^2 + n^2 + 2 \sqrt{3} m n = k \cdot \frac{\sqrt{3}}{2}, \]i.e., \[ 6m^2 + 2n^2 = (k - 4mn) \sqrt{3}. \tag{3} \] Because \( 3m^2 + n^2 \neq 0 \) and from equality (3), it follows that \( 4mn \neq k \). Using (3) again, we get \[ \sqrt{3} = \frac{6m^2 + 2n^2}{k - 4mn}, \]which contradicts the fact that \( \sqrt{3} \) is irrational, because \( \frac{6m^2 + 2n^2}{k - 4mn} \) is a rational number. Finally, we have obtained a contradiction, which proves that the desired covering is impossible.