Problem 597

Using the Pythagorean Theorem, we get $x^2+x^2=\sqrt{8}^2$ or $2x^2=8$. So, $x^2=4$ or $x=2$ (since $x>0$).

Thus, the right-angled isosceles triangle has area $\$\dfrac12\times 2\text{ cm}\times2\text{ cm}=2\text{}$ cm}^2$.

A square with side length $\$8\text{}$
cm}$has area$$8\text{ cm}\times8\text{ cm}=64\text{}$ cm}^2$.

Thus, the smallest number of these triangles needed to completely cover
the square is $\$\dfrac{64\text{ cm}^2}{2\text{}}$
cm}^2}=32$.