Answer :
From the text of the task we have:
smaller leg: x
longer leg: x+3
hypotenuse: x + 6
Using the Pythagorean Theorema:
[tex](x+6)^2=x^2+(x+3)^2\\ \\ x^2+12x+36=x^2+x^2+6x+9\\ \\ x^2-6x-27=0\\ \\ \Delta=(-6)^2-4.1.(-27)=36+108=144\\ \\ x=\frac{6 \pm \sqrt{144}}{2}=\frac{6 \pm 12}{2}\\ \\ x_1=-3\\ \\ x_2=9[/tex]
Obviously -3 is not possible, so, smaller leg is 9 cm
smaller leg: x
longer leg: x+3
hypotenuse: x + 6
Using the Pythagorean Theorema:
[tex](x+6)^2=x^2+(x+3)^2\\ \\ x^2+12x+36=x^2+x^2+6x+9\\ \\ x^2-6x-27=0\\ \\ \Delta=(-6)^2-4.1.(-27)=36+108=144\\ \\ x=\frac{6 \pm \sqrt{144}}{2}=\frac{6 \pm 12}{2}\\ \\ x_1=-3\\ \\ x_2=9[/tex]
Obviously -3 is not possible, so, smaller leg is 9 cm