Solution:
Given:
Let the length of the shorter leg be represented by x.
Hence, using Pythagoras theorem,
![\begin{gathered} \text{hypotenuse}^2=\text{adjacent}^2+\text{opposite}^2 \\ where; \\ \text{hypotenuse}=3+2x \\ \text{adjacent}=x \\ \text{opposite}=2+2x_{} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/pefot1c5l1a626weqnk083s1dgihlnq0no.png)
Hence,
![\begin{gathered} \text{hypotenuse}^2=\text{adjacent}^2+\text{opposite}^2 \\ (3+2x)^2=x^2+(2+2x)^2 \\ \text{Expanding the bracket,} \\ 9+12x+4x^2=x^2+4+8x+4x^2 \\ 9+12x+4x^2=5x^2+8x+4 \\ \text{Collecting the like terms to one side of the equation to form a quadratic equation,} \\ 0=5x^2-4x^2+8x-12x+4-9 \\ 0=x^2-4x-5 \\ x^2-4x-5=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/mhcbm3tc4py97ltsdqmlqbkwqg3h8gvdy0.png)
Factorizing the quadratic equation,
![\begin{gathered} x^2-4x-5=0 \\ x(x-5)+1(x-5)=0 \\ (x+1)(x-5)=0 \\ x+1=0\text{ OR x - 5 = 0} \\ x=0-1\text{ OR x = 0+5} \\ x=-1\text{ OR x = 5} \\ \\ Si\text{ nce the length of the side of a triangle can not be negative, then we ignore the negative value of x. Thus,} \\ x=5 \end{gathered}]()
Hence, the length of the three sides are;
![\begin{gathered} x=5\text{feet} \\ 2+2x=2+2(5)=2+10=12\text{feet} \\ 3+2x=3+2(5)=3+10=13\text{feet} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/kw72nke3k82mqv921gwatehgw5ftmuqthq.png)
Therefore, the length of each of the three sides of the right triangle are;
5feet, 12 feet, and 13 feet.