Given
It is given that the longer leg is 4 cm longer than the shorter leg.
It is also given that the hypotenuse is 8 cm longer than the shorter leg.
Step-by-step explanation
Let the shorter leg be x.
The longer leg be y.
Then the relation formed is
![y=x+4](https://img.qammunity.org/2023/formulas/mathematics/high-school/ls0mnu85f8s2uzc02n7uo809jwxrnnlfm6.png)
It is also given that the hypotenuse is 8 cm longer than the shorter leg.
Let the hypotenuse is z.
![z=8+x](https://img.qammunity.org/2023/formulas/mathematics/high-school/3lhqa72u7ewwbom5rgqbl2unyxomaakhin.png)
Then the perimeter of the right triangle is the sum of all the sides.
![x+y+z](https://img.qammunity.org/2023/formulas/mathematics/high-school/ssrluinedfoxs6fqed9qk0ppjmebllllxd.png)
But to find the value of x , use the Pythagoras theorem.
![z^2=y^2+x^2](https://img.qammunity.org/2023/formulas/mathematics/high-school/35f9q6zrom8nnskjaenv8nqcejm46g520d.png)
Substitute the values.
![\begin{gathered} (8+x)^2=(4+x)^2+x^2 \\ 64+x^2+16x=16+x^2+8x+x^2 \\ 64+16x=16+8x+x^2 \\ 64-16+16x-8x-x^2=0 \\ -x^2+8x+48=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/agozwinimc3q3vc6zrcypp4vl96ubu5f7t.png)
Solve the quadratic equation to find the value of x.
![\begin{gathered} (x-12)(x+4)=0 \\ x=12,-4 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/ehadd7lnfda1ykqklzton778o9siet5fje.png)
As x cannot be negative , then the value of x is 12.
The length of shorter leg is 12.
The length of longer leg is 12+4=16.
The hypotenuse is 12+8=20.
Now , the perimeter of right triangle is
![\begin{gathered} P=12+16+20 \\ P=48cm \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/94y8dxp726kl087fi0o9qyzph9pjgl0p4i.png)
Answer
The perimeter of the triangle is 48 cm.