the perimeter of triangle is equal to the sum of all the side of the triangle,
The perimeter of given triangle is x+y+9+3
Solve for x and y,
Apply pythagoras theorem in triangle ADC,
[tex]\begin{gathered} AC^2=DC^2+AD^2 \\ y^2=81+AD^2^{} \\ AD^2=y^2-81 \end{gathered}[/tex]
Now, in triangle ADB
Apply pythagoras,
[tex]\begin{gathered} AB^2^{}=BD^2+AD^2 \\ x^2=9+AD^2 \\ AD^2=x^2-9 \end{gathered}[/tex]
Compare the value of AD from both the equation,
[tex]\begin{gathered} x^2-9=y^2-81 \\ x^2-y^2=-72 \end{gathered}[/tex]
Now, in triangle ABC
[tex]\begin{gathered} BC^2=AB^2+AC^2 \\ 12^2=x^2+y^2 \\ x^2+y^2=144 \end{gathered}[/tex]
Add the equation
[tex]\begin{gathered} x^2+x^2-y^2+y^2=-72+144 \\ 2x^2=72 \\ x^2=36 \\ x=6 \end{gathered}[/tex]
Substitute x=6 in the above equation,
[tex]\begin{gathered} x^2_{}+y^2=144 \\ 36+y^2=144 \\ y^2=144-36 \\ y^2=108 \\ y=10.39 \end{gathered}[/tex]
So, the value of the other side of triangle are 6, 10.39
Perimeter of triangle =sum of all sides of triangle
[tex]\begin{gathered} \text{Perimeter}=AB+BC+AC \\ \text{Perimeter}=AB+BD+CD+AC \\ \text{Perimeter}=6+3+9+10.39_{} \\ \text{Perimeter}=28.39 \end{gathered}[/tex]
ANSWER : Perimeter is 28.39
