Solution
First we will label the triangle
In triangle ACD , we will use pythagoras theorem
[tex]\begin{gathered} (hyp)^2=(opp)^2+(adj)^2 \\ x^2=4^2+y^2 \\ x^2=16+y^2 \end{gathered}[/tex]
In triangle ACB, Using pythagoras theorem
[tex]\begin{gathered} (9+4)^2=x^2+z^2 \\ 13^2=x^2+z^2 \\ 169=x^2+z^2 \end{gathered}[/tex]
In triangle ABD , we will also use pythagoras theorem
[tex]\begin{gathered} z^2=y^2+9^2 \\ z^2=y^2+81 \end{gathered}[/tex]
Now we have three equations
Let us find a replacement for y^2 in the first equation
From equ(3),
[tex]z^2-81=y^2[/tex][tex]\begin{gathered} x^2=16+y^2 \\ x^2=16+z^2-81 \\ x^2=z^2-65 \end{gathered}[/tex]
Now we will find a replacement for z^2 in the new equation
We will recall that
[tex]\begin{gathered} 169=x^2+z^2 \\ 169-x^2=z^2 \\ z^2=169-x^2 \end{gathered}[/tex]
We will substitute for z^2 in x^2=z^2 - 65
[tex]\begin{gathered} x^2=z^2-65 \\ x^2=169-x^2-65 \\ x^2+x^2=104 \\ 2x^2=104 \\ x^2=\frac{104}{2} \\ \\ x^2=52 \\ x=\sqrt{52} \\ \\ x=2\sqrt{13} \end{gathered}[/tex]
