Answer:
[tex]x=4\sqrt{6}\ units[/tex]
[tex]y=4\sqrt{2}\ units[/tex]
[tex]z=4\sqrt{3}\ units[/tex]
Step-by-step explanation:
The picture of the question in the attached figure
step 1
In the right triangle ABD
Applying the Pythagorean Theorem
[tex]x^2=y^2+(12-4)^2[/tex]
[tex]x^2=y^2+64[/tex]
[tex]y^2=x^2-64[/tex] ----> equation A
step 2
In the right triangle BDC
Applying the Pythagorean Theorem
[tex]z^2=y^2+4^2[/tex]
[tex]z^2=y^2+16[/tex]
[tex]y^2=z^2-16[/tex] ----> equation B
step 3
In the right triangle ABC
Applying the Pythagorean Theorem
[tex]12^2=x^2+z^2[/tex]
[tex]144=x^2+z^2[/tex] ----> equation C
step 4
Equate equation A and equation B
[tex]x^2-64=z^2-16[/tex]
[tex]x^2=z^2+48[/tex] -----> equation D
step 5
substitute equation D in equation C
[tex]144=z^2+48+z^2[/tex]
solve for z
[tex]2z^2=144-48[/tex]
[tex]2z^2=96[/tex]
[tex]z^2=48[/tex]
[tex]z=\sqrt{48}\ units[/tex]
simplify
[tex]z=4\sqrt{3}\ units[/tex]
Find the value of x
[tex]x^2=z^2+48[/tex]
[tex]x^2=48+48=96[/tex]
[tex]x=\sqrt{96}\ units[/tex]
[tex]x=4\sqrt{6}\ units[/tex]
Find the value of y
[tex]y^2=z^2-16[/tex]
[tex]y^2=48-16[/tex]
[tex]y^2=32[/tex]
[tex]y=\sqrt{32}\ units[/tex]
[tex]y=4\sqrt{2}\ units[/tex]