Respuesta :

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]

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