Answer:
Part 1) [tex]z=4\sqrt{2}\ units[/tex]
Part 2) [tex]x=18\ units[/tex]
Part 3) [tex]y=12\sqrt{2}\ units[/tex]
Step-by-step explanation:
step 1
Find the value of z
In the smaller right triangle IFG of the figure
Applying the Pythagoras Theorem
[tex]6^{2}=2^{2}+z^{2}[/tex]
[tex]z^{2}=6^{2}-2^{2}[/tex]
[tex]z^{2}=32[/tex]
[tex]z=4\sqrt{2}\ units[/tex]
step 2
In the right triangle HFG
Applying the Pythagoras Theorem
[tex]x^{2}=y^{2}+6^{2}[/tex]
[tex]y^{2}=x^{2}-36[/tex] -----> equation A
step 3
In the right triangle HIG
Applying the Pythagoras Theorem
[tex]y^{2}=z^{2}+(x-2)^{2}[/tex]
[tex]y^{2}=32+(x-2)^{2}[/tex] -----> equation B
step 4
equate equation A and equation B
[tex]x^{2}-36=32+(x-2)^{2}\\x^{2}-36=32+x^{2}-4x+4\\4x=36+36\\4x=72\\x=18\ units[/tex]
step 5
Find the value of y
Substitute the value of x in the equation B
we have
[tex]x=18\ units[/tex]
[tex]y^{2}=32+(18-2)^{2}[/tex]
[tex]y^{2}=32+(16)^{2}[/tex]
[tex]y^{2}=288[/tex]
[tex]y=12\sqrt{2}\ units[/tex]
