Answer:
x = 9/2
Step-by-step explanation:
The outside triangle is also a right triangle, with hypotenuse 8 and side 6.
The missing side can be calculated with pythagoras as:
[tex]\sqrt{8^2-6^2} = \sqrt{28}[/tex]
Now for the vertical line, lets call it y, which is a side for two "back to back" triangles, you can do pythagoras twice:
left triangle: [tex]y^2 + (8-x)^2 = 28[/tex]
right triangle: [tex]y^2 + x^2 = 6^2[/tex]
Eliminating y², you can solve:
[tex]28 - (8-x)^2 = 36 - x^2[/tex]
[tex]28 -x^2 + 16x - 64 = 36 - x^2[/tex]
[tex]16x = 36 + 64 - 28[/tex]
[tex]x = \frac92[/tex]
