Answer:
x = 2[tex]\sqrt{13}[/tex]
Step-by-step explanation:
Using Pythagoras' identity on each of the 3 right triangles
The square on the hypotenuse is equal to the sum of the squares on the other 2 sides.
Outer right triangle
x² + z² = (9 + 4)² = 13² = 169, that is
x² + z² = 169 → (1)
Left right triangle
y² + 4² = x²
y² + 16 = x² ⇒ y² = x² - 16 → (2)
Right right triangle
z² = y² + 9²
z² = y² + 81 → (3)
Substitute (3) into (1)
x² + y² + 81 = 169 ← substitute (2) into here
x² + x² - 16 + 81 = 169
2x² + 65 = 169 ( subtract 65 from both sides )
2x² = 104 ( divide both sides by 2 )
x² = 52 ( take the square root of both sides )
x = [tex]\sqrt{52}[/tex]
= [tex]\sqrt{4(13)}[/tex]
= [tex]\sqrt{4}[/tex] × [tex]\sqrt{13}[/tex] = 2[tex]\sqrt{13}[/tex]
