We have three different ways to the solution.
1.
It's a right triangle and isosceles triangle.
The formula of the hypotenuse of this triangle:
[tex]h=x\sqrt2[/tex]
therefore
[tex]x\sqrt2=210\ \ \ |\cdot\sqrt2\\\\2x=210\sqrt2\ \ \ |:2\\\\\boxed{x=105\sqrt2}[/tex]
2.
Use the Pythagorean theorem:
[tex]x^2+x^2=210^2\\\\2x^2=44,100\ \ \ \ |:2\\\\x^2=22,050\to x=\sqrt{22,050}\\\\x=\sqrt{11,025\cdot2}\\\\x=\sqrt{11,025}\cdot\sqrt2\\\\\boxed{x=105\sqrt2}[/tex]
3.
Use trigonometric function:
[tex]\sin45^o=\dfrac{x}{210}\\\\\sin45^o=\dfrac{\sqrt2}{2}\\\\\dfrac{x}{210}=\dfrac{\sqrt2}{2}\ \ \ |cross\ multiply\\\\2x=210\sqrt2\ \ \ \ |:2\\\\\boxed{x=105\sqrt2}[/tex]
Answer: b. 105√2
