I don't know how to solve this problem. What I've done so far is to construct 3 equations. However, I don't know how to solve those 3 equations:
Let y be the distance between the box and the ladder, and z be the length of the portion of the ladder that is beneath the top of the box:
$(x-1)^2+1=(10-z)^2$
$(y)^2+1=z^2$
$(x)^2+(1+y)^2=(10)^2$
I don't know how to proceed from here, however.
By Pythagoras and similarity we obtain: $$\frac{x}{x-1}=\frac{\sqrt{100-x^2}}{1}$$ or $$x^4-2x^3-98x^2+200x-100=0,$$ where $1<x<10,$ or $$(x^2-x+1)^2-101(x-1)^2=0,$$ which after factoring gives: $$x=\frac{1}{2}(1+\sqrt{101}-\sqrt{98-2\sqrt{101}})$$ or $$x=\frac{1}{2}(1+\sqrt{101}+\sqrt{98-2\sqrt{101}}).$$