Answer:
[tex]AB=2\sqrt{ab}[/tex]
Step-by-step explanation:
From figure,
[tex]OA=a, \quad \quad O'B=b\\\Rightarrow OO'= (a+b) \quad \quad \text{and}\quad OD=(a-b)[/tex]
In triangle [tex]OO'D[/tex]
[tex](OO')^2=(OD)^2+(O' D)^2[/tex]
[tex]\Rightarrow (a+b)^2=(a-b)^2+(O' D)^2\\\Rightarrow a^2+b^2+2ab-a^2-b^2+2ab=(O' D)^2\\\Rightarrow 4ab=(O' D)^2\\\Rightarrow O'D=2\sqrt{ab} \\\Rightarrow O' D=2\sqrt{ab}=AB \quad \quad [\because O' DAB\;\; \text{is a rectangle.}][/tex]
Hence, [tex]AB=2\sqrt{ab}[/tex]
