[tex]\it R=circle's \ radius;\ \ D=diameter;\ \ D=2R\\ \\ \mathcal{C}=2\pi R\\ \\ \mathcal{A}=\pi R^2\\ \\ \mathcal{C}=\dfrac{2}{5}\cdot\mathcal{A} \Rightarrow 2\pi R=\dfrac{2}{5}\pi R^2|_{\cdot\frac{5}{2}} \Rightarrow 5\pi R=\pi R^2|_{:(\pi R)} \Rightarrow 5=R|_{\cdot2} \Rightarrow 10=2R\\ \\ But,\ 2R=D,\ so\ \ D=10\ in[/tex]
