imc

Draw the center of the large circle, and connect it with two adjacent centers of the smaller circles. Draw the line between the centers. Then, let the line connecting the center of the smaller circle to the center of the larger circle be $x.$ By the Pythagorean Theorem, we have that $$x^2+x^2=(1+1{)}^2$$ Simplifying, we have $$\begin{array}{rl}x& =\sqrt{\frac{(1+1{)}^2}{2}}\\ & =\sqrt{\frac{4}{2}}\\ & =\sqrt{2}\end{array}$$ Thus, the radius is $$1+x=1+\sqrt{2}$$ The answer we desire is $$\frac{4\cdot \left(\pi \cdot 1^2\right)}{\pi \cdot (1+\sqrt{2}{)}^2}$$ Simplifying, we have that $$\begin{array}{rl}\frac{4\cdot \left(\pi \cdot 1^2\right)}{\pi \cdot (1+\sqrt{2}{)}^2}& =\frac{4\cdot \pi }{\pi \cdot (1+\sqrt{2}{)}^2}\\ & =\frac{\pi \cdot 4}{\pi \cdot (1+\sqrt{2}{)}^2}\\ & =\frac{4}{(1+\sqrt{2}{)}^2}\\ & =\frac{4}{1^2+2\cdot 1\cdot \sqrt{2}+(\sqrt{2}{)}^2}\\ & =\frac{4}{1+2\sqrt{2}+2}\\ & =\frac{4}{3+2\sqrt{2}}\\ & =\frac{4(3-2\sqrt{2})}{(3+2\sqrt{2})(3-2\sqrt{2})}\\ & =\frac{4(3-2\sqrt{2})}{3^2-(2\sqrt{2}{)}^2}\\ & =\frac{4(3-2\sqrt{2})}{9-8}\\ & =\frac{4(3-2\sqrt{2})}{1}\\ & =\boxed{4(3-2\sqrt{2})}\end{array}$$