7 non-overlapping unit circles are inscribed inside a large circle as shown below. What is the area of the shaded region?

Round your answer to 3 decimal places.

The answer is 0.886.

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Relevant wiki: Length and Area - Composite FiguresFor this problem, we set the following:

Since all smallest circles have radius of $1$ , then $|AO| = |AA'| + |A'O| = 1 + 2 = 3$ , which shows that the radius of the largest circle is $3$ . Likewise, due to symmetry, $|BO| = 3$ . Because these mutually tangent circles have same radii, $\angle O = 60^{\circ}$ . In this case, the area of the large circular sector is $A_{L} = \dfrac{\pi}{6} \cdot (3)^2 = \dfrac{3\pi}{2}$ .

Since we are interested in determining the area of the shaded region, the next step is to determine the area of the white region within the dashed region. Here are two areas we need to determine: (1) area of two sectors and (2) dashed equilateral triangle.

Equilateral Triangle: Since the segments between their centers form the equilateral triangle of side length $2$ , the area of the indicated equilateral triangle is $A_{\Delta} = \dfrac{\sqrt{3}}{4} \cdot 2^2 = \sqrt{3}$ .Sector: Angle chasing, we see that $\angle AA'B' = \angle BB'A' = 120^{\circ}$ . Since we are then interested in determining the area of the two sectors that enclose the shaded region, the remaining area to beremovedis $A_c = \dfrac{\pi}{3}$ .Therefore, the area of the shaded region is $\begin{array}{rl} A_{\text{shaded}} &= A_L - \left(A_{\Delta} + 2A_c\right)\\ &= \dfrac{3\pi}{2} - \left(\sqrt{3} + \dfrac{2\pi}{3}\right)\\ &= \dfrac{5\pi}{6} - \sqrt{3} \approx \boxed{0.886} \end{array}$