7 little circles

Relevant wiki: Length and Area - Composite Figures

For this problem, we set the following:

• Points $A$ and $B$ are points on the circumference of both circles with centers $A'$ and $B$ .
• Point $O$ is the center of the largest center, which is also the center of the central circle.

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 be removed is $A_c = \dfrac{\pi}{3}$ .

Therefore, the area of the shaded region is $\begin{array}{rl} A_{\text{shaded}} &= A_L - \left(A_{\Delta} + 2A_c\right)\\ &= \dfrac{3\pi}{2} - \left(\sqrt{3} + \dfrac{2\pi}{3}\right)\\ &= \dfrac{5\pi}{6} - \sqrt{3} \approx \boxed{0.886} \end{array}$

Area of the large circle: $\pi\times3^2=9\pi$

Minus areas of 7 small circles: $9\pi-7\pi=2\pi$

Minus 6 areas between small circles (calculated below): $2\pi-6(\sqrt{3}-\frac{\pi}{2})=5\pi-6\sqrt{3}$

Divided by 6: $\frac{5}{6}\pi-\sqrt{3}\approx0.886$

Calculating area between three unit circles:

Area of an equilateral triangle side $2$ , vertices centers of circles: $\frac{1}{2}bh=\frac{1}{2}\times2\times2\frac{\sqrt{3}}{2}=\sqrt{3}$

Minus $\frac{3}{6}$ of a small circle: $\sqrt{3}-\frac{\pi}{2}$

Here is the plan:
1.Find 1/6 of the big circle's area (R=3);
2.Subtract the area of equilateral triangle (side=2);
3.Subtract the area of 2/3 of the small circle (r=1).
The result is 0.886.