$a+\frac{b}{\sqrt{c}}-d\pi$ (such that $a,~b,~c,~\text{and}~d$ are integers and $c$ is square-free), then find $\frac{a+b+c}{d}$ .

The figure to the right shows 5 unit circles that are tangent to each other and the edges of the trapezoid. If the yellow area can be expressed as
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Note:
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I did not create this problem; I simply solved it and decided to post it on here. Credit goes to my math teacher.

The answer is 6.0.

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The area of the yellow region $A_{\color{#CEBB00}\text{yellow}}$ is given by the area of the trapezoid $A_{\text{trapezoid}}$ minus the area of the 5 circles $A_{\text{circle}}$ .

From the figure, we note that the length of the base of the trapezoid $a_b = 4+2a = 4+\dfrac 2{\tan 30^\circ} = 4+2\sqrt 3$ . The length of its top $a_t = 2+2b = 2 + 2\tan 30^\circ = 2 + \dfrac 2{\sqrt 3}$ . The height of the trapezoid $h = 2+c = 2+2 \sin 60^\circ = 2+\sqrt 3$ .

Therefore, we have:

$\begin{aligned} A_{\color{#CEBB00}\text{yellow}} & = A_{\text{trapezoid}} - 5A_{\text{circle}} \\ & = \frac {(a_b+a_t)h}2 - 5\pi 1^2 \\ & = \frac 12\left(4+2\sqrt 3+2+\frac 2{\sqrt 3}\right)(2+\sqrt 3) - 5\pi \\ & = 10 + \frac {17}{\sqrt 3} - 5\pi \end{aligned}$

Then $\dfrac {a+b+c}d = \dfrac {10+17+3}5 = \boxed 6$ .