24 of 100: Roundabout

$\angle{ABC}=360^\circ-90^\circ-60^\circ-60^\circ=150^\circ$

Total Red Length $=4\times\frac{150}{360}\times2\pi\times6=\boxed{20\pi}$

Although the picture is drawn to look as if the centres of the four circles form the corners of a square, this is not necessary. They must, however, form the vertices of a rhombus. If we consider the effect of the four equilateral triangles and the rhombus, the sum of the four exterior angles (marked in red - they do not have to be all equal, only equal in pairs) is $4\times360^\circ - 8\times60^\circ - 360^\circ = 600^\circ$ or $\tfrac{10}{3}\pi$ radians. Thus the total exterior arc length is $\tfrac{10}{3}\pi \times 6 = \boxed{20\pi}$ cm.

Because there are 4 circles, we can just look at one of the circles, and try to find the angle of the red arc, then to find the length of the arc using the circumference.

Since we are looking for the red outer part, we can calculate the arc angle of the rest of the circle(the black inner part). When we connect two adjacent blue points, and the black dot 'tip,' we form a equilateral triangle, and since every circle has two of those, we already have 120 degrees.

Then, it is obvious that the center of the circles create a square, or another 90 degrees. From this, we know that the inner angle of the circle(not the red part) will be 210 degrees, so the outer black part will have an angle of 150.

Finally, we can calculate the actual length: Every circle will have a red part of $\frac{150}{360} \times 2\pi r (r = 6) \longrightarrow \frac{5}{12} \times 12\pi=5\pi$ per circle. Thus, the red outer rim of the 4 circles combined, and the answer to the problem would be $\boxed{20\pi }$

The given options are: 16π, 18π, 20π, 22π, 24π.

I will give a solution which includes more logic rather than math.

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Radius = 6cm

Therefore,

circumference of each circle = 2πr = 12π

All 4 circles are identical

Thus, Circumference of all 4 circles = 4 • 12π = 48π

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Now, out of the given options, only 16π, 20π and 24π are divisible by 4

Hence, 18π and 22π are not the correct answers; we strike them out

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Half of 48π is 24π -- for 24π to be the correct answer, the red arc should be a semicircle which it is not as seen in the figure

Hence we strike out 24π.

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If we observe the figure carefully, we find that red arc isn't even 1/3rd of the whole circumference.

Hence, we strike out 16π as it is 1/3rd of 48π

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SO , the option which remains is 20π

Answer = 20π

Consider the triangle formed by the centers of two adjacent circles and their point of intersection on the outside of the shape. This is an equilateral triangle, with side length $6$ cm. Therefore the central angle here is $60^{\circ}$ above horizontal. That means its complement is $30^{\circ}$ from vertical, as is the corresponding arc segment. By symmetry each circle contributes two copies of such a $30^{\circ}$ arc to the perimeter of the shape, along with the main $90^{\circ}$ contribution. Thus each circle contributes $\frac{5}{6}$ th of a semicircle, i.e. $\frac{5}{6}*\pi*6$ cm. Summing over the four corners gives us $4*\frac{5}{6}*6*\pi$ cm, i.e. $20\pi$ cm.

Divide the shape into eight identical segments.

Draw segment DO, which is length 6cm

CO is perpendicular to DO

Triangle BOD is equilateral so Angle BOD = 60°

Therefore, angle BOC = 30°

Angle AOC = 45°

Total angle of segment AB = 45° + 30° = 75°

The total perimeter is made of eight segments, so 8 x 75° = 600°

Total perimeter = 2π r x 600° / 360° = 12 π x 600° / 360° = 20 π

LaTeX: $\text {The perimeter of one circle is :} \; \Large P = 2 \times \pi \times R \; \text {then} \; P = 2\times6\times\pi = \color{#D61F06}{ 12\pi}$

LaTeX: $\text {As you can see, the radius defines an equilateral or regular triangle (EKB),}$

LaTeX: $\text {that is to say, each angle is equal to}\; \frac{\pi}{3} \; \text {and the good news is that 6 equilateral triangles }$

LaTeX: $\text {can be inscribed into a cercle ! We get it ! The hexagone can now appear in the circle.}$

LaTeX: $\text {Let's calculate the parts of the perimeter that has to be deduced of}\ {\color{#D61F06}{\text {the main perimeter :}}}$

LaTeX: $\text {We have :} \;\Huge 3 \times \frac{\pi}{3} \; \text {and a } \; \frac{1}{2} \times \frac{\pi}{3}$

LaTeX: $\text {We get :} \; \Huge \frac{3\pi}{3} + \frac{\pi}{6} = \frac{6+1}{6} = \frac{7}{6}$

LaTeX: $\text {Now we have the total of all the angles added, we have to take into account the radius to}$

LaTeX: $\text {calculate the part of the perimeter to be deduced of the main perimeter :}$ LaTeX: $\Huge \frac{7\pi}{6} \times 6 = 7 \pi$

LaTeX: $\Large \color{#D61F06} {12 \pi} - \color{#3D99F6} {7 \pi} = \color{#333333} 5 \pi, \normalsize \; {\text {but that's not quite over, this is just the perimeter for one circle !}}$

LaTeX: $\text {We just have to multiply by the 4 circles to get the overall perimeter required :}$

LaTeX: $\Huge 5\pi \times 4 = \color{#D61F06} \boxed {20\pi}$

1. The point of intersection of circles is forming an equilateral triangle when joined with centres of intersecting circles.
2. The centres of four circles is forming a square.
3. Take any circle. The black arc has three portions, black dot to blue dot, blue dot to blue dot and blue to black dot. The middle portion of black arc (blue dot to blue dot) is covering a 90° angle at one corner of square and two remaining portions of black arc (blue dot to black dot) are each covering a 60° angle in equilateral triangle on either side of 90° angle. The total angle of black arc in each circle is = 90+60+60 =210°.
   The angle of red arc = 360 - 210 = 150°.
   The total angle of four red arcs = 4×150=600°.
   
   
4. The perimeter of red figure.
   = (600/360)×(2πr)
   = (5/3)(12π) = 20π.