It is not a Beach Ball

Consider my diagram.

The circumference of a circle is given by $c = \pi d$ , where $d$ is the diameter. Substituting, we have

$120 = \pi d$

$d=\dfrac{120}{\pi}$

It follows that the radius is $r=\dfrac{60}{\pi}$

The diameter of each blue semicircular arc $a$ is $\dfrac{60}{\pi}$ . So

$2a=2\left(\dfrac{1}{2}\right)(\pi)\left(\dfrac{60}{\pi}\right)=60$

It follows that the length of $b$ is

$b=\dfrac{1}{4}(\pi)\left(\dfrac{120}{\pi}\right)=30$

So the required perimeter is

$P=2a+b=60+30=\large{\boxed{90}}$

Each region is bordered by one quarter of the large circle and two semicircles that have half the radius the large circle. Since the large circle has a circumference of $120$ , the quarter curve has a length of $\frac{1}{4}120 = 30$ and the two semicircle curves have a length of $\frac{1}{2}120 = 60$ , and $30 + 60 = 90$ .

Notice in each region, there are two circumferences of semi circles and one circumference of a quarter of a circle.

The perimeter or circumference of a quarter of the large circle is = $\frac{120}{4} = 30.$

The radius of the large circle is $\frac{120}{2 \pi}$ .

The radius of one small semi circle is half the radius of the large circle: $\frac{120}{4 \pi}.$

$\implies \text { the circumference of one semi-circle is } \frac{60 \pi}{2 \pi} = 30 \implies \text{ total perimeter is: } 30+30+30=\boxed {90}$

120/4 = long part = 30

Each small circle is 1/2 (because the radios is the half) of the big one, so is the perimeter too. 120/2 is the perimeter of the small circle = 60. So, two halves = 2*(60/2) = 60.

30+60=90

The radius of the big circle is 60/pi

The edges of the blue region are made of:

One quarter of the circumference of the big circle = 30

And (in two parts) circumference of a circle with radius 30/pi

This gives a total of 90

The small half-circles can be rearranged to form two full circles, each having half the diameter of the large circle. The total of all perimeters is therefore the large circle, plus TWICE the total circumference of the two small circles, as the small circles, being part of the perimeter of TWO neighboring shapes everywhere, need to be counted twice. (The large circle only counts once, as it is not on a shared boundary.)

The small circles have half the circumference of the large circle, as their diameter is half as large. So 120 + 2 x 2 x (120/2) = 360. Divide by four, as there are four identical shapes, and you get the answer: 360/4 = 90. (No Pi() required!!!)

A quarter circle is .25 x 120 = 30 The semi-circle is equivalent to 1/4 of the large circle and therefore also .25 x 120 = 30 You have one quarter circle and two semi-circles in each color region, so .75 x 120 = 90