$1$ big circle formed by$20$ smaller circles

Question

Hello i need to make a circle out of 20 smaller ones. The smaller circles radius is r=9.3cm heres what i wanna do:

Answer 1

Let the radius of the large circle be $R$ CM

So, the perimeter of the concentric Circle with radius $=R+9.3 $ CM

will be $$2\pi(R+9.3)$$ which will be same as $$20\cdot2\cdot9.3$$

Answer 2

If you take the centres of the smaller circles and draw lines between them, you get a 20 sided polygon (icosagon) with each side being twice the radius of a small circle. If you draw a line from the centre of each small circle to the centre of your large circle, it's $360^\circ/20 = 18^\circ$ for each slice.

If you take the point where two of the small circles touch $ = A$, the centre of one of those small circles $ = B$, and the centre of the large circle $ = C$, you get a triangle $ABC$.

$\overline{AB} = 9.3cm$

$\angle{ACB} = 9^\circ$

$\sin{9^\circ} = \frac{9cm}{h}$

$h \approx 57.53 cm$

So that's the radius of a circle made from the centres of the small circles. The radius you desire is $$\frac{9cm}{\sin{9^\circ}} - 9.3cm \approx 48.23 cm$$

Answer 3

Draw a triangle between the center of the big circle and the centers of two adjacent small circles, and bisect it to make two narrow right triangles. (Sorry about the bad picture, but you get the idea.)

If you take one of the narrow right triangles, its small angle is $9^\circ$ (one-half of one-twentieth of $360^\circ$), the short side is $9.3$cm, and the hypotenuse is $R+9.3$cm, where $R$ is the radius of the large circle.

So $\sin(9^\circ)=\frac{9.3}{R+9.3}$. And $\sin 9^\circ\approx0.1564$ and you can solve for $R$.