1. ## Geometry problem

Hi everyone my name is Rukkk and I am new to this forum I have been trying to solve the below problem but not succeeding.
Can anyone help me?
Working outs would also be greatly appreciated.
Thanks for the help,
Rukkk

2. if you draw lines down to form a square within the circle you'll see the diameter is 80 and radius therefore 40 then use length of arc formulas?

3. (r-8)^2 + 32^2 = r^2
a = arcsin 32/r
circumference = r*(2*pi - 2*a)

sorry I'm in a hurry

4. Hi! Now I have more time:

a = half the angle AOB

(r-8)^2 + 32^2 = r^2 (Theorem of Pythagorah)
sin a = 32/r => a = arcsin 32/r
circumference = r*2*a (Sorry, wrong last time)

Kind regards

5. Hello, rukkk!

Welcome aboard!
Code:
C
* * *
*     |8    *
A* - - - + - - - *B
*  *     |D    *  *
*   |   *R
*       * | *       *
*         *         *
*         O         *

*                 *
*               *
*           *
* * *
We have a circle with center $\displaystyle O.$
The radius is: .$\displaystyle OA = OB = OC = R$
Chord $\displaystyle AB = 64\quad\Rightarrow\quad DB = 32$

Since $\displaystyle CD = 8$, then $\displaystyle DO = R-8$

In right triangle $\displaystyle BDO\!:\;\;DO^2 + DB^2 \:=\:BO^2 \quad\Rightarrow\quad (R-8)^2 + 32^2 \:=\:R^2$

. . $\displaystyle R^2 - 16R + 64 + 1024 \:=\:R^2\quad\Rightarrow\quad 16R \:=\:1088\quad\Rightarrow\quad\boxed{ R \:=\:68}$