Hello,
The formula for finding the area of a circular region is: [tex]A= \frac{ \alpha *r^{2} }{2} [/tex]
then:
[tex]A_{1} = \frac{80*r^{2} }{2} [/tex]
With the two radius it is formed an isosceles triangle, so, we must obtain its area, but first we obtain the height and the base.
[tex]cos(40)= \frac{h}{r} \\ \\ h= r*cos(40)\\ \\ \\ sen(40)= \frac{b}{r} \\ \\ b=r*sen(40)[/tex]
Now we can find its area:
[tex]A_{2}=2* \frac{b*h}{2} \\ \\ A_{2}= [r*sen(40)][r*cos(40)]\\ \\A_{2}= r^{2}*sen(40)*cos(40)[/tex]
The subtraction of the two areas is 100cm^2, then:
[tex]A_{1}-A_{2}=100cm^{2} \\ (40*r^{2})-(r^{2}*sen(40)*cos(40) )=100cm^{2} \\ 39.51r^{2}=100cm^{2} \\ r^{2}=2.53cm^{2} \\ r=1.59cm[/tex]
Answer: r= 1.59cm