Answer :

Illuminati750
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
AndyBryer
Ok so we need to subtract the area of the triangle from the area of the segment and this will equal 100.
We know that the area of the segment is:
[tex] \frac{80}{360} * \pi r^{2} [/tex]
And that the area of the triangle is:
[tex] \frac{1}{2} r^{2} sin(80)[/tex]
Therefore:
[tex] \frac{80}{360} * \pi r^{2} - \frac{1}{2} r^{2} sin(80)=100[/tex]
We can simplify it through these steps:
[tex] \frac{80}{360} * \pi r^{2} - \frac{1}{2} r^{2} sin(80)=100[/tex]
[tex]4 \pi r^{2} - 9 r^{2} sin(80)=1800[/tex]
[tex] r^{2}(4 \pi -9sin(80))=1800 [/tex]
[tex] r^{2} = \frac{1800}{4 \pi -9sin(80)} [/tex]
[tex]r= \sqrt{\frac{1800}{4 \pi -9sin(80)} } [/tex]
Therefore r=22.04cm (4sf)