1. ## circle problem

I'm having trouble with this question. For the first part Im guessing that the radius is 2 times larger for the bigger circle and therefore its area is four times as large, but I can't seem to prove that the radius is twice as large.

I'd appreciate it if anyone could help me out on this question

2. If c is the length of side of triangle
Height is $\sqrt{\frac{3}{2}} c$ by pythagore theorem
Height is $r_{small} + r_{large}$
Draw a rectangle triangle which hypothenuse is $r_{large}$ and one of small side $r_{small}$
You find that $r_{large} = \frac{\sqrt{3} \cdot c}{3}$
$r_{small} = \frac{\sqrt{3} \cdot c}{6}$
And you were right.
Ok for the rest?

3. thanks for the help.

I'm also a bit stuck on the second part, any ideas on what to do?

4. Of course, the first question is $\frac{4}{1}$ because
$A = \pi r^2$ if r is twice as large than area is four time bigger.
$area_{curve triangle} = \frac{area_{triangle} - area_{small circle}}{3}$
$area_{arc of circle} = \frac{area_{large circle } - area_{triangle}}{3}$
You know all you need to calculate these area with what I told you
Answer is $\frac{2 \cdot(3\cdot\sqrt(3)-2\cdot \pi)}{\pi-6\cdot\sqrt(3)}$

5. ## circle problem

posted by david 18 and answered by vincesonfire

this problem is simplified by taking unity as the radius of the large circle.Following vinci formulars iget an answer as follows

3/4xradical 3-pi/4 dividedby pi-3/4 xradical 3 and numerically slightly lower than the previous answer and does not generate negative numbers

bj