1. PF Insights is off to a great start! Fresh and interesting articles on all things science and math. Here: PF Insights

# Inscribed Circle Geometry

1. ### Wildcat

116
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
I worked this but I'm not sure if its correct. I looked at the first quadrant so a quarter of a circle with radius 5. I drew the radius that would bisect the angle at the origin into 45° angles. I found the midpoint of the radius and dropped a line to the x-axis from that midpoint. This forms a 45-45-90 triangle which would make the radius of the inscribed circle 2.5/√2. Is this correct??

2. ### Villyer

294
Your method seems correct, but I got $\frac{5}{1+\sqrt{2}}$.

Is it possible you can show a picture of where you are making your triangle?

3. ### Wildcat

116
I am looking at the first quadrant. Center of the circle with radius 5 is at the origin, so I draw my radius from the origin and bisecting the 90° angle. Then from the midpoint of that radius (2.5)(which I think will be the center of the inscribed circle) I drop the segment to the x-axis which will be a leg of the 45-45-90 and also the radius of the inscribed circle.?? so the triangle has a hypotenuse of 2.5 and dividing by √2 would be the leg/radius ??

4. ### Villyer

294
Is the center circle here the one you are trying to find the radius of?

5. ### Wildcat

116
Yes! I love the drawing!

6. ### Villyer

294
The problem with your work I think was assuming that the line dropped down from the center of that circle bisected the bottom line, which isn't true.

7. ### Wildcat

116
No I wasn't assuming that. The first part of the bottom line would be 2.5/√2 as well since the triangle formed is a 45-45-90.

8. ### Villyer

294
Oh oh oh, I see now. But if that were true, then wouldnt the radius also have to be 2.5? Since the segment from the center to the outside of the larger circle is 2.5 and that is another radius of the circle.

9. ### Wildcat

116
I don't think the midpoint of the 5 radius is the center of the inscribed circle anymore because you are correct about the 2.5 but the perpendicular line would also have to be a radius and it is 2.5√2 so that must not be the center

10. ### Villyer

294
If it helps, I used the diagonal line in order to solve for the radius.

I don't want to say too much however.

11. ### Wildcat

116
Are you talking about the radius drawn in that bisects the 90° at the origin??

12. ### Villyer

294
Yeah, the one that is 45 degrees above the horizontal.

13. ### Wildcat

116
I'm assuming you didn't use 2.5,2.5 as your center??

14. ### Villyer

294
Right.

What is the length of that segment? How can you label its parts in terms of r?

15. ### Wildcat

116
I'm not sure I need to look at it some more :(

16. ### Wildcat

116
Ok now I got (10-√2)/4 its close to your answer but still not the same. I need to go back and see if I made a calculation error.

17. ### Villyer

294
Explain how you got to that.

18. ### Wildcat

116
Its wrong I was assuming that perpendicular segment was the radius of the inscribed circle. I need to add 2.5 to that piece that I keep thinking is a radius (but not) to get the diameter but I don't know how to get it

19. ### Villyer

294
The perpindicular segment is a radius though.

And you never use the number 2.5 here.

20. ### Wildcat

116
Ok so 2 of those segments make the diameter of the inscribed circle. and that little piece outside the circle would need to be subtracted from 5. Should I be able to find that segment length first??