Radius of Inscribed Circle in a Quadrant of a Circle

In summary, you found the radius of a circle inscribed in a quadrant of a circle with radius 5. The radius of the inscribed circle is 2.5√2.
  • #1
Wildcat
116
0

Homework Statement



Find the radius of a circle inscribed in a quadrant of a circle with radius 5

Homework Equations





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??
 
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  • #2
Your method seems correct, but I got [itex]\frac{5}{1+\sqrt{2}}[/itex].

Is it possible you can show a picture of where you are making your triangle?
 
  • #3
Villyer said:
Your method seems correct, but I got [itex]\frac{5}{1+\sqrt{2}}[/itex].

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

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

2iM6G.png
 
  • #5
Villyer said:
Is the center circle here the one you are trying to find the radius of?

1wZYA.png

Yes! I love the drawing!
 
  • #6
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
Villyer said:
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.

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
Oh oh oh, I see now. But if that were true, then wouldn't 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
Villyer said:
Oh oh oh, I see now. But if that were true, then wouldn't 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.

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

I don't want to say too much however.

Are you talking about the radius drawn in that bisects the 90° at the origin??
 
  • #12
Yeah, the one that is 45 degrees above the horizontal.
 
  • #13
I'm assuming you didn't use 2.5,2.5 as your center??
 
  • #14
Right.

What is the length of that segment? How can you label its parts in terms of r?
 
  • #15
I'm not sure I need to look at it some more :(
 
  • #16
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
Explain how you got to that.
 
  • #18
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
The perpindicular segment is a radius though.

And you never use the number 2.5 here.
 
  • #20
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??
 
  • #21
No, that little segment doesn't have to be solved for. It's easier to solve for the segment from the center of the small center to the center of the large circle.
 
  • #22
So the length from the center of the small circle to the center of the large circle is r√2?
 
  • #23
which means r√2 +r=5 then r=5/(√2 +1) Right!
 
  • #24
Yep, that's what I got!

If you rationalize the denominator the answer gets a little bit simpler, I'm not sure if this is for a class where you have to do that.
 
  • #25
Thank you so Much!
 

1. What is an inscribed circle in geometry?

An inscribed circle in geometry is a circle that is drawn inside a polygon such that the circle touches each side of the polygon at exactly one point. This circle is also known as an inscribed incircle.

2. How is the inscribed circle related to the polygon it is inscribed in?

The inscribed circle is always tangent to each side of the polygon it is inscribed in. This means that the radius of the circle is perpendicular to the side of the polygon at the point of tangency.

3. What is the relationship between the radius of the inscribed circle and the sides of the polygon?

The radius of the inscribed circle is always half of the length of the shortest side of the polygon. This relationship can be used to find the radius of the inscribed circle given the length of the shortest side of the polygon.

4. Can an inscribed circle exist in any polygon?

Yes, an inscribed circle can exist in any polygon as long as the polygon is not self-intersecting and has at least three sides. However, the inscribed circle may not always be unique for a given polygon.

5. How is the area of the inscribed circle related to the area of the polygon?

The area of the inscribed circle is always smaller than the area of the polygon. The ratio of the area of the inscribed circle to the area of the polygon is known as the apothem ratio and is equal to the radius of the inscribed circle divided by the length of the shortest side of the polygon.

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