# Find the area of this larger circle

• grandpa2390
In summary: I think the answer is that ##R = 2*sqrt(2+sqrt(2))##yes, it's very clear now. thanks!In summary, my colleagues and I can't figure out how to come to that answer. It's probably something simple.
grandpa2390
Homework Statement
I'm helping to run a math competition at my school and we pulled a problem from this site that we are having trouble solving.

Eight circles of radius 1 have centers on a larger common circle and adjacent circles are tangent. Find the area of the common circle. See the illustration below.
Relevant Equations
Not sure. It has been a long time since I've done this kind of math.

if it helps, the answer is supposed to be

my colleagues and I can't figure out how to come to that answer. It's probably something simple.

edit: I tried to solve it by inscribing an octagon, and then finding the distance from the center of the octagon to the side of the octagon. but I got 1 + sqrt(2) as the radius... rather than the 2*sqrt(2+sqrt(2))

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grandpa2390 said:
if it helps, the answer is supposed to be
The ##\pi## shouldn't be in there -- as you later seem to know...

##\theta = \pi/8\quad\& \quad R = 1/\sin\theta ## so: Area = ##\pi \over \sin^2\theta##

##\cos 2\theta =
\cos^2 {\theta} - \sin^2 {\theta}=1- 2\sin^2 {\theta} \quad \Rightarrow \sin^2 {\theta} = {1 - \cos2\theta\over 2}## and with ##\cos {\pi\over 4}= {1\over 2}\sqrt 2## we get ##{1/ \sin^2 {\pi\over 8}} = {\displaystyle {4\over 2-\sqrt 2}} = 2(2+\sqrt 2)##

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Delta2
BvU said:
The ##\pi## shouldn't be in there -- as you later seem to know...

View attachment 250517

##cos 2\theta =
\cos^2 {\theta} - \sin^2 {\theta}=1- 2\sin^2 {\theta} \quad \Rightarrow \sin {\theta} = {1 - \cos2\theta\over 2}## and with ##\cos {\pi\over 8}= {1\over 2}\sqrt 2## we get ##{1/ \sin {\pi\over 8}} = {\displaystyle {4\over 2-\sqrt 2}} = 2(2+\sqrt 2)##

why shouldn't the pi be there? Area = pi*r^2

Oh sorry, fixed on determining R

Have to edit my ##\TeX## some more, patience ...

grandpa2390
BvU said:
Oh sorry, fixed on determining R

Have to edit my ##\TeX## some more, patience ...
that's ok. I solved it with sin(22.5) = 1/h. for some reason I didn't get it earlier. but now it's working out.

no way I could have gotten it in the form you got it though. the form the solution wants.

edit: I see my mistake. I thought the radius would be 2*sqrt(2+sqrt(2)), but it is sqrt(2*(2+sqrt(2))

Post #2 fixed now, I hope...

Half-angle formula clear ?

grandpa2390
BvU said:
Post #2 fixed now, I hope...

Half-angle formula clear ?
you helped me see my error. I miscalculated what the radius was supposed to be.
I also inscribed my octagon improperly. I should have been looking for the distance to an angle rather than the distance to a side.
thanks!

BvU
BvU said:
The ##\pi## shouldn't be in there -- as you later seem to know...

View attachment 250517
##\theta = \pi/8\quad\& \quad R = 1/\sin\theta ## so: Area = ##\pi \over \sin^2\theta##

##\cos 2\theta =
\cos^2 {\theta} - \sin^2 {\theta}=1- 2\sin^2 {\theta} \quad \Rightarrow \sin^2 {\theta} = {1 - \cos2\theta\over 2}## and with ##\cos {\pi\over 4}= {1\over 2}\sqrt 2## we get ##{1/ \sin^2 {\pi\over 8}} = {\displaystyle {4\over 2-\sqrt 2}} = 2(2+\sqrt 2)##
I can't see how ##R## is the radius of the big circle, it looks like just a portion of it, could you explain more please?

##R## is what the sketch starts out with. Its relation to the radius of the small circles is the object of the analysis. Is it clear to you where the ##\pi/8## comes from ?

BvU said:
##R## is what the sketch starts out with. Its relation to the radius of the small circles is the object of the analysis. Is it clear to you where the ##\pi/8## comes from ?
My bad, for some reason, it wasn't clear to me in your picture that the intersection point was the center of the common circle.

## 1. How do you find the area of a larger circle?

To find the area of a larger circle, you can use the formula A = πr², where A is the area and r is the radius of the circle. First, measure the radius of the circle using a ruler or measuring tape. Then, plug the value of the radius into the formula and solve for the area.

## 2. What is the formula for finding the area of a larger circle?

The formula for finding the area of a larger circle is A = πr², where A is the area and r is the radius of the circle. This formula is derived from the mathematical constant pi (π) and the relationship between a circle's radius and its area.

## 3. Can you use the same formula to find the area of any circle?

Yes, the formula A = πr² can be used to find the area of any circle, regardless of its size. It is a universal formula that applies to all circles, including larger ones.

## 4. What units are used to measure the area of a larger circle?

The area of a larger circle is typically measured in square units, such as square inches, square feet, or square meters. This is because area is a measure of 2-dimensional space, so it is represented by a unit squared.

## 5. How can finding the area of a larger circle be useful?

Finding the area of a larger circle can be useful in many real-world applications. For example, if you are a builder or landscaper, you may need to know the area of a larger circle to determine how much material is needed. In geometry and engineering, the area of a larger circle can also be used to solve various problems and design structures.

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