Hexagon In-Out
Imagine a hexagon with a circle inside it. Six points, on the circle, each lie on the center of a different side of the hexagon ... thus it is the largest circle which will fit inside the hexagon. Now imagine another hexagon inside the circle ... each vertex of the hexagon lies on the circle ... making it the largest hexagon which will fit inside the circle.
The inner hexagon has an area of three square units.
What is the area of the outer hexagon?
The inner hexagon has an area of three square units.
What is the area of the outer hexagon?
Labels: geometrick
posted by Zaux at
1:10 PM
10 Comments:
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Sorry about the deletes, my silly errors and typos were unacceptable. If I've not goofed this time, the answer is 4 units squared.
As far as I can see, the radius of the containing hexagon is 2/Sqrt(3) of the circle (very geometry/trig). The radius of the inner hexagon is the same as the radius of the circle (measuring to the vertices).
As area goes up with the square of the linear dimensions, we have area of outer hexagon = 3*(2/Sqrt(3))^2 = 4.
I make it 3.75.
If an arm of the smaller hexagon has length r, this is also the distance from the centre to the mid-point of a side of the larger hexagon. This forms one side of a 30-degree right angled tiangle whose other side is r/2. So the radius of the larger hexagon, which is the hypotenuse of this triangle is sqrt(5)*r/2.
So, scaling up the area as Chris has done, we find the ratio of the two areas is 5/4, hence the area of the outer hexagon is 15/4 or 3.75.
I got an area of 4
The circle area = 3.6276
Hi WIz. Your r/2 expression is wrong. It should be r*sqrt(3)/2, but I've cheated to determine that. I used the fact that we have a 30º angle => cos(30º) = √3/2.
I'll put that another way: your r/2 should have been R/2, where R is the radius of the larger hexagon.
Hex problem
Given:
- hexagon in circle with corners on circle
-that circle is is in hexagon with circle on centre of sides
-Area of inner hexagon is 3 units
Calculate area of a hexagon:
6 similar segments
360/6=60 degrees
If we align the flat edges of the hexagon with the x axis we can see that the segment on top with the flat edge can be formed via 2 30 degree right angle triangles, with line from centre to vertex being length R.
A of 30 degree segment=1/2*b*h
b=length to edge= R*cos30= R*sqrt(3)/2
h=length of edge= R*sin 30= R*1/2
A=sqrt(3)/8*R^2
360/30=12 similar segments
12*At= Ahex
Ahex= 3/2*sqrt(3)*R^2
For inner hex , its radius = radius of the circle, Rhi = Rc
Ahexi=3/2*sqrt(3)*Rc^2
For outer hex
distance to edge=Rcircle
b=Rho*cos30=Rcircle
Rho=Rcircle/cos30
Ahexo=3/2*sqrt(3)*Rho^2
Ahexo/Ahexi=Rho^2/Rc^2=1/(cos30)^2=1/(sqrt(3)/2)^2=1/(3/4)=4/3
If Ahexi=3
Ahexo=4/3*3=4
Hopefully not mistakes.
Answer:
Outer hex has area of 4 units
Cam
I'm not sure it's a correct demonstration, but I found 4, and as the problem is given you can think it will always be 4. (or you just forgot to mention it is a REGULAR hexagon)
Turn the inside hexagon such as each vertex is on a side of the outside hexagon (that's possible when you have a regular one since there is a point of each side on the circle).
Then, trace the star of David from the inside hexagon. the star can be divided as :
- the inside hexagon.
- 6 outside triangles, which total area is the same as the inside hexagon, 3.
These triangles, as you can see, each contain a part of the outside hexagon, and you easily show that each part's area is a third of a triangle.
The outside pentagons area is 3(from the inside pentagon) plus 1 (3, from the triangles, divided by 3) : 4.
As I said, it's good as long as we speak of a regular hexagon.
Karys (how do you create an account on this blog ?)
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