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IntMath forum | Differential Equations

polygons [Solved!]

My question

Trying to find the area of a honey comb. My answer does not equal the online answer. The actual question is Question 7 at the url I entered below.

Relevant page

Polygon Area Practice - MathBitsNotebook(Geo - CCSS Math)

What I've done so far

From vertex to vertex the distance is 5. Therefore the radius is 2.5.

a = apothem

Using r = a/cos (180/n) the apothem then is 2.17.

Constructing a right triangle, the unknown side is 1.24.

Multiply this by 2 to get the length of one side: 2.48

perimeter = 2.48 x 6 = 14.88

Area = .5 x 14.88 x 2.17 = 16.1448

16.1448 x 36 1/2 cells = 589.2852 `mm^2`

X

Trying to find the area of a honey comb.  My answer does not equal the online answer.  The actual question is Question 7 at the url I entered below.
Relevant page

<a href="https://mathbitsnotebook.com/Geometry/Polygons/POpolygonAreaPractice.html">Polygon Area Practice - MathBitsNotebook(Geo - CCSS Math)</a>

What I've done so far

From vertex to vertex the distance is 5.  Therefore the radius is 2.5.  

a = apothem

Using r = a/cos (180/n) the apothem then is 2.17.

Constructing a right triangle, the unknown side is 1.24.

Multiply this by 2 to get the length of one side: 2.48

perimeter = 2.48 x 6 = 14.88


Area = .5 x 14.88 x 2.17 = 16.1448


 16.1448 x 36 1/2 cells = 589.2852 `mm^2`

Re: polygons

@phinah: I (roughly) agree with your answer.

But I think there is an easier way to find it. If the vertex-to-vertex length is 5 mm, then the side of each triangle (they are equilateral triangles) is 2.5 mm.

Using the formula

`"Area hexagon" = (3sqrt(3))/2 s^2,`

where `s` is the side length, we obtain for one cell:

`"Area" = (3sqrt(3))/2 (2.5)^2 ~~ 16.23798 "mm"^2`

Since there are 36.5 cells in the illustration, the total area would be:

`"Area" = 36.5 xx 16.23798 ~~ 592.687 "mm"^2`, which is closest to their second answer.

(Their accepted answer, 2371, would mean each cell has area `2371/36.5 = 64.96`. However, if we take a rough estimate for the hexagon's area via the circle through the vertices, we get `"Area" = pi r^2 = pi(2.5)^2 = 19.63`. The actual area has to be less than this for each cell.)

X

@phinah: I (roughly) agree with your answer.

But I think there is an easier way to find it. If the vertex-to-vertex length is 5 mm, then the side of each triangle (they are equilateral triangles) is 2.5 mm.

Using the formula 

`"Area hexagon" = (3sqrt(3))/2 s^2,` 

where `s` is the side length, we obtain for one cell:

`"Area" = (3sqrt(3))/2 (2.5)^2 ~~ 16.23798 "mm"^2`

Since there are 36.5 cells in the illustration, the total area would be:

`"Area" = 36.5 xx 16.23798 ~~ 592.687 "mm"^2`, which is closest to their second answer.

(Their accepted answer, 2371, would mean each cell has area `2371/36.5 = 64.96`. However, if we take a rough estimate for the hexagon's area via the circle through the vertices, we get `"Area" = pi r^2 = pi(2.5)^2 = 19.63`. The actual area has to be less than this for each cell.)

Re: polygons

We took "vertex-to-vertex" to mean non-adjacent vertices as well as adjacent vertices.

So we measured from the furthest left vertex to the furthest right vertex to be 5 mm. That is how we arrived at 589. We reworked the exercise identifying the segments between adjacent vertices to be = to 5 mm.

So this is how we went forward:

apothem = `s/[2 tan (180/n)]` = `5/[2 tan 30]` = 4.33

So Area =` 1/2`(apothem)(perimeter) = (1/2)(4.33)(30) = 64.95 `times` 36.5 = 2370.6 = 2371.

Thanks for helping.

X

We took "vertex-to-vertex" to mean non-adjacent vertices as well as adjacent vertices.  

So we measured from the furthest left vertex to the furthest right vertex to be 5 mm. That is how we arrived at 589. We reworked the exercise identifying the segments between adjacent vertices to be = to 5 mm.

So this is how we went forward:

apothem = `s/[2 tan (180/n)]` = `5/[2 tan 30]` = 4.33

So Area =` 1/2`(apothem)(perimeter) = (1/2)(4.33)(30) = 64.95 `times` 36.5 = 2370.6 = 2371.

Thanks for helping.

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