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I need some help on designing a box packaging for 3 chocolate balls to fit in. The...

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beablanca123 | eNoter

Posted June 10, 2013 at 6:13 AM via web

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I need some help on designing a box packaging for 3 chocolate balls to fit in.

The first picture tells about the dimension of the chocolate ball which is 30mm in diametre and what the box would look like.

The second picture tells about my ideas of boxing for the chocolate ball to fit in. Idea 1 is like in a way toblerone chocolate boxes are like. So, it would be 3 chocolate balls in a row. The length is 110 just to allow spaces for the chocolate balls to fit in. I don't know the dimensions would be for the triangle shape on the end side tho. (that's what I need help for)

Idea 1 is just like puting the chocolate balls in a pyramid form sort of way. And I also don't know the dimension would be for the triangle. This one is a prism. Not a pyramid.

 

 

Please help me. Thank youuuu!!

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1 Answer | Add Yours

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted June 10, 2013 at 2:09 PM (Answer #1)

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In both cases the box is a prism:

(1) In the first case the height of the prism is 110 (given -- leaves extra 20mm for spacing: why?), and the base is an equilateral triangle. We can model the base as an equilateral triangle circumscribed about a circle of diameter 30 or as a circle with diamter 30 inscribed in an equilateral traingle.

Now the center of a circle inscribed in an equilateral triangle is 2/3 the distance from a vertex of the triangle and the side opposite that vertex. So the height of the triangle is 45. Then the base is `30sqrt(3)` .

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The box can be modeled as a prism with equilateral triangles as bases. The triangles have side length `30sqrt(3)` mm `(~~52"mm")`, and height 45mm . The height of the prism is 110mm.

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(2) In the second case the bases are also equilateral triangles. The height will be 35mm (given for spacing purposes.)

We will have 3 circles of diameter 30 mutually tangent with an equilateral triangle circumscribed about the system as in the picture.

Take one side of the triangle and label the vertices A and B. Let the centers of the circles tangent to AB be P and Q. Drop a radius from P and one from Q to AB; label the intersection points M and N respectively.

Now MN=30 (The sum of the radii of circles P and Q). Triangle APM is a 30-60-90 right triangle (radii meet tangents at right angles, the angles of an equilateral triangle measure 60 degrees -- ray AP bisects angle A.) So PM=15 (radius of the circle); since it is the side opposite the 30 degree angle AP=30 and `AM=BN=15sqrt(3)`

Now the side length of the equilateral trianle is `15sqrt(3)+30+15sqrt(3)=30(1+sqrt(3))`

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The prism in this case has a height of 35mm (given) and the bases are equilateral triangles with side length `30(1+sqrt(3))"mm"~~82"mm"`

The height of the triangular base is `(15sqrt(3)+45)"mm"~~71"mm"`

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