Math friends... HELP!!

[Mom2Em]

A friend of mines daughter wants to know if it is possible to find the area of a Hexacontagon (60 sided figure) knowing that 7/60 of it forms a 42 degree angle in the middle...

Any help would be greatly appreciated!!! See her problem of the month...

http://www.springfield.k12.il.us/teachers/jankocis/projects/231/Problem_of_the_Month.pdf

 

[highland3]

OMG!!!!! And I thought my 10 year old DD's math homework was tough!!!

 

[Vivianne]

Let's see...

I'll get back to you on that.

 

[Kimberly]

7 out of 60 hexagons in a straight line or clumped together in a "circular" formation. It is possible... but I think you need more information.

 

[Mom2Em]

Kimberly,
That was what I was thinking... It looks that 7/60th of the hexacontagon is shaded and 14/120th or 7/60th of the rectangle is shaded... So how does she tell if the overall area is the same? I took 4 years of Calculus but always hated the Geometry...

 

[Kimberly]

What age is this problem for? I don't even have work this difficult and I'm in medical school...

If it is shaded that 7 of them are shaded in a circular formation, just take the distance across the shaded area to get the arc length (or in any pattern that they are shaded, take the angle at which the average radius of the shaded formation covers for the arc length), and then the circumference can be figured out. Use that to find the diameter, and then you can use that and the radius along with pi to find the surface area.

It won't be exact, but it will work!

 

[Kimberly]

Ah, just saw the adobe file... not as simple as i orginally thought. How irritating...

I just don't have the time to solve this problem, I'm really sorry. Anyone else wish to help!?

 

[Mom2Em]

This problem is for 7th grade basic math... I know where you are coming from!!!

Thanks so much for taking the time to look!!

 

[Gone Disney]

I'm not a professor and have been out of school for quite awhile but my 2 cents is this


both have the same shaded area.

You would need additional information such as a diameter of the circle or length of the rectangle to find the actual area, however that information is not supplied. You could assume the rectangle was 15 square inches and then deduce from there but that would be mere speculation. That's my guess


My take would be that you already answered the questions asked. In the problem it reads "IF the overall area of the hexacontagon is EQUAL to that of the large rectangle" which figure has the larger shaded area. I take it to mean that we are to base the information off of the two figures having the same area. If she is trying to figure out the actual area of the two figures for additional problem solving I doubt she would be able to do so with the information supplied.

But, like I said I'm no professor so I could be mistaken here

 

[grlpwrd]

I have no idea. ...lol

I showed this to my 12yo and asked her how she would set it up. She said you need to figure out the areas using angles, like A=(pi)(r squared) for the hexacontagon since you know the angle is 42 degrees. It's about triangles/angles and fractions.

I would look online.

Oh my - that is really difficult. Then again I don't know if you should go by fractions just by looking at it or if you should as far as using sin and cos values. lol

 

[Mom2Em]

Gone Disney,
Thanks so much!! That makes perfect sense!! I knew I could can't on my DIS friends!!!

 

[almacdonald]

Going out on a limb here...

Hexacontagon has 42 of 360 degrees shaded = 7/60 = 0.116666666 of the is total area shaded.

For the Large Rectange, there are 15 sub-divide squares. Add up the total amount of shading from each square (from top left to bottom right) 1/8 + 1/8 + 2/8 + 1/4 + 2/8 +2/4 + 2/8 = 1.75shaded/15total squares = 7/60 total area is shaded.

So.... If both the Hexacontagon and the Large rectangle have the same total area (regardless of what the area is quantitatively) the shaded area of both shapes is the same (7/60). Just a simple fractions problem.