Real life area of triangle

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Leah5467 said:
Hello,this is the question i don't know:
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If you drew vertical and horizontal lines joining through the touching points of the coins - what would be the shape (rectangle, square, triangle, etc.) that would surround each coin?

Assuming the width of the $10 note = w and the length = 2w, what would be the area of each coin as a function of 'w'?
 
Thank you for your detailed explanation!
12 triangles i think.
Let area of each coin be x
the function would be=(w)(2w)-8x

I don't really know how to find x.

It gave us the formula:1/2ab(sinc),but the sides are unknown.

I am guessing the sides of each triangle are the same length because the angles are the same(150/2)((150 is from 12-2)x180/12)
then i tried to set up formulas:
1/2(ab)sin75=1/2(a^2)sin30

But there are two unknowns...so i can't solve....
But are my previous steps correct? And what should i do next?
 
Let the "radius" of the coin be r. Then the length of the note is 8r and the width is 4r.

Now, pka's formula for the area of a dodecahedron is just 12 lots of \(\displaystyle \frac{1}{2}r^2 cos 30^o = 12 * \frac{1}{2}r^2 * \frac{1}{2} = 3r^2\)

So, in terms of r the total area of the note is …….???

The total area of the coins is ……..??

The area NOT covered by coins is ..........??

The fraction of the note NOT covered is ………??
 
the total area is (8r)(4r)

total area of the coins=(3r^2)12

The area NOT covered by coins=(8r)(4r)-(3r^2)12

The fraction of the note NOT covered=(8r)(4r)-(3r^2)12/(8r)(4r)

Thank you so much! Are my steps correct?
 
ah one more question please:why use 1/2(cos30)(r^2) but not 1/2(sin30)(r^2) for the area of triangle? Will there be any difference if i use it?
 
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