http://www.mathopenref.com/annulus.html
Sorry it's after midnight... I was a dumb assumtion. Go see the page
However much it does look like it, assuming is bad according to my Geometry teacher
we should look for a relation to tangents, radii, and surface area.
however, just a quick question: if we are doing surface area, are we doing the whole coin? or just one flat surface? because they say surface area...which would imply two faces and the outer ridge/rim part.
Sorry it's after midnight... I was a dumb assumtion. Go see the page
~x~
no prob.. just first thought that comes to mind... just checked page, btw
ah, annulus. we did a whole section on those in calc. hold on while i check my book
Let's see...
We can't get the radius of either the smallest or the largest circle, because we're missing someting (like the distance between the tangent and the end of the largest circle), so I guess we can't do it that way. I tried drawing many circles with the same 23 mm tangent, and all were possible, so what I asked myself was:
What if the area is always the same? Because the larger the inner, the larger the outer, and so on.
So I made it so the inner circle would be 0,00....1 mm diameter (zero), so the diameter of the outer circle would be 23.
Then:
A = 3,1416 x (23/2)^2
And then, the number they want is 2A, for the coin has two sides, and they ask you for the total area in the coin, not in that circle.
i worked on an alike problem...
i cracked up using annulus, but then i found an easier solution, my math teacher was like
but i can't remember how did
D'oh!
so the radius IS 23?
weird o.o
and is the @A for just the gold section of the coin?
if the diameter was in fact 23, it would be too easy...
http://www.vias.org/comp_geometry/geom_circle_ring.html
Ok, if someone can understand what the formula on the above page means we have it
It goes like this:
It can be shown that the area of the annulus is equal to the area of the circle whose diameter is tangent to the inner circle (point C) and whose radius is the line segment between point C and the intersection of the tangent and the outer circle (points A and B).
The diameter of the coin is NOT 23, that is just something I made supposing the Area of the outer circle would be always the same given a chord tangent to an inner circle.
If not, then we're missing something and it is impossible to do it.
Excuse my bad English, first time I use it for something related to maths.
so the area of the annulus is a circle with a diameter of 23.
which would be pi(23/2)^2
emily1213 wrote:
however, just a quick question: if we are doing surface area, are we doing the whole coin? or just one flat surface? because they say surface area...which would imply two faces and the outer ridge/rim part.
The question says "what is the surface area, excluding the edge" so I would assume they mean both faces, but not the ridge/rim/edge/side/whatever you want to call it.
that's what I was thinking, since the angle is tangent, you can draw a square about the circle, 23mmx23mm. QED, the diameter is 23, radius is 11.5.
That is it. I have just realised Anneska's possible answer is the same as mine, only that we arrived there from different paths.
Don't forget to multiply by two, keep in mind that the coin has two sides.
Thank you for submitting your answer
Good luck!!
i have to go to work
yay for educated guesses
good luck to you all
OK... if I understand the following correctly
It can be shown that the area of the annulus is equal to the area of the circle whose diameter is tangent to the inner circle (point C) and whose radius is the line segment between point C and the intersection of the tangent and the outer circle (points A and B)
The answer would be
(3.14 x 17.5 squered) x 2
= 1923.25 mm
How does that sound
Anneska...
for my understanding..
it says
S(A) of Annulus = Area of Bigger circle - Area of smaller circle
S(A)= (pi)R^2 - (pi)r^2
S(A)= (pi)x(R^2 - r^2)
not sure if that was what you were asking...
(i used to compete in math olympiads but i paused for a year, so i'm a lil out of practice)
Anneska, the radius is 11.5, not 17,5...
It is okay :wink:.
Oops so the the answer is
830.53?
And that would be 831.
That is what I put.
I hope I get a golden trophy this time...
