im having conundrum trouble. please help me?

Question

A Coin is unique in that the outer edge is gold while the inside is silver. Assuming both sides of the coin are identical, and a line tangent to the inner circle is 23 mm from edge to edge of the coin, and excluding the edge of the coin in the calculation, what is the surface area of gold on the coin, in square millimetres? Please round up to the nearest square millimetre.

Answer 1

I think you forget to copy the diagram.
http://i14.photobucket.com/albums/a344/allstarhunk69/221_dubloon2.gif

R = outer radius
r = inner radius

You'll find a right angled triangle involving 11.5, r and hypothenuse R

R^2 - r^2 = 11.5^2
π(R^2 - r^2) = π * 11.5^2 = area of gold
= 415.48 mm^2
On both sides it's double this value = 830.95 ~= 831 mm^2

Answer 2

This is a poorly worded question, but after re-reading it a couple of times I think I finally get it.

Basically we have two concentric circles. If you draw a line tangent to the inner circle, it defines a chord of the outer circle whose length is 23 mm.

Let "r" be the radius of the inner circle; and let "R" be the radius of the outer circle (variables are case-sensitive here!)

Draw a line segment from the center of the coin, to the point on the inner circle where the tangent line touches it. The length of this segment is "r".

Now draw another line segment from the center of the coin, to (one of) the points on the outer circle where the chord hits the outer circle. The length of this segment is "R".

If you're doing this on a piece of paper, you should see that the two segments you drew, plus half of the chord, form a right triangle. Then, by the pythagorean theorem:

r² + 11.5² = R²

Remember that: we'll come back to it.

Now, the question asks for the area of just the GOLD part (the outer part). This is the area of the outer circle minus the area of the inner circle:

Area of gold = πR² – πr² = π(R² – r²)

But from the previous equation:

R² – r² = 11.5²

So, combining these equations we get:

Area of gold = π(11.5²) = 415.48 mm²

So that's your answer. An interesting point is that we still don't know the values of "r" or "R", nor the surface area of the WHOLE coin! This means there are multiple possible coin sizes which would make this work; but in every case the area of the gold part is 415.48 mm².

Answer 3

This problem leads to the interesting theorem:

"The area of a ring formed by concentric circles is given

A = (π/4)T²,

where T is the length of a chord of the outer circle tangent to the inner circle."

I'm sure the other contributors have clearly demonstrated this fact beforehand.....

Answer 4

Lenny conundrum answers can always be seen here

http://www.able2know.com/forums/viewtopic.php?t=42093&postdays=0&postorder=asc&start=4730&sid=a7b24b4a6b55035c47a705be4be76079

This weeks answer, and the answer to your question is 831. To see how I got it go to the forum

Answer 5

note
if you are having conundrum trouble then you are having trouble being confused