Help with surface area????

Question

You need to view the picture here:
http://i14.photobucket.com/albums/a344/allstarhunk69/221_dubloon2.gif




A Two Hundred Dubloon 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

Take the midpoint of the tangent line shown there, call that A. Call the center of the coin O. Call either end of the tangent line B. Notice the right triangle OAB has the following side lengths:
R -- total radius (hypotenuse)
r -- silver radius
11.5 --- half of 23

that means:

R² = r² + 11.5²

R² - r² = 11.5² = 132.25

So:
gold area =
total area - silver area=
π R² - π r²=
π (R^2 - r^2) = 132.25 π

actually, double it, since there's two sides of the coin, to get the final answer:
264.5 π mm²

which is approximately:
830.951257 mm²

Answer 2

First, make 23mm tangent lines around the other 3 "sides" of the circle. This will form a square, and it will be easy to see how to find the radii of the silver part and entire coin.

The radius of the silver part:
23/2 = 11.5

The radius of the entire coin
23^2 + 23^2 = d^2
d = sqrt(1058)
r = [sqrt(1058)] / 2


Surface area of the gold part:
2 * (Area of entire coin - Area of silver part)
=2 [(Pi* [[sqrt(1058)] / 2]^2) - (Pi * 11.5^2)]
=2 [(Pi* (1058/4)) - (Pi * 11.5^2)]
=2 (Pi (264.5 - 132.25))
=2 * Pi * 132.25
=264.5 * Pi mm^2
=830.95 mm^2

Answer 3

Two Hundred Dubloon Coin

Answer 4

How many of you guys are in the same class? I answered this identical question earlier today. Here I'm pasting my solution.
---
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 5

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 6

Let r_1 be the radius of the silver circle and r_2 be the radius of the coin. Then

r_1^2+(23/2)^2=r_2^2

so
pi*r_1^2+pi*(23/2)^2=pi*r_2^2

so the area of the gold ring is

pi*r_2^2-pi*r_1^2

=pi*(23/2)^2 mm^2

=415.475628 mm^2