Surface area of a Circle problem?

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?




How do I solve this?

Answer 1

I answered it here:

http://answers.yahoo.com/question/index;_ylt=AhULOTC7cvzLX1jrg1wKXuvty6IX?qid=20070706055853AA2OLcc&show=7#profile-info-oaFyKJn9aa

but I forgot that the coin has two sides.

Double that to get the correct answer. So the correct value is 831 mm².

Answer 2

R = larger radius
r = smaller radius

The area A of the gold surface is given by
A = pi*R^2 - pi * r^2
A = pi(R^2 - r^2)

Then, by drawing the two concentric circles, the 23 mm line segment, and connecting one endpoint of this 23 mm line segment to the center of the circles, you will have a right triangle using just half of the 23 mm line segment. (Draw it to see it.)

Use the Pythagorean theorem to get:
r^2 + 11.5^2 = R^2
So 132.25 = R^2 - r^2

Use this in the equation for the area to get:
A = pi(R^2 - r^2)
A = pi(132.25)
A = approx 415.48

I hope this helps!

Answer 3

Area of circle is pi R squared, or pi D squared over 4.

If d=23mm then
A=3.1416 x 23 squared over 4 in mm

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