Math question.?

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? Round up to the nearest square millimetre.

Answer 1

Yes! You can find R and r. Indeed, you must be able to find them in order to calculate a numerical answer. You just have to understand the nature of tangent lines to a circle; how they behave; and what sorts of angles they create with respect to the various parts of the circle. The arc containing the tangent line subtends a 90° angle with the center of both circles. I will leave it up to you to prove why this is true. If you draw a careful diagram of the coin and the line tangent to the inner circle, you will understand why, and you will also understand why all of my calculations follow. If you need more explanation, you can e-mail me through Yahoo! Answers.

Using the above assertion, we can calculate that the radius of the outside circle is:

2R² = (23)²
2R² = 529
R² = 264.5
R = 16.2635

Now that we know the radius of the outer circle, we can use that to find the radius of the inner circle:

2r² = R²
r² = R²/2
r = (√2 R) / 2
r = [√2 (16.2635)] / 2
r = 23/2
r = 11.5

The surface area of gold on one side is:

Area of larger circle - area of smaller circle

A = π R² - π r²
A = π (R² - r²)
A = π [(16.2635)² - (11.5)²]
A = π [132.25] mm²
A ~ 415.5 mm²

Since there are two sides to the coin, we multiply the last result above by 2.

A (t) = 2 A ~ 2 (415.5 mm²) ~ 831 mm²

Answer 2

Let R be the overall radius.
Let r be the radius of the silver circle.
You can't find R or r, but it doesn't matter.
Draw the segment from the center to one end of the tangent segment. That has length R.
Draw the segment from the center to the tangent point. That has length r.
And you just formed a right triangle whose hypotenuse is R, with legs r and 11.5
So:
11.5^2 + r^2 = R^2
Or:
R^2 - r^2 = 11.5^2
The area of the outside ring is:
A = piR^2 - pir^2
A = pi(R^2 - r^2)
Just substitute and don't forget to double it because there are two sides to the coin.

Answer 3

I`ve been playing with this for a while, and I can`t find a solution. The only thing I`ve come up with is that if the radius of the larger circle is R, snd that of the smaller is r, then by Pythag, R^2 - r^2 = [11.5]^2. but without any further information, I`m lost. Looking forward to seeing a solution !

Answer 4

ahh yes this weeks Lenny Conundrum!! tricky one isnt it!

i believe the answer is 831mm.

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