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#840691 - 10/06/12 03:33 PM Math Fun
Pandora Offline

Sonic Boomer

Registered: 08/06/00
Posts: 46048
Loc: Upstate NY
A 24cm horizontal line is drawn at a tangent to the inside circle, as shown. Calculate the shaded area and show your work, please.

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#840707 - 10/06/12 05:01 PM Re: Math Fun [Re: Pandora]
CanukDenis Offline
Graduate Boomer

Registered: 11/26/00
Posts: 15500
Loc: Ottawa Ontario Canada
Both circles have same center: make that point C.
Let A = halfway (tangent) point of tangent line.
Let B = right end of tangent line (could be left end).
Let r = radius of inner circle.
Let r + x = radius of outer circle.

We now have right triangle ABC: AB = 12, AC = r, BC = r + x.
Per Big Pete Pythagoras: (r + x)^2 = r^2 + 12^2.
Simplified: x^2 + 2rx = 144 [1]

(P = pi): Area outer circle = P(r + x)^2
Area inner circle = Pr^2

Let K = shaded area; then:
K = area outer - area inner
K = P(r + x)^2 - Pr^2
K/P = (r + x)^2 - r^2 ; simplified:
K/P = x^2 + 2rx [2]

Substitute [1] in [2]:
K/P = 144
K = ~3.14 * 144 = ~452.16

What's interesting(?!) here is r can be any value;
x will decrease proportionally as r increases.



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#840710 - 10/06/12 05:17 PM Re: Math Fun [Re: Pandora]
Pandora Offline

Sonic Boomer

Registered: 08/06/00
Posts: 46048
Loc: Upstate NY
An excellent solve, Denis, and well explained. smile smile smile
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