Posted by: Pandora

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

A 24cm horizontal line is drawn at a tangent to the inside circle, as shown. Calculate the shaded area and show your work, please.

Posted by: Pandora
## Math Fun - 10/06/12 03:33 PM

A 24cm horizontal line is drawn at a tangent to the inside circle, as shown. Calculate the shaded area and show your work, please.

Posted by: CanukDenis
## Re: Math Fun - 10/06/12 05:01 PM

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.

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.

Posted by: Pandora
## Re: Math Fun - 10/06/12 05:17 PM

An excellent solve, Denis, and well explained.