Stand back, I'm doing MATH
Posted 8th January 2016 at 11:23 AM by Scuba Ben
Apropos of nothing, I recently remembered a math problem in high school. Or more to the point, I remembered that I was unable to explain why my solution worked.
Given two concentric circles; the larger has radius r1 = 1,000,000 (units are optional), and the smaller has radius r2 of a similarly large value. Also given is the length of a chord line of the larger circle that is tangent to the smaller circle. Find the area between the two circles.
The intended solution is to calculate the circles' areas and subtract. I had recalled a shortcut from some book on logic insights (possibly called Aha!) that used only the length of the chord-tangent line: Calculate the area of a circle whose diameter is that of the chord-tangent line; this is the answer to the problem.
When I presented the shortcut, I was unable to explain beyond "it works!" This sort of explanation is insufficient in exact science. So when the memory resurfaced, I set out to explain it.
(Caveat lector: Math begins here.)
Start by drawing the diagram: Large circle, small circle, and chord-tangent line. Make it easy on your eyes by drawing the chord-tangent line horizontally. Draw r1 from the center to an endpoint of the line, and r2 from the center to the tangent point. We'll come back to this diagram in a bit.
The area of the larger circle is of course A1 = π * r12. (1)
And the area of the smaller circle is A2 = π * r22. (2)
So the desired solution is A = π * r12 - π * r22. (3)
Factoring out the constant,
A = π * (r12 - r22) (4)
Ready to see why the shortcut works? Look back at that diagram. By the definition of a tangent line, r2 meets the chord-tangent at a right angle; by definition of the circles being concentric, that tangent point is the midpoint of the chord-tangent. The diagram should now look familiar: It's a right triangle. So half the length of the chord-tangent ("hTC") can be found by the Pythagorean theorem.
hTC = SQRT( r12 - r22 ). (5)
Square both sides.
hTC2 = r12 - r22. (6)
Use line 6 to simplify line 4:
A = π * hTC2
Quod erat demonstrandum, amen, selah.
Given two concentric circles; the larger has radius r1 = 1,000,000 (units are optional), and the smaller has radius r2 of a similarly large value. Also given is the length of a chord line of the larger circle that is tangent to the smaller circle. Find the area between the two circles.
The intended solution is to calculate the circles' areas and subtract. I had recalled a shortcut from some book on logic insights (possibly called Aha!) that used only the length of the chord-tangent line: Calculate the area of a circle whose diameter is that of the chord-tangent line; this is the answer to the problem.
When I presented the shortcut, I was unable to explain beyond "it works!" This sort of explanation is insufficient in exact science. So when the memory resurfaced, I set out to explain it.
(Caveat lector: Math begins here.)
Start by drawing the diagram: Large circle, small circle, and chord-tangent line. Make it easy on your eyes by drawing the chord-tangent line horizontally. Draw r1 from the center to an endpoint of the line, and r2 from the center to the tangent point. We'll come back to this diagram in a bit.
The area of the larger circle is of course A1 = π * r12. (1)
And the area of the smaller circle is A2 = π * r22. (2)
So the desired solution is A = π * r12 - π * r22. (3)
Factoring out the constant,
A = π * (r12 - r22) (4)
Ready to see why the shortcut works? Look back at that diagram. By the definition of a tangent line, r2 meets the chord-tangent at a right angle; by definition of the circles being concentric, that tangent point is the midpoint of the chord-tangent. The diagram should now look familiar: It's a right triangle. So half the length of the chord-tangent ("hTC") can be found by the Pythagorean theorem.
hTC = SQRT( r12 - r22 ). (5)
Square both sides.
hTC2 = r12 - r22. (6)
Use line 6 to simplify line 4:
A = π * hTC2
Quod erat demonstrandum, amen, selah.
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