I teach direction vector math on my Dataflow Geometry channel. How would it be used in this problem? The key unknown we'll solve for is the direction from the circle center to the circle point shared by both chords. Call it d, with components < d.x, d.y> There's a theorem I developed about chord lengths inside a quarter circle that can be applied, based on unknown r and unknown d: 1 = r sqrt(2 -2d.x) and √2 = r sqrt(2 - 2d.y) start by eliminating r sqrt(2 - 2d.x) = sqrt(1 - d.y) , then square both sides 2 - 2d.x = 1 - d.y we have the direction vector identity d.x^2 + d.y^2 = 1 , and use it to eliminate d.y 2 - 2d.x = 1 -sqrt (1 - d.x^2) this simplifies down to 5 * d.x = 4 --> d.x = 4/5 substitute for d.x in the 1st chord equation, r = √ (5/2) 1/4 π r^2 = A = 5π/8
@@mrtalbotmaths Fair enough, must've been YT deleting it, as they do sometimes. Looks like that X account has lifted it from other peoples channels without credit then: watch?v=kkyPephRj7U
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I teach direction vector math on my Dataflow Geometry channel. How would it be used in this problem?
The key unknown we'll solve for is the direction from the circle center to the circle point shared by both chords. Call it d, with components < d.x, d.y>
There's a theorem I developed about chord lengths inside a quarter circle that can be applied, based on unknown r and unknown d:
1 = r sqrt(2 -2d.x) and √2 = r sqrt(2 - 2d.y) start by eliminating r
sqrt(2 - 2d.x) = sqrt(1 - d.y) , then square both sides
2 - 2d.x = 1 - d.y we have the direction vector identity d.x^2 + d.y^2 = 1 , and use it to eliminate d.y
2 - 2d.x = 1 -sqrt (1 - d.x^2) this simplifies down to
5 * d.x = 4 --> d.x = 4/5 substitute for d.x in the 1st chord equation,
r = √ (5/2) 1/4 π r^2 = A = 5π/8
respectively
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So I see my previous comment about this problem being directly lifted from PreMath without credit has been deleted...
I haven’t deleted any comments and I credited where I got the problem from in the description :)
@@mrtalbotmaths Fair enough, must've been YT deleting it, as they do sometimes.
Looks like that X account has lifted it from other peoples channels without credit then: watch?v=kkyPephRj7U
Thanks for highlighting. If I know the problem is from another channel in future I will credit them for sure 👍🏼