Points $(2, 4)$ $(2 ,4 )$ and $(4, 1)$ $(4 ,1 )$ are $\frac{3 π}{4}$ $(3 pi)/4$ radians apart on a circle. What is the shortest arc length between the points?

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Points $(2, 4)$ $(2 ,4 )$ and $(4, 1)$ $(4 ,1 )$ are $\frac{3 π}{4}$ $(3 pi)/4$ radians apart on a circle. What is the shortest arc length between the points?

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Answer 1 · 2016-02-15T11:40:20

$(\frac{3 π}{4}) \cdot \sqrt{\frac{13 (2 + \sqrt{2})}{2}}$ $((3pi)/4)*sqrt{frac{13(2+sqrt2)}{2}}$

Explanation:

Let the radius of the circle be $R$ $R$ . The length of the shorter arc between $(2, 4)$ $(2,4)$ and $(4, 1)$ $(4,1)$ would simply be $R (\frac{3 π}{4})$ $R({3pi}/4)$ . Therefore, the key to solving this problem, and many other geometry problems is to find $R$ $R$ .

We denote the center of the circle as $O$ $O$ .

The line segment from $O$ $O$ to any point on the circle has length $R$ $R$ . This means that the line segment from $O$ $O$ to $(2, 4)$ $(2,4)$ and the line segment from $O$ $O$ to $(4, 1)$ $(4,1)$ both have length $R$ $R$ .

Consider the triangle formed by $O$ $O$ , $(2, 4)$ $(2,4)$ and $(4, 1)$ $(4,1)$ . Using the Law of Cosines, we can find show that the length of the line segment from $(2, 4)$ $(2,4)$ to $(4, 1)$ $(4,1)$ is

$\sqrt{R^{2} + R^{2} - 2 (R) (R) cos (\frac{3 π}{4})}$ $sqrt{R^2+R^2-2(R)(R)cos({3pi}/4)}$

$= \sqrt{2} R \sqrt{1 - cos (\frac{3 π}{4})}$ $= sqrt2 R sqrt{1-cos({3pi}/4)}$

$= \sqrt{2} R \sqrt{1 - \frac{\sqrt{2}}{2}}$ $= sqrt2 R sqrt{1-sqrt2/2}$

$= \sqrt{2} R \frac{\sqrt{2 - \sqrt{2}}}{\sqrt{2}}$ $= sqrt2 R sqrt{2-sqrt2}/sqrt2$

$= \sqrt{2 - \sqrt{2}} R$ $= sqrt{2-sqrt2}R$

We can also use Pythagorean Theorem to find the exact length between $(2, 4)$ $(2,4)$ and $(4, 1)$ $(4,1)$ .

$\sqrt{{(4 - 2)}^{2} + {(1 - 4)}^{2}} = \sqrt{13}$ $sqrt((4-2)^2 + (1-4)^2) = sqrt13$

To find $R$ $R$ , we just need to equate the 2 independent expressions for the length of the line segment from $(2, 4)$ $(2,4)$ to $(4, 1)$ $(4,1)$ .

$\sqrt{2 - \sqrt{2}} R = \sqrt{13}$ $sqrt{2-sqrt2}R = sqrt13$

Solving the above equation gives

$R = \sqrt{\frac{13}{2 - \sqrt{2}}}$ $R = sqrt{frac{13}{2-sqrt2}}$

$= \sqrt{\frac{13 (2 + \sqrt{2})}{(2 - \sqrt{2}) (2 + \sqrt{2})}}$ $= sqrt{frac{13(2+sqrt2)}{(2-sqrt2)(2+sqrt2)}}$

$= \sqrt{\frac{13 (2 + \sqrt{2})}{2}}$ $= sqrt{frac{13(2+sqrt2)}{2}}$

The arc length is given by

$R (\frac{3 π}{4}) = \sqrt{\frac{13 (2 + \sqrt{2})}{2}} (\frac{3 π}{4})$ $R((3pi)/4) = sqrt{frac{13(2+sqrt2)}{2}}((3pi)/4)$

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Points $(2, 4)$ $(2 ,4 )$ and $(4, 1)$ $(4 ,1 )$ are $\frac{3 π}{4}$ $(3 pi)/4$ radians apart on a circle. What is the shortest arc length between the points?

1 Answer

Explanation:

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Points (2 ,4 )(2,4)(2 ,4 ) and (4 ,1 )(4,1)(4 ,1 ) are (3 pi)/4 3π4(3 pi)/4 radians apart on a circle. What is the shortest arc length between the points?

1 Answer

Explanation:

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Points $(2, 4)$ $(2 ,4 )$ and $(4, 1)$ $(4 ,1 )$ are $\frac{3 π}{4}$ $(3 pi)/4$ radians apart on a circle. What is the shortest arc length between the points?