A circle has a chord that goes from $\frac{π}{12}$ $pi/12$ to $\frac{π}{4}$ $pi/4$ radians on the circle. If the area of the circle is $8 π$ $8 pi$ , what is the length of the chord?

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A circle has a chord that goes from $\frac{π}{12}$ $pi/12$ to $\frac{π}{4}$ $pi/4$ radians on the circle. If the area of the circle is $8 π$ $8 pi$ , what is the length of the chord?

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Answer 1 · 2016-12-16T09:16:31

$2 \sqrt{3} - 2$ $2sqrt(3)-2$

Explanation:

The area of a circle is given by the formula:

$area = π r^{2}$ $"area" = pi r^2$

where $r$ $r$ is the radius.

So in our example:

$8 π = π r^{2}$ $8pi = pi r^2$

Hence:

$r = \sqrt{8} = 2 \sqrt{2}$ $r = sqrt(8) = 2sqrt(2)$

So the chord is the base of an isoceles triangle with the other two sides of length $2 \sqrt{2}$ $2sqrt(2)$ .

Bisecting the chord with a line joining the origin to the midpoint at $\frac{π}{6}$ $pi/6$ we obtain two right angled triangles, each with hypotenuse of length $2 \sqrt{2}$ $2sqrt(2)$ and acute angle $\frac{π}{12}$ $pi/12$ .

Hence the length of half of the chord is the length of the smaller leg of the right angled triangle:

$2 \sqrt{2} \cdot sin (\frac{π}{12}) = 2 \sqrt{2} \cdot \frac{1}{4} (\sqrt{6} - \sqrt{2}) = \sqrt{3} - 1$ $2sqrt(2)*sin(pi/12) = 2sqrt(2)*1/4(sqrt(6)-sqrt(2)) = sqrt(3)-1$

So the length of the chord is $2 \sqrt{3} - 2$ $2sqrt(3)-2$

graph{(x^2+y^2-8)(y-x)(y - (1/4(sqrt(6)-sqrt(2)))x)(y-1/(sqrt(3))x)((y-2)+tan(pi/3)(x-2)) = 0 [-1.055, 3.945, -0.26, 2.24]}

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A circle has a chord that goes from $\frac{π}{12}$ $pi/12$ to $\frac{π}{4}$ $pi/4$ radians on the circle. If the area of the circle is $8 π$ $8 pi$ , what is the length of the chord?

1 Answer

Explanation:

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A circle has a chord that goes from $\frac{π}{12}$ $pi/12$ to $\frac{π}{4}$ $pi/4$ radians on the circle. If the area of the circle is $8 π$ $8 pi$ , what is the length of the chord?