A circle has a chord that goes from $\frac{π}{4}$ $( pi)/4$ to $\frac{13 π}{8}$ $(13 pi) / 8$ radians on the circle. If the area of the circle is $32 π$ $32 pi$ , what is the length of the chord?

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Answer 1 · 2017-04-27T20:13:58

$c \approx 9.4 \leftarrow$ $c ~~ 9.4" "larr$ the length of the chord.

We can use the area of the circle, $32 π$ $32pi$ , to compute the radius:

$Area = π r^{2}$ $"Area" = pir^2$

$32 π = π r^{2}$ $32pi = pir^2$

$r = \sqrt{32} = 4 \sqrt{2}$ $r = sqrt(32) = 4sqrt2$

Two radii and the chord form an isosceles triangle where "c" is the unknown length of the chord.

The lengths of the other two sides are:

$a = b = r = 4 \sqrt{2}$ $a = b = r = 4sqrt(2)$

The angle between the radii is:

$θ = \frac{13 π}{8} - \frac{π}{4} = \frac{11 π}{8}$ $theta = (13pi)/8-pi/4 = (11pi)/8$

We can use the Law of Cosines to find the length of the chord:

$c = \sqrt{a^{2} + b^{2} - 2 (a) (b) cos (θ)}$ $c = sqrt(a^2+b^2-2(a)(b)cos(theta)$

$c = \sqrt{32 + 32 - 64 cos (\frac{11 π}{8})}$ $c = sqrt(32+32-64cos((11pi)/8)$

$c \approx 9.4 \leftarrow$ $c ~~ 9.4" "larr$ the length of the chord.

A circle has a chord that goes from π4( pi)/4 to 13π8(13 pi) / 8 radians on the circle. If the area of the circle is 32π32 pi , what is the length of the chord?