A circle has a chord that goes from $\frac{π}{8}$ $( pi)/8$ to $\frac{4 π}{3}$ $(4 pi) / 3$ radians on the circle. If the area of the circle is $64 π$ $64 pi$ , what is the length of the chord?

Question

A circle has a chord that goes from $\frac{π}{8}$ $( pi)/8$ to $\frac{4 π}{3}$ $(4 pi) / 3$ radians on the circle. If the area of the circle is $64 π$ $64 pi$ , what is the length of the chord?

1 Answer

Impact of this question

1685 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-31T19:39:56

$c = \sqrt{{(8 cos (\frac{π}{8}) + 4)}^{2} + {(8 sin (\frac{π}{8}) + 4 \sqrt{3})}^{2}}$ $c=sqrt{(8cos(pi/8)+4)^2+(8sin(pi/8)+4sqrt3)^2}$
$c \approx 15.15088$ $c approx 15.15088$

Explanation:

Radius:
Area $= 64 π$ $=64pi$
$π r^{2} = 64 π$ $pir^2=64pi$
$r^{2} = 64$ $r^2=64$
$r = 8$ $r=8$

Chord length:
The first angle, $\frac{π}{8}$ $pi/8$ , is in quadrant I, and the second angle, $\frac{4 π}{3}$ $(4pi)/3$ , is in quadrant III. Imagine a right triangle where the hypotenuse is the chord from polar coordinates $(8, \frac{π}{8})$ $(8,pi/8)$ to $(8, \frac{4 π}{3})$ $(8,(4pi)/3)$ . The horizontal and vertical distances can be found by adding the absolute values of the cosine and sine values, respectively.

enter image source here
(Drawing not to scale)

To solve for chord length:
$x^{2} + y^{2} = c^{2} (Pythagorean theorem)$ $x^2+y^2=c^2" (Pythagorean theorem)"$

Substitute in values for x and y (shown in drawing above):
$c^{2} = {[∣ ∣ 8 cos (\frac{π}{8}) ∣ ∣ + ∣ ∣ ∣ 8 cos (\frac{4 π}{3}) ∣ ∣ ∣]}^{2} + {[∣ ∣ 8 sin (\frac{π}{8}) ∣ ∣ + ∣ ∣ ∣ 8 sin (\frac{4 π}{3}) ∣ ∣ ∣]}^{2}$ $c^2 = [|8cos(pi/8)|+|8cos((4pi)/3)|]^2+[|8sin(pi/8)|+|8sin((4pi)/3)|]^2$

Simplify (get rid of absolute values by determining their sign)
$c^{2} = {[8 cos (\frac{π}{8}) - 8 cos (\frac{4 π}{3})]}^{2} + {[8 sin (\frac{π}{8}) - 8 sin (\frac{4 π}{3})]}^{2}$ $c^2=[8cos(pi/8)-8cos((4pi)/3)]^2+[8sin(pi/8)-8sin((4pi)/3)]^2$

$c^{2} = {[8 cos (\frac{π}{8}) - 8 (- \frac{1}{2})]}^{2} + {[8 sin (\frac{π}{8}) - 8 (- \frac{\sqrt{3}}{2})]}^{2}$ $c^2=[8cos(pi/8)-8(-1/2)]^2+[8sin(pi/8)-8(-sqrt3/2)]^2$

$c^{2} = {(8 cos (\frac{π}{8}) + 4)}^{2} + {(8 sin (\frac{π}{8}) + 4 \sqrt{3})}^{2}$ $c^2=(8cos(pi/8)+4)^2+(8sin(pi/8)+4sqrt3)^2$

$c = \sqrt{{(8 cos (\frac{π}{8}) + 4)}^{2} + {(8 sin (\frac{π}{8}) + 4 \sqrt{3})}^{2}}$ $c=sqrt{(8cos(pi/8)+4)^2+(8sin(pi/8)+4sqrt3)^2}$

$c \approx 15.15088$ $c approx 15.15088$

This answer appears correct because if the radius is 8, then the diameter is 16, and this given chord is just less than 16 because it is not the diameter.

Science

Math

Humanities

... and beyond

A circle has a chord that goes from $\frac{π}{8}$ $( pi)/8$ to $\frac{4 π}{3}$ $(4 pi) / 3$ radians on the circle. If the area of the circle is $64 π$ $64 pi$ , what is the length of the chord?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

A circle has a chord that goes from π8( pi)/8 to 4π3(4 pi) / 3 radians on the circle. If the area of the circle is 64π64 pi , what is the length of the chord?

1 Answer

Explanation:

Related questions

Impact of this question

A circle has a chord that goes from $\frac{π}{8}$ $( pi)/8$ to $\frac{4 π}{3}$ $(4 pi) / 3$ radians on the circle. If the area of the circle is $64 π$ $64 pi$ , what is the length of the chord?