A circle has a chord that goes from $\frac{3 π}{2}$ $( 3 pi)/2$ to $\frac{7 π}{4}$ $(7 pi) / 4$ radians on the circle. If the area of the circle is $121 π$ $121 pi$ , what is the length of the chord?

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A circle has a chord that goes from $\frac{3 π}{2}$ $( 3 pi)/2$ to $\frac{7 π}{4}$ $(7 pi) / 4$ radians on the circle. If the area of the circle is $121 π$ $121 pi$ , what is the length of the chord?

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Answer 1 · 2016-03-24T17:02:20

See geometric figure:
Chord, $¯ ¯¯¯¯ ¯ A B = 11 \sqrt{2}$ $bar(AB) = 11sqrt(2)$

enter image source here

Explanation:

This straight forward problem:
A) Determiner the radius from $C_{A} = π R^{2}$ $C_A=piR^2$ ; $R = 11$ $R =11$
Look at the angular displacement between $A \to B$ $A ->B$ it form
a right angle at the center so trangle AOB is an "isosceles right angle". Thus the ratio of side of an isosceles right triangle is:
$s_{1} : s_{2} : h = 1 : 1 : \sqrt{2}$ $s_1:s_2:h=1:1:sqrt(2)$ So for triangle AOD $\Rightarrow a : f : ¯ ¯¯¯¯ ¯ A D = 1 : 1 : \sqrt{2}$ $=> a:f:bar(AD)=1:1:sqrt(2)$
$A D = 11 \frac{\sqrt{2}}{2}$ $AD=11sqrt(2)/2$ , thus the chord, $A B = 11 \sqrt{2}$ $AB=11sqrt(2)$

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A circle has a chord that goes from $\frac{3 π}{2}$ $( 3 pi)/2$ to $\frac{7 π}{4}$ $(7 pi) / 4$ radians on the circle. If the area of the circle is $121 π$ $121 pi$ , what is the length of the chord?

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A circle has a chord that goes from $\frac{3 π}{2}$ $( 3 pi)/2$ to $\frac{7 π}{4}$ $(7 pi) / 4$ radians on the circle. If the area of the circle is $121 π$ $121 pi$ , what is the length of the chord?