A circle has a chord that goes from #( pi)/3 # to #(2 pi) / 3 # radians on the circle. If the area of the circle is #4 pi #, what is the length of the chord?

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Answer 1 · 2016-04-23T11:08:41

Length of the chord is 1

Tony B

Given:#" "/_A = (2pi)/3-pi/3 = pi/3#

Known: #/_C=/_B#

#/_A+/_B+/_C = pi" radians" -> 180^o#

Thus #color(blue)(/_C=/_B = 1/2( pi-pi/3) = pi/3" radians" ->60^o)#

Thus #Delta ABC ->" equilateral triangle"#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Given:#" area"=4pi#

But #" area"=pir^2#

Equate through area

#" "4pi="area"=pir^2#

Divide both sides by #pi#

#=>4=r^2#

#=> r=2#

Side #b=c=2#
'~~~~~~~~~~~~~~~~~~~~~~~~
Known: #cos(/_C)= a/2 -:r=a/(2r)#

#=>2rcos(/_C)=a#

#=>2(2)cos(pi/3)=a#

but #cos(pi/3)=1/2#

#=>2(2)(1/2)=a#

#=>"chord "= a=1#