A circle has a chord that goes from #( pi)/3 # to #(15 pi) / 8 # radians on the circle. If the area of the circle is #48 pi #, what is the length of the chord?

Question

A circle has a chord that goes from #(  pi)/3  # to #(15  pi) /  8  # radians on the circle.  If the area of the circle is #48   pi  #, what is the length of the chord?

Mason Adams · Accepted Answer

#8sqrt(3)sin(37/48 pi) ~~9.14# to 2 decimal places

Area of a circle is #pir^2#
#=> pir^2=48pi#

#=> r^2=48#

#=>r=sqrt(2^2xx2^2xx3)#

#=>r=4sqrt(3)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The angle of the arc is
#(15pi)/8-pi/3" " =" " pi((45-8)/24)" "=" "37/24 pi#

So 1/2 of this angle is #37/48pi#

The length of the chord is: #2(rsin(37/48 pi))#

#=8sqrt(3)sin(37/48 pi) ~~9.14# to 2 decimal places

Savannah Aguilar · Answer

undefined To find the length of the chord, use the formula $L = 2r\sin(\frac{\theta}{2})$, where $L$ is the length of the chord, $r$ is the radius of the circle, and $\theta$ is the angle subtended by the chord at the center of the circle. Given that the area of the circle is $48\pi$, you can solve for the radius $r$ using the formula for the area of a circle, $A = \pi r^2$. Then, substitute the given angle $\theta$ into the formula for the length of the chord to find its length.