A circle has a chord that goes from #( 5 pi)/4 # to #(4 pi) / 3 # radians on the circle. If the area of the circle is #42 pi #, what is the length of the chord?

Question

A circle has a chord that goes from #( 5 pi)/4  # to #(4  pi) /  3  # radians on the circle.  If the area of the circle is #42   pi  #, what is the length of the chord?

Aurora Ayala · Accepted Answer

#=1.7# Area of the circle#=pir^2=42pi# or #r^2=42# or #r=sqrt42# or #r=6.48# Circumference=#2pir=2pi(6.48)=40.72# Chord goes from #5pi/4# to #4pi/3# that is moves#(4/3-5/4)pi=pi/12# Therefore we can write #2pi# corresponds to #40.72# #pi/12#corresponds to #40.72/(2pi)(pi/12)# #=1.7#

Theodore Abad · Answer

undefined To find the length of the chord, we need to use the formula for the length of a chord in a circle, which is $2r \sin(\frac{\theta}{2})$, where $r$ 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 $42\pi$, we can use the formula for the area of a circle, $A = \pi r^2$, to find the radius of the circle.

$42\pi = \pi r^2$  
$r^2 = 42$  
$r = \sqrt{42}$

Now, we need to find the angle $\theta$ subtended by the chord at the center of the circle. The chord spans from $\frac{5\pi}{4}$ to $\frac{4\pi}{3}$ radians.

To find the angle subtended by the chord, we subtract the smaller angle from the larger angle:

$\theta = \frac{4\pi}{3} - \frac{5\pi}{4}$  
$\theta = \frac{16\pi}{12} - \frac{15\pi}{12}$  
$\theta = \frac{\pi}{12}$

Now, we can use the chord length formula:

$2r \sin\left(\frac{\theta}{2}\right) = 2\sqrt{42} \sin\left(\frac{\pi}{24}\right)$

Using a calculator to evaluate $\sin\left(\frac{\pi}{24}\right)$, we find:

$\sin\left(\frac{\pi}{24}\right) \approx 0.1305$

Substituting this value into the formula:

$2\sqrt{42} \times 0.1305 \approx 5.5003$

So, the length of the chord is approximately $5.5003$ units.