A chord with a length of #3 # runs from #pi/8 # to #pi/6 # radians on a circle. What is the area of the circle?

Answer 1

It is #22.93#.

In a circle, the length of a chord is

#c=2rsin(theta/2)#

and so the radius is

#r=c/(2sin(theta/2)#

In our case theta is given by

#theta=pi/6-pi/8=pi/24#.

then

#r=3/(2sin(pi/48))\approx22.93#
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Answer 2

To find the area of the circle, we need to determine the radius.

First, let's find the central angle subtended by the chord. The central angle is ( \frac{\pi}{6} - \frac{\pi}{8} = \frac{\pi}{24} ) radians.

Now, using the formula for the length of a chord in terms of the radius and central angle, which is ( 2r \sin(\frac{\theta}{2}) ), where ( r ) is the radius and ( \theta ) is the central angle, we can solve for the radius.

Given that the length of the chord is 3 and the central angle is ( \frac{\pi}{24} ), we have:

[ 3 = 2r \sin\left(\frac{\frac{\pi}{24}}{2}\right) ]

[ 3 = 2r \sin\left(\frac{\pi}{48}\right) ]

[ r = \frac{3}{2\sin\left(\frac{\pi}{48}\right)} ]

Now, we can use the formula for the area of a circle, ( A = \pi r^2 ), to find the area:

[ A = \pi \left(\frac{3}{2\sin\left(\frac{\pi}{48}\right)}\right)^2 ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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