Question

A chord with a length of `4` runs from `pi/8` to `pi/2` radians on a circle. What is the area of the circle?

Answer 1

`(16pi)/(2-sqrt(2-sqrt2))`

Explanation:

Here we have a depiction of what the circle could look like:

• The blue line shows the circle's radius at a measure of `pi/8` radians.
• The green line shows the circle's radius at a measure of `pi/2` radians.
• The purple line is the chord of length `4`.
• The angle between the green and blue radius is `pi/2-pi/8=(3pi)/8`.

We know that the triangle formed here is isosceles, since the radius forms two of its sides. Thus the triangle can be bisected to form two right triangles:

Let's examine just the triangle formed by the blue radius, part of the purple chord, and the orange bisector:

• The blue chord, the radius, has length `r`.
• The part of the purple chord, now bisected, has length `2`.
• The angle between the blue radius and orange chord is half of the previous angle measure, meaning its measure is now `(3pi)/16`.

Examine the right triangle. We have:

• An angle of `(3pi)/16`.
• Opposite the angle, a leg of length `2`.
• A hypotenuse of length `r`.

Through trigonometry, this translates into:

`sin((3pi)/16)=2/r`

Solving for `r`:

`r=2/sin((3pi)/16)`

We could use a calculator to determine that

`rapprox3.5999`

However, we could also find the exact value of `(3pi)/16` using half-angle identities, since `(3pi)/4xx1/2xx1/2=(3pi)/16`.

The sine half-angle identity:

`sin(x/2)=+-sqrt((1-cos(x))/2)`

So, finding `sin((3pi)/16)` will require finding `cos((3pi)/8)`, since:

`sin((3pi)/16)=sqrt((1-cos((3pi)/8))/2)`

To find `cos((3pi)/8)`, we will use the cosine half-angle formula:

`cos(x/2)=+-sqrt((1+cos(x))/2)`

So, finding `cos((3pi)/8)` by taking the positive root since `(3pi)/8` is in the first quadrant:

`cos((3pi)/8)=sqrt((1+cos((3pi)/4))/2)=sqrt((1+(-sqrt2/2))/2)=sqrt(1/2-sqrt2/4)`

`=sqrt((2-sqrt2)/4)=sqrt(2-sqrt2)/2`

Now using this value to find `sin((3pi)/16)`:

`sin((3pi)/16)=sqrt((1-cos((3pi)/8))/2)=sqrt((1-sqrt(2-sqrt2)/2)/2)=sqrt(1/2-sqrt(2-sqrt2)/4)`

`=sqrt((2-sqrt(2-sqrt2))/4)=sqrt(2-sqrt(2-sqrt2))/2`

We can now substitute this into `r=2/sin((3pi)/16):`

`r=2/(sqrt(2-sqrt(2-sqrt2))/2)=4/sqrt(2-sqrt(2-sqrt2))`

To find the circle's area, use `A=pir^2`:

`A=pi(4/sqrt(2-sqrt(2-sqrt2)))^2=(16pi)/(2-sqrt(2-sqrt2))`