Question

Express answer in exact form. Given a circle with an 8" radius, find the area of the smaller segment whose chord is 8" long

Answer 1

Answer:

see 1st picture below

Explanation:

to solve for a segment you you subtract the area of the sector by the area of the triangle. First find the area of the sector. The equation would look like this:

$\frac{60}{360} \pi (8)^2$  then this gets simplified into this: $(\frac{1}{6}) \pi (64)$ which equals: $\frac{32}{3} \pi$

Now that you have found the area of the sector, you have to find the area of the triangle. The triangle is an equilateral triangle with an 8" side. This means that the angles in the triangle are equal to 60 degrees. In order to find the area of the triangle you need to know the base and height of the triangle.

The base is clearly 8 but the height still needs to be determined. If you split the triangle in half you get a right  triangle with angles equivalent to 30, 60, and 90. In a 30-60-90 triangle, the long leg $\sqrt{3}$ multiplied by the length of the short leg. In this case, the short leg is half of the base of the whole triangle. So the height is $4\sqrt{3}$. The equation to find the area of this triangle looks like this:

$\frac{1}{2}(8)(4\sqrt{3})$ This equals: $16\sqrt{3}$

Now that you know the area of the sector and the area of the triangle all you need to do is subtract and you get:

$\frac{32}{3}\pi-16\sqrt{3}$

Sorry for the bad camera quality on the second picture.

Answer 2

Anyway... check the picture below

the circle has a radius of 8, the "red" chord has a length of 8, the "blue" segment made by chord is there then

now.... if the radius and the chord are all 8, that's makes an equilateral triangle, "inside the sector of the circle", bear in mind that, for an equilateral triangle, all angles are equal as well, and thus all are 60°.

now, if you just take the area of the "sector" of the circle, with that radius and that angle, and then subtract the equilateral triangle, what's leftover is the area of the segment made by the chord.

the are of the sector of the circle, includes the segment and the triangle, so subtracting the triangle from it, leaves only the area of the segment.