Question

Express answer in exact form.
Find the area of the larger segment whose chord is 8" long in a circle with an 8" radius. (Hint: A chord divides a circle into two segments. In problem 1, you found the area of the smaller segment.)

Answer 1

So, first off, you draw two radius to the end of your 8" chord

Now, we have a piece of our circle and an equilateral triangle, so
for the whole circle you have
\[A=\pi*R^2\]
That means that for part of your circle you have
\[A=\pi*R^2*\frac{ \angle "CO" }{ 360 }\]
where angle "CO" is the angle between the radius we have just draw, as they form an equilateral triangle, it is equal to...

Answer 2

\[A=\sqrt{?}\times \pi \times radius \sqrt\]

Answer 3

can you put the answer in this form?

Answer 4

\[Total Area=\frac{ 60 }{ 360 } *\pi*R^2+ triangl||area\], i won't put the answer, the idea is that you solve it

Answer 5

yes i understand the formula and i got \[A=\frac{ ? }{ ? } \pi + 16\sqrt{3}\] i just cannot determine how to get the first part of the answer @Umangiasd

Answer 6

oh, okay, you're searching for this angle

why? because of center angle theorem it is equal to the measure of the correspondant arc

Now, (I got a thing wrong up there, you should calculate this area

So you've got to take a part of your circle equivalent to (360-60/360)
why is that? because you divide your circle in 360 parts and take the 300 (360-60) parts that we should take (according to the first diagram of this post)
You got it now?