Geometry question w/ a Circle. find area of a shaded region. (Harder problem)

- anonymous

Geometry question w/ a Circle. find area of a shaded region. (Harder problem)

- jamiebookeater

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

|dw:1345779977023:dw|

- anonymous

I think what you have to do is to bisect the triangle into 60 degree up top and then the bottom left and right are still 30 degrees. then you do the 30 60 90 triangle thing.

- anonymous

i am stuck from there

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- alexwee123

i did it a different way...

- anonymous

well explain.

- alexwee123

with integrals...

- anonymous

well. it shouldn't be that complicated im only doing geometry.

- Mimi_x3

Area of minor segment \[A = \frac{1}{2} r^2 \left(\theta-\sin\theta\right) \]
But the only problem is \(r\) is not given..

- Hero

I think I meant to post the area of a sector of a circle

- anonymous

you guys are thinking too hard its only like subtracting a sector

- Mimi_x3

Area of sector - area of triangle

- Hero

Yeah that

- Mimi_x3

But \(r\) is not given!!!
Im thinking of using the cosine rule and sine rule to find \(r\) would it work?

- anonymous

here ill take it a bit further.
|dw:1345780578744:dw|
does that help?

- Hero

It will not be hard to find \(r\)

- Hero

Is the vertex of angle 120 is the center of the circle?

- anonymous

yep

- Mimi_x3

|dw:1345780720301:dw|
looks like you can use the sine rule..

- Hero

Okay, so yeah just do
Area of Sector - Area of Triangle like what @Mimi_x3 suggested

- anonymous

soo all we need is the area of the circle and then we can use the 120/360 * pi r2 or something and subtract the area from that.

- Mimi_x3

lets find \(r\) first

- anonymous

|dw:1345780842471:dw|

- Mimi_x3

I got 8 as well using the sine rule

- Mimi_x3

Now its easy..you can do it yourself

- anonymous

ok so 64 pi is the area of the circle

- anonymous

and the area of the triangle left side is 1/2 * 4sqroot3 * 4

- anonymous

*2 to make up for the other side.

- Mimi_x3

I will just use this formula\[A = \frac{1}{2} r^2 \left(\theta-\sin\theta\right) \]
No need to derive it

- anonymous

ok

- Mimi_x3

and it must be in radians..so convert it..

- anonymous

oh heck nah jose

- alexwee123

you guys are thinking too hard o.0

- anonymous

oh wait i got the answer. just do 120/360 which is 1/3
and then i did 1/3 * 64pi which is the area of the circle and it is
64pi/3 and now subtract it by the area of the triangle which is

- anonymous

1/2 4sqrt3 * 4 or
8sqrt3 * 2 for the other side

- alexwee123

|dw:1345781211630:dw|
|dw:1345781273195:dw|

- alexwee123

|dw:1345781325118:dw|

- anonymous

so the final answer is \[64\pi/3 - 16\sqrt{3}\]

- alexwee123

yes good job @Lethal

- anonymous

yep, i would best response both of u but i can only do one. thx alex.

Looking for something else?

Not the answer you are looking for? Search for more explanations.