What is the area of a segment bounded by a chord and an arc in a circle?

[Heathcoat]

This one is kicking my butt. In a circle, find the exact area of a segment bounded by chord $$\overline{CD}$$ and $$\stackrel{\frown}{CD}$$. $$\overline{CD}=12\sqrt{3}$$ and the $$\measuredangle \stackrel{\frown}{CD}=120^{\circ}$$.

 

[Evgeny.Makarov]

Hi, and welcome to the forum.

Using the law of cosines, find the radius $r$. Then find the area of the $120^\circ$ sector. Finally, find the area of the triangle: $\frac12r^2\sin120^\circ$ and subtract it from the area of the sector.

Not sure why the symbols didn't work

You should enclose LaTeX code in dollar signs or $$...$$ tags (the $\Sigma$ button in the toolbar above the text box where you type your post).

 

[MarkFL]

Hello and welcome to MHB, Heathcoat! :D

I have enclosed your $\LaTeX$ codes in MATH tags, and added a few symbols.

Here is a sketch of the problem:

View attachment 3564

Can you use the above suggestions to find the green area?

 

[Heathcoat]

Hello and welcome to MHB, Heathcoat! :D

I have enclosed your $\LaTeX$ codes in MATH tags, and added a few symbols.

Here is a sketch of the problem:

View attachment 3564

Can you use the above suggestions to find the green area?

I am getting the radius as 6rad6. using this I can find the area of the circle, and the segment is 1/3rd the area of the circle. I then need to subtract the area of the triangle from the area of the segment. I am struggling with finding the area of the triangle. The formula I know is A=1/2bh. My base is 12rad3 but I don't remember how to find the height of the triangle to use this formula. Cosine and sin is well above what I have learned so far (basic geometry level, no calculus)

 

[Evgeny.Makarov]

I am getting the radius as 6rad6.

Please write \$6\sqrt{6}\$ instead of 6rad6. It produces $6\sqrt{6}$.

If you draw an altitude $OH$ from the center $O$ of the circle to $CD$, then you get a right triangle $OCH$ with angles $90^\circ$, $60^\circ$ and $30^\circ$. You know the long leg: $6\sqrt{3}$. and you probably know that the short leg, which faces $30^\circ$, is half the hypotenuse, or the radius $r$. Using the Pythagoras theorem, you should be able to find $r$. I get $r=12$. Knowing $OH=r/2$, you can find the area of $\triangle OCD$.

 

[Heathcoat]

Please write \$6\sqrt{6}\$ instead of 6rad6. It produces $6\sqrt{6}$.

If you draw an altitude $OH$ from the center $O$ of the circle to $CD$, then you get a right triangle $OCH$ with angles $90^\circ$, $60^\circ$ and $30^\circ$. You know the long leg: $6\sqrt{3}$. and you probably know that the short leg, which faces $30^\circ$, is half the hypotenuse, or the radius $r$. Using the Pythagoras theorem, you should be able to find $r$. I get $r=12$. Knowing $OH=r/2$, you can find the area of $\triangle OCD$.

Honestly, I am having enough trouble figuring out the geometry without also trying to figure out how to use the proper html or coding process for this site. My brain is thoroughly taxed trying to figure out this problem without trying to tack on new things to learn. You knew what I meant by 6rad6, right? It seems to me that both legs from the triangle will be radii, so how can this be a 30,60,90? Perhaps this site isn't equipped to deal with people at lower levels of understanding of math, because your answer makes me more confused than when I started.

 

[Evgeny.Makarov]

It seems to me that both legs from the triangle will be radii

Which triangle are you talking about? I was describing $\triangle OCH$ where $O$ is the center of the circle and $OH$ is altitude in $\triangle OCD$. That is, $H$ lies on $CD$ and $OH$ is perpendicular to $CD$. Again, the triangle with $90^\circ$, $60^\circ$ and $30^\circ$ is $\triangle OCH$.

 

[Heathcoat]

Which triangle are you talking about? I was describing $\triangle OCH$ where $O$ is the center of the circle and $OH$ is altitude in $\triangle OCD$. That is, $H$ lies on $CD$ and $OH$ is perpendicular to $CD$. Again, the triangle with $90^\circ$, $60^\circ$ and $30^\circ$ is $\triangle OCH$.

Ok, sorry for my confusion. The triangle I was referring to is triangle OCD, which needs to be subtracted from the segment of the circle OCD to get the answer I am looking for. Thanks for your help, I am just not getting this, it's all way over my head.

 

[MarkFL]

But, $\triangle OCH$ is just the bisection of $\triangle OCD$, that is the area of the smaller is half the area of the larger. :D