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## Homework Statement

I know that the radius of the circle is 7 and the angle of the segment is 150°

## Homework Equations

Area of a circle: [tex]A = \pi{r}^2[/tex]

Area of the sector of the circle: [tex]A = \frac{n}{360}\pi r^{2}[/tex]

Area of a triangle: [tex]A = \frac{n}{360}\pi r^{2}[/tex]

## The Attempt at a Solution

I know to get the final solution to get the area of that particular part of the circle, I need to subtract the whole area of the circle with the segment of the circle (the triangular portion).

To get the area of the triangle portion, I subtract the area of the sector of the circle with the area of the triangle of that portion multiplied by 2 (since there are essentially two triangles). Also, to get the area of the triangular portion, I can use the 30-60-90 rule. To get this:

Short leg:

[tex]A_s = \frac{1}{2}H = \frac{1}{2}*7 = \frac{7}{2}[/tex]

Long leg:

[tex]A_l=\frac{1}{2}H\sqrt{3} = \frac{7}{2}\sqrt{3}[/tex]

Area of whole triangle:

[tex]A_t=\frac{1}{2}*\frac{7}{2}*\frac{7}{2}\sqrt{3}*2=\frac{49}{4}\sqrt{3}[/tex]

Now the area of the whole sector is:

[tex]A_t = \frac{150}{360}49\pi-\frac{49}{4}\sqrt{3}[/tex]

To get the area of just that portion of the circle the question is asking, subtract the area of the whole circle with the answer from above:

[tex]A = 49\pi-(\frac{150}{360}49\pi-\frac{49}{4}\sqrt{3})[/tex]

I was wondering if I am doing this problem correctly? I am not getting any of the answers in the possible solutions. Thanks!