MHB Find the areas of segment in circle

[Etrujillo]

So far i have.
14) area of sector is πr²/3 = 12π
length of chord. that triangle has two sides of 6 and angle of 120º
split the triangle in two right triangles with angle of 120/2 = 60 and hyp=6. other (longer) side is:
sin 60 = x/6
s = 6 sin 60 = 6(√3/2) = 3√3
third side is
s = 3 cos 60 = 3
area =(1/2)(3)(3√3) = 4.5√3, double for both triangles
subtract that from the sector to get (12π) – (9√3)

15) similar to above.
find the area of the 270º sector and add the area of the triangle
area of sector is (270/360)(π9²) or (3/4)81π
area of triangle is (1/2)81

Is this correct?
What can i do differently?

View attachment 8705

 

[MarkFL]

Re: Find the area of the shared region.

14.) I would take the area of the circular sector, and subtract from it the area of the triangle to get the shaded area \(A\):

$$A=\frac{1}{2}r^2\theta-\frac{1}{2}r^2\sin(\theta)=\frac{r^2}{2}(\theta-\sin(\theta))$$

Next, we identify:

$$r=6\text{ cm}$$

$$\theta=\frac{2\pi}{3}$$

Hence:

$$A=\frac{(6\text{ cm})^2}{2}\left(\frac{2\pi}{3}-\sin\left(\frac{2\pi}{3}\right)\right)=3\left(4\pi-3\sqrt{3}\right)\text{ cm}^2\quad\checkmark$$

This is equivalent to the area you stated. :D

15.) I would find the sum of 3/4 of the area of the circle and the right isosceles triangle:

$$A=\frac{3}{4}\pi r^2+\frac{1}{2}r^2=\frac{r^2}{4}(3\pi+2)$$

We identify:

$$r=9\text{ in}$$

And so:

$$A=\frac{(9\text{ in})^2}{4}(3\pi+2)=\frac{81}{4}(3\pi+2)\text{ in}^2\quad\checkmark$$

This is equivalent to what you would get when you add the two areas you found.