MHB Finding the Area of a shaded region (two shapes)

[Coder74]

Hello,
I've done something similar to this before but this question is really different because it contains two shapes. Now I'm really confused and I really appreciate the help~! View attachment 6004

-Cheers

 

[MarkFL]

I would take the area of the circular sector $A_S$ and subtract from that the area of the isosceles triangle $A_T$ to get the shaded area $A$:

$$A=A_S-A_T$$

where:

$$A_S=\frac{1}{2}r^2\theta$$ where $\theta$ is in radians.

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

Can you proceed?

 

[Coder74]

Thanks for the reply, Mark I really appreciate it!
However, I'm unfamiliar with " \theta " I haven't seen that before.

 

[MarkFL]

Thanks for the reply, Mark I really appreciate it!
However, I'm unfamiliar with " \theta " I haven't seen that before.

It is a Greek letter usually used to represent angles. In this problem, we have:

$$\theta=150^{\circ}=\frac{5}{6}\pi$$

 

[Coder74]

Thanks again, Mark!

Triangle AT=193.21
Sphere AS=1,011.64
Shaded Area=818.43

This is what my final answers came up to be.

 

[MarkFL]

I get:

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

Now plug in the given values for $r$ and $\theta$:

$$A=\frac{1}{2}(27.8\text{ in})^2\left(\frac{5}{6}\pi-\frac{1}{2}\right)=\frac{1}{12}(27.8\text{ in})^2\left(5\pi-3\right)\approx818.4\text{ in}^2\quad\checkmark$$