MHB Proving area theorem within semi-circle

[Milly]

The question is attached in the attachment below.

View attachment 3332

 

[I like Serena]

The question is attached in the attachment below.

Hi Milly! Welcome to MHB! :)

The area of a full circle is $\pi^{} r^2$.
If we take a segment out of it with angle $\theta$, its area is:
$$\text{Area of segment with angle }\theta = \pi^{} r^2 \cdot \frac \theta {2\pi} = \frac 1 2 r^2 \theta^{}$$

Can you find the area of the segment $POC$ using this formula?
And how about the area of the shaded area between $P$ and $A$?

 

[MarkFL]

The area $A_1$ of the sector $POB$ is:

$$A_1=\frac{1}{2}r^2\theta$$

Now, the shaded area $A_2$ will be the area of sector $POA$ minus the area of triangle $POA$.

Can you use the formula for the area of a sector and the area of a triangle (where 2 sides are known along with the angle subtended by those two sides) to give an expression for $A_2$?

 

[Milly]

But the angel POA is pi - theta which i have to then convert into degree to find area of triangle.

 

[MarkFL]

But the angel POA is pi - theta which i have to then convert into degree to find area of triangle.

The area of the triangle is:

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

Can you think of an identity for the sine function to simplify this?

 

[Milly]

I thought for sin (pi - theta) the angle should be in degrees but the angle given is in radians.

 

[MarkFL]

I thought for sin (pi - theta) the angle should be in degrees but the angle given is in radians.

If the angle $\theta$ is in radians, then we use for the supplementary angle $$\pi-\theta$$.

 

[Milly]

Ya but I don't know how to separate the ($/pi$−θ) in sin function

 

[MarkFL]

Ya but I don't know how to separate the ($/pi$−θ) in sin function

You could use the angle-difference identity, or simply use the identity:

$$\sin(\pi-\theta)=\sin(\theta)$$

Think of the unit circle...we have one point at (1,0) and one point at (-1,0). Now these two points begin moving along the top half of the circle at the same speed, towards one another...do you see that their $y$-coordinates will always be the same?