MHB What is the Area of the Shaded Region in Trigonometry Problem?

[Alaba27]

Hi again,

I'm completing my tutoring booklet for trigonometry and I have a question that is quite difficult. I've never done anything like it and my solutions page only has the final answer. I'd like to see how the result is found step-by-step, so if anyone can help me out or at least guide me to the solution it would be much appreciated.

http://speedcap.net/sharing/files/57/af/57afa49f650622a931c668f649477a83.png
http://speedcap.net/sharing/files/6a/e5/6ae5a228d615fb949ba020cd1a1c725d.png

If you cannot read the question, it is:

sin(x) = (√3)/2 in the circle with centre O, shown on the right. Furthermore AB // OP and m OA = 1 unit. What is the area of the shaded region?

Thanks in advance!

 

[MarkFL]

I would suggest finding the area of the circular sector $$OAB$$ (you know or can find the radius and the subtended angle), then subtract away the area of $$\Delta OAB$$.

I notice you have $$\overline{OA}$$ marked as 2, but it should be 1, making $$\overline{AB}=\sqrt{3}$$.

 

[Alaba27]

I would suggest finding the area of the circular sector $$OAB$$ (you know or can find the radius and the subtended angle), then subtract away the area of $$\Delta OAB$$.

I notice you have $$\overline{OA}$$ marked as 2, but it should be 1, making $$\overline{AB}=\sqrt{3}$$.

Thanks for the reply! So would OA be identical to OB, then?

 

[Alaba27]

Alright, I managed to answer my own question. OA and OB are identical. But how can I find the radius of the circle? I guess that is basically what the entire issue is, seeing as how I cannot get the radius from the diagram (I think).

Thanks in advance!

 

[MarkFL]

The radius is $$1=\overline{OA}=\overline{OB}$$. Any line segment from the center to a point on the circle is a radius.

 

[Alaba27]

The radius is $$1=\overline{OA}=\overline{OB}$$. Any line segment from the center to a point on the circle is a radius.

OK, then. But the height of the triangle AOB is still 1/2, am I correct? I have gotten this:

Area of Circle
A = (pi)(r)^2
= (pi)(1)^2
= (pi) units

Area of Triangle AOB
A = (BxH)/2
= (√3 x 0.5)/2
= (√3)/4 units

So I would subtract the Area of the Triangle AOB from the Area of the Circle, and I would get the shaded area?

Thanks in advance!

 

[MarkFL]

You want to subtract the area of the triangle (which you have correctly computed) from the area of the circular sector, not the entire circle. Do you see why?

 

[Alaba27]

You want to subtract the area of the triangle (which you have correctly computed) from the area of the circular sector, not the entire circle. Do you see why?

My apologies, I've never done a question like this before and I don't see anything from my notes or previous questions that resembles something along these lines. So I am going to guess that I have to do this:

> Divide area of entire circle by 2
> Divide segments by 2 both times (because the top right and bottom right segments are broken in half)
> Add together

So, something like this:

Area of entire circle = π units
(The circle has 4 equal segments)
Area of 1 segment = π/4
(2 segments of the same size, so multiply by 2)
Area of both full segments = 0.5π

(For segments that are cut in half, divide π/4 by 2)
Area of half-segment = 1/8π
(Multiply the half-segment by 2 because there are 2 of them)
Area of both half-segments = 0.25π

Add them together = 0.75π

Is this correct?

 

[MarkFL]

Using your method, you could get the correct result, but the picture is misleading you with regards to the angle. The angle is actually larger than it is drawn.

The formula for the area $A_S$ of a circular sector is given by:

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

Now, we already know $r=1$, and we know $$\theta=2x$$, and we know:

$$\sin(x)=\frac{\sqrt{3}}{2}$$

So, can you find $x$, then substitute for $r$ and $\theta$ in the above formula for the area of the sector?

 

[Alaba27]

Using your method, you could get the correct result, but the picture is misleading you with regards to the angle. The angle is actually larger than it is drawn.

The formula for the area $A_S$ of a circular sector is given by:

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

Now, we already know $r=1$, and we know $$\theta=2x$$, and we know:

$$\sin(x)=\frac{\sqrt{3}}{2}$$

So, can you find $x$, then substitute for $r$ and $\theta$ in the above formula for the area of the sector?

OK, so I followed what you said and got this:

A = (0.5)(r)^2(θ)
= (0.5)(1)^2([√3]/2 x 2)
= (√3)/2 units

Is this correct?

 

[MarkFL]

No, you are using $$\theta=2\sin(x)$$, but what we need is $$\theta=2x$$

If $$\sin(x)=\frac{\sqrt{3}}{2}$$ and $$0<x<\frac{\pi}{2}$$ then what is $x$?