Finding area

[DavidE]

I've been working on this problem for several days and cannot come up with a solution.

Find the area of the shaded (light green) area as a function of r (radius of the circle) and h (distance to C from the circumference of the circle). h is the extension of a radius. Lines AC and BC are tangent to the circle.

 

[Deleted member 4993]

I've been working on this problem for several days and cannot come up with a solution.

Find the area of the shaded (light green) area as a function of r (radius of the circle) and h (distance to C from the circumference of the circle). h is the extension of a radius. Lines AC and BC are tangent to the circle.


View attachment 15595

What are your thoughts?
Please share your work with us ...even if you know it is wrong.
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Hint:

Let AOC = \(\displaystyle \theta\)
cos(\(\displaystyle \theta) = \frac{r}{r+h}\)
AC = r * tan(\(\displaystyle \theta\))
The "kite area" = 2 * r * AC
The "pie slice" area = \(\displaystyle \theta * r^2\)

Continue.....

 

[DavidE]

Well, I just came up with a bunch of things that I know are true but I cannot seem to piece them together to create an area formula as a function of r and h.

I know that OAB and OAC are right triangles since a radius is always 90-degrees to a tangent.

I thought I could calculate the area of these two triangles then subtract out the sector of the circle within this area. That would leave the shaded area that I seek.

The area of triangle OAC can be found using 1/2(base)(height)= 1/2 (r)(AC)
We need to find AC expressed as function of r & h. That can be done using the Pythagorean Theorem.
For triangle OAC: (r+h)^2 =r^2 + AC^2. Therefore: AC = sqrt((r+h)^2 - r^2)
This leads to the area of triangle OAC = 1/2(r)sqrt((r+h)^2 - r^2)

I can multiple this by 2. This would give me the area of quadrilateral OACB as only a function of r & h.

Now, I need to figure out how to subtract out the OAB sector of the circle. The area of this sector would be r^2*(angle AOB in radians)

This is where I have problems, trying to figure how to express the "angle of AOB in radians" as a function of r & h.

So far, I have: The shaded area = (the area of quadrilateral OACB) minus (the area of the sector OAB.
This gives: [r*sqrt((r+h)^2-r^2)] minus r^2(the area of the sector OAB)

Close. But stuck on how to express "the area of the sector OAB" as a function of r + h. It seems it would involve finding an expression for the angle AOB.

Stuck.

 

[DavidE]

The area sought is "kite area" minus "pie slice".

You helped me determine that the "kite area" equals 2*r*AC.

I need to find AC in terms of r and h.

AC = r*tan(AOC)=r*[sin(AOC)/cos(AOC)]

I can express the numerator sin(AOC) in terms of cos(AOC) which I already know equals r/(r+h)
Now I have AC in terms of r and h. That will give me the area of "kite area" in terms of r & h.

Now I have to subtract out the "pie slice" which is equal to angleAOC time r.

The only way I can think of expressing angleAOC is arccos(r/r+h). That should do it, right?

 

[DavidE]

You say the "kite area" is 2*r*AC. But wouldn't it be simply r*AC?

The reason I say that is that the triangle AOC is 1/2(base)(height) = 1/2(r)(AC)

Yet, the area of triangle AOC equals the area of triangle BOC. The combined area of these two triangles is the area of the "kite area".

The area of triangle BOC is 1/2(base)(height) = 1/2(r)(BC)
But BC equals AC which means the area of triangle BOC could be expressed exactly as the area of triangle AOC is expressed.
When you add them together you get r*AC, not 2*r*AC. Right?

 

[DavidE]

OK, using your hints, I think I found the answer. I came up with a formula for the "kite area" and the "pie slice" only in terms of r & h. The hard part was trying to eliminate the angle AOC (theta) and only express it in terms of r & h.

Here's what I got:

 

[pka]

I've been working on this problem for several days and cannot come up with a solution.

Find the area of the shaded (light green) area as a function of r (radius of the circle) and h (distance to C from the circumference of the circle). h is the extension of a radius. Lines AC and BC are tangent to the circle.


View attachment 15595

First a comment on your use on geometric terminology. A circle is a set of points. Circumference in a number, not a set so h cannot be distance from C to the circumference. However, h is distance of C to the circle.
\(\displaystyle \Delta OAC\cong\Delta OBC\) can you justify that?
Thus the \(\displaystyle \text{Area}(\Delta AOC)=\text{Area}(\Delta BOC)=\frac{r\cdot AC}{2}\). So the area of the Kite is \(\displaystyle 2\text{Area}(\Delta AOC)\)
Here is a valuable reference. If the length of arc \(\displaystyle AB=L\) the the area of the sector is \(\displaystyle A=\dfrac{r\cdot L}{2}.\)
So the green shaded area is \(\displaystyle 2\cdot\frac{r\cdot AC}{2}-\dfrac{r\cdot L}{2}\).

 

[DavidE]

First a comment on your use on geometric terminology. A circle is a set of points. Circumference in a number, not a set so h cannot be distance from C to the circumference. However, h is distance of C to the circle.
\(\displaystyle \Delta OAC\cong\Delta OBC\) can you justify that?
Thus the \(\displaystyle \text{Area}(\Delta AOC)=\text{Area}(\Delta BOC)=\frac{r\cdot AC}{2}\). So the area of the Kite is \(\displaystyle 2\text{Area}(\Delta AOC)\)
Here is a valuable reference. If the length of arc \(\displaystyle AB=L\) the the area of the sector is \(\displaystyle A=\dfrac{r\cdot L}{2}.\)
So the green shaded area is \(\displaystyle 2\cdot\frac{r\cdot AC}{2}-\dfrac{r\cdot L}{2}\).

Thanks, all I have to do is convert your L and AC to expressions that only include r and h. I think I may be able to do that.

Also, I think I made a mistake in my earlier calculation.

In any case, you've been very helpful. I think the key is realizing that the angle AOC is definitely part of the solution. As the value of r and h change, that angle is going to change. An inverse trigonometric function simply must be part of the solution if the restriction is to only use r and h as variables.

 

[pka]

Thanks, all I have to do is convert your L and AC to expressions that only include r and h. I think I may be able to do that.
Also, I think I made a mistake in my earlier calculation. In any case, you've been very helpful. I think the key is realizing that the angle AOC is definitely part of the solution. As the value of r and h change, that angle is going to change. An inverse trigonometric function simply must be part of the solution if the restriction is to only use r and h as variables.

Go to that provided link. Anyone working with circles must know the formulae on that page.

 

[DavidE]

Here's what I've finally come up with, using only r and h.