Area in Circle with two chords

[bfischer]

Hi Forum:

I need help with a problem that I'm helping my step-son with in Geometry (I'm teaching him this subject).

There's a problem that asks for the "area" (the shaded portion) on each side of two chords in a circle.

Here's the problem attached in a picture.

Any help is greatly appreciated... I've been stuck for two days now trying to figure this out!

Thanks,



Brett

 

[Deleted member 4993]

Hi Forum:

I need help with a problem that I'm helping my step-son with in Geometry (I'm teaching him this subject).

There's a problem that asks for the "area" (the shaded portion) on each side of two chords in a circle.

Here's the problem attached in a picture.

Any help is greatly appreciated... I've been stuck for two days now trying to figure this out!

Thanks,



Brett

What are your thoughts?

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/announcement.php?f=33

 

[bfischer]

Area

Hi:

I basically have the area of the circle (pie x r-square)... and divided by 2 to find the one side.

I'm stuck on the other more complicated side. Since the chord is to the inside, I can't use a portion of the circle equation... and it's offset so the radius curves and it's not a regular square.

I'm sure it's solved with triangles or something like that... not a sure what angles to use (30, 60, 90) and how to solve the area that is approximated by the radius.

There's about 10 different problems in his assignment (all the other problems I've figured-out), and this seems to be the only one I can't find any information on the internet which helps.

I could cheat and get the answer in seconds using my CAD program... but, I like the challenge!

Thanks,


Brett

 

[stapel]

I basically have the area of the circle (pie x r-square)... and divided by 2 to find the one side.

I will guess that the variable "x" in the above is meant to indicate multiplication, and that the product, "$\displaystyle \pi\, e$", is meant to be just the first factor, $\displaystyle \pi.$ So you actually have:

. . . . .$\displaystyle A_{circle}\, =\, \pi\, r^2$

When you say that you "divided by 2 to find the one side", I will guess that you mean that you divided the expression for the area A by 2, in order to find the value of the area of the lower semicircular portion.

If any of this is not correct, please reply with clarification.

By the way, is the "12" the value of the diameter, all the way across, or does that value indicate the length from the left-hand side of the circle to the "3" vertical line?

I'm stuck on the other more complicated side. Since the chord is to the inside, I can't use a portion of the circle equation... and it's offset so the radius curves and it's not a regular square.

I'm sorry, but I don't know what this means...? For instance, what do you mean when you say that the radius, defined to be a straight line, actually "curves"? What is the "it" that is not a "regular square"? What is a "regular square"? What is the "a portion of the circle equation"?

I'm sure it's solved with triangles or something like that... not a sure what angles to use (30, 60, 90) and how to solve the area that is approximated by the radius.

There's about 10 different problems in his assignment (all the other problems I've figured-out), and this seems to be the only one I can't find any information on the internet which helps.

Draw the line from the center of the circle to the midpoint of the chord. Draw the radius lines from the center of the circle to the endpoints of the chord. Label the vertical line as "3", the radius lines as "r", and the half-chord lines (being the halves of the chord, after being split by the vertical) as "c/2" (for "chord-length c, divided in half").

You should have something that looks like this:

triangle:

    c/2     c/2
*--------*--------*
  \      |      /
    \    |3   /r
    r \  |  /
        \|/
         *

Take either half of this figure, and flip right-side up, so you have a right triangle like this:

right triangle:

         *
        /|
    r /  |3
    /    |
  /      |
*--------*
    c/2

Where does this lead?

 

[bfischer]

Area

Hi:

I think the 12 is the diameter of the circle... from what I'm guessing.

Yes... I see that splitting an equilateral triangle makes for a 30, 60, 90 triangle. However, I'm not sure how to solve this problem to find the area in the smaller portion since the arc of the circle on the sides curves around.

On other problems, the chord runs along one-side of the circle and not in the middle. What I used to solve those problems was to take:
[FONT=STIXGeneral, Georgia]
?/360 x [/FONT]π

r
2






Next, I found the area of the equilateral triangle:

A
=

3





4
​

a
2







Then I subtracted the two to find the area of the chord.

Kind of confused how the triangle can be used to solve this problem... since there's arcs on both sides of the area to subtract from the portion of the semi-circle.

Thanks!


Brett

 

[mmm4444bot]

What I used to solve those problems was to take:

?/360 x π

r
2

Next, I found the area of the equilateral triangle:

A
=

3





4
​

a
2

If you need help typing math expressions as text, you can learn how at this site.

Also, please use the Preview Post button, to see how your typing will render, before submitting your posts. This way, you can fix issues like those above before you post. :cool:

 

[bfischer]

Area

These are the formulas I used to solve for sectors with chord on one side of circle. I solved for the area in an equilateral triangle and then subtracted it from the sector area of a circle... to get the remaining area.

However, on this problem... there's two arcs on each side of the area in the middle... not sure how a triangle will solve this issues?

That's what I need help with... how does the triangle solve for the arc on each side???

Thanks!


Brett

 

[mmm4444bot]

I solved for the area in an equilateral triangle and then subtracted it from the sector area of a circle... to get the remaining area.

You meant to say isosceles, not equilateral. Yes?

You do not need to consider the small chords at each "end" of the unshaded region. Instead, use the base of the upper-shaded region as a chord. You mentioned 30-60-90 triangles earlier, so I'm assuming you've already determined that the central angle for this chord is 120°.

Using stapel's suggestion, I obtained the chord's length (c) via the Pythagorean Theorem. (See formulas below, for another way.)

Subtracting the area of the isosceles triangle from the area of the sector yields the area of the segment (i.e., the upper-shaded region in your diagram). Add that to half the circle's area.

I get 78.6593 square units (rounded). Am I correct? :cool:

In the diagram below, S is the segment height, L is half the chord's length, and R is the circle's radius.

If you know any two of these three measurements, you can find the third.

S = R - √[R^2 - L^2]

L = √[2SR - S^2]

R = (S^2 + L^2) / (2S)

 

[bfischer]

Area

Hi:

Not sure how you got your answers... but this is where I'm at with this problem.

Thanks,


Brett

 

[mmm4444bot]

I cannot read your latest image. Please crop images, before posting. About two-thirds of your image is blank, white space. The system reduced the size of your image, so it could include the entire image.

 

[mmm4444bot]

Not sure how you got your [answer] …

Do you have any questions about what I wrote below?

Subtracting the area of the isosceles triangle from the area of the sector yields the area of the segment (i.e., the upper-shaded region in your diagram). Add that to half the circle's area.

 

[Deleted member 4993]

Hi:

Not sure how you got your answers... but this is where I'm at with this problem.

Thanks,


Brett

Flip the triangle around its base (within the circle) and look at the resulting figure....

 

[bfischer]

Area

Hi:

Here's the picture revised.

Thanks!


Brett

 

[bfischer]

Got it!

Hello:

Thank you!

Finally, I had the "Aah-Ha" moment.

For beginners... I think math authors think we know what certain angles are going to be of shapes (like the triangle) when they give us dimensions of objects "not to scale".

In all, I'm very appreciated of your help.

How easy was this!!!

Bye,


Brett