Section of a circle












0












$begingroup$


I need some help with some of my homework, I can't figure it out.



I have a radius on a circle and a height from the circle to the chord.



I found this formula
$$
h=r left(1-cos frac{v}{2} right)
$$
And isolated it to
$$
v = arccos left( frac{h/r -1}{2}right)
$$
Not sure if that's correct. Then I input it in this formula
$$
A=r^2((pi v)/360-(sin v)/2)
$$
I have tried with a radius of 1000 and a height from the circle to the chord of 1000.



But it gives a wrong result.










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$endgroup$












  • $begingroup$
    What is given, what do you search for? So given: radius r, and distance to (some) chord h. What is $v$?
    $endgroup$
    – Simon Markett
    Jun 28 '12 at 11:57










  • $begingroup$
    I want to find the areal, of the circle section. R=1000, Distinace to chord=1000. Then i calculate v in the first formula.
    $endgroup$
    – NikolajSvendsen
    Jun 28 '12 at 12:07










  • $begingroup$
    When the radius and distance to the chord is exactly the same, it should be the half of the circle. But it gives a wrong result, i am not sure whetever i mixup radians and degress, or i isolated it wrong.
    $endgroup$
    – NikolajSvendsen
    Jun 28 '12 at 12:29
















0












$begingroup$


I need some help with some of my homework, I can't figure it out.



I have a radius on a circle and a height from the circle to the chord.



I found this formula
$$
h=r left(1-cos frac{v}{2} right)
$$
And isolated it to
$$
v = arccos left( frac{h/r -1}{2}right)
$$
Not sure if that's correct. Then I input it in this formula
$$
A=r^2((pi v)/360-(sin v)/2)
$$
I have tried with a radius of 1000 and a height from the circle to the chord of 1000.



But it gives a wrong result.










share|cite|improve this question











$endgroup$












  • $begingroup$
    What is given, what do you search for? So given: radius r, and distance to (some) chord h. What is $v$?
    $endgroup$
    – Simon Markett
    Jun 28 '12 at 11:57










  • $begingroup$
    I want to find the areal, of the circle section. R=1000, Distinace to chord=1000. Then i calculate v in the first formula.
    $endgroup$
    – NikolajSvendsen
    Jun 28 '12 at 12:07










  • $begingroup$
    When the radius and distance to the chord is exactly the same, it should be the half of the circle. But it gives a wrong result, i am not sure whetever i mixup radians and degress, or i isolated it wrong.
    $endgroup$
    – NikolajSvendsen
    Jun 28 '12 at 12:29














0












0








0





$begingroup$


I need some help with some of my homework, I can't figure it out.



I have a radius on a circle and a height from the circle to the chord.



I found this formula
$$
h=r left(1-cos frac{v}{2} right)
$$
And isolated it to
$$
v = arccos left( frac{h/r -1}{2}right)
$$
Not sure if that's correct. Then I input it in this formula
$$
A=r^2((pi v)/360-(sin v)/2)
$$
I have tried with a radius of 1000 and a height from the circle to the chord of 1000.



But it gives a wrong result.










share|cite|improve this question











$endgroup$




I need some help with some of my homework, I can't figure it out.



I have a radius on a circle and a height from the circle to the chord.



I found this formula
$$
h=r left(1-cos frac{v}{2} right)
$$
And isolated it to
$$
v = arccos left( frac{h/r -1}{2}right)
$$
Not sure if that's correct. Then I input it in this formula
$$
A=r^2((pi v)/360-(sin v)/2)
$$
I have tried with a radius of 1000 and a height from the circle to the chord of 1000.



But it gives a wrong result.







geometry algebra-precalculus






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jun 28 '12 at 19:52









Arturo Magidin

266k34590920




266k34590920










asked Jun 28 '12 at 11:35









NikolajSvendsenNikolajSvendsen

32




32












  • $begingroup$
    What is given, what do you search for? So given: radius r, and distance to (some) chord h. What is $v$?
    $endgroup$
    – Simon Markett
    Jun 28 '12 at 11:57










  • $begingroup$
    I want to find the areal, of the circle section. R=1000, Distinace to chord=1000. Then i calculate v in the first formula.
    $endgroup$
    – NikolajSvendsen
    Jun 28 '12 at 12:07










  • $begingroup$
    When the radius and distance to the chord is exactly the same, it should be the half of the circle. But it gives a wrong result, i am not sure whetever i mixup radians and degress, or i isolated it wrong.
    $endgroup$
    – NikolajSvendsen
    Jun 28 '12 at 12:29


















  • $begingroup$
    What is given, what do you search for? So given: radius r, and distance to (some) chord h. What is $v$?
    $endgroup$
    – Simon Markett
    Jun 28 '12 at 11:57










  • $begingroup$
    I want to find the areal, of the circle section. R=1000, Distinace to chord=1000. Then i calculate v in the first formula.
    $endgroup$
    – NikolajSvendsen
    Jun 28 '12 at 12:07










  • $begingroup$
    When the radius and distance to the chord is exactly the same, it should be the half of the circle. But it gives a wrong result, i am not sure whetever i mixup radians and degress, or i isolated it wrong.
    $endgroup$
    – NikolajSvendsen
    Jun 28 '12 at 12:29
















$begingroup$
What is given, what do you search for? So given: radius r, and distance to (some) chord h. What is $v$?
$endgroup$
– Simon Markett
Jun 28 '12 at 11:57




$begingroup$
What is given, what do you search for? So given: radius r, and distance to (some) chord h. What is $v$?
$endgroup$
– Simon Markett
Jun 28 '12 at 11:57












$begingroup$
I want to find the areal, of the circle section. R=1000, Distinace to chord=1000. Then i calculate v in the first formula.
$endgroup$
– NikolajSvendsen
Jun 28 '12 at 12:07




$begingroup$
I want to find the areal, of the circle section. R=1000, Distinace to chord=1000. Then i calculate v in the first formula.
$endgroup$
– NikolajSvendsen
Jun 28 '12 at 12:07












$begingroup$
When the radius and distance to the chord is exactly the same, it should be the half of the circle. But it gives a wrong result, i am not sure whetever i mixup radians and degress, or i isolated it wrong.
$endgroup$
– NikolajSvendsen
Jun 28 '12 at 12:29




$begingroup$
When the radius and distance to the chord is exactly the same, it should be the half of the circle. But it gives a wrong result, i am not sure whetever i mixup radians and degress, or i isolated it wrong.
$endgroup$
– NikolajSvendsen
Jun 28 '12 at 12:29










2 Answers
2






active

oldest

votes


















0












$begingroup$

The problem is that your solution is incorrect. You have the 2 factor in the wrong place. the correct solution is:



                    v=2arccos((1-h/r)





share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks that works. But can you explain me how v, can be 180 degress. Isn't it a triangle like on this image matb1htx.systime.dk/fileadmin/indhold/ISBNXXXXXXXXXXXXX/…
    $endgroup$
    – NikolajSvendsen
    Jun 28 '12 at 12:59












  • $begingroup$
    In your example, the height to the chord is equal to the radius. Therefore the chord becomes a diameter and the angle subtended by a diameter is 180 degrees.
    $endgroup$
    – Barry
    Jun 28 '12 at 13:12



















0












$begingroup$

Yep, that is correct:

$v=2arccos(1-frac{h}{r})$



And this is how we get it:



$cos(frac{v}{2}) = frac{r-h}{r}$

$cos(frac{v}{2}) = 1-frac{h}{r}$

$frac{v}{2} = arccos(1-frac{h}{r})$

$v=2arccos(1-frac{h}{r})$



See this homework help resource for more.






share|cite|improve this answer











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    The problem is that your solution is incorrect. You have the 2 factor in the wrong place. the correct solution is:



                        v=2arccos((1-h/r)





    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks that works. But can you explain me how v, can be 180 degress. Isn't it a triangle like on this image matb1htx.systime.dk/fileadmin/indhold/ISBNXXXXXXXXXXXXX/…
      $endgroup$
      – NikolajSvendsen
      Jun 28 '12 at 12:59












    • $begingroup$
      In your example, the height to the chord is equal to the radius. Therefore the chord becomes a diameter and the angle subtended by a diameter is 180 degrees.
      $endgroup$
      – Barry
      Jun 28 '12 at 13:12
















    0












    $begingroup$

    The problem is that your solution is incorrect. You have the 2 factor in the wrong place. the correct solution is:



                        v=2arccos((1-h/r)





    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks that works. But can you explain me how v, can be 180 degress. Isn't it a triangle like on this image matb1htx.systime.dk/fileadmin/indhold/ISBNXXXXXXXXXXXXX/…
      $endgroup$
      – NikolajSvendsen
      Jun 28 '12 at 12:59












    • $begingroup$
      In your example, the height to the chord is equal to the radius. Therefore the chord becomes a diameter and the angle subtended by a diameter is 180 degrees.
      $endgroup$
      – Barry
      Jun 28 '12 at 13:12














    0












    0








    0





    $begingroup$

    The problem is that your solution is incorrect. You have the 2 factor in the wrong place. the correct solution is:



                        v=2arccos((1-h/r)





    share|cite|improve this answer









    $endgroup$



    The problem is that your solution is incorrect. You have the 2 factor in the wrong place. the correct solution is:



                        v=2arccos((1-h/r)






    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jun 28 '12 at 12:45









    BarryBarry

    25612




    25612












    • $begingroup$
      Thanks that works. But can you explain me how v, can be 180 degress. Isn't it a triangle like on this image matb1htx.systime.dk/fileadmin/indhold/ISBNXXXXXXXXXXXXX/…
      $endgroup$
      – NikolajSvendsen
      Jun 28 '12 at 12:59












    • $begingroup$
      In your example, the height to the chord is equal to the radius. Therefore the chord becomes a diameter and the angle subtended by a diameter is 180 degrees.
      $endgroup$
      – Barry
      Jun 28 '12 at 13:12


















    • $begingroup$
      Thanks that works. But can you explain me how v, can be 180 degress. Isn't it a triangle like on this image matb1htx.systime.dk/fileadmin/indhold/ISBNXXXXXXXXXXXXX/…
      $endgroup$
      – NikolajSvendsen
      Jun 28 '12 at 12:59












    • $begingroup$
      In your example, the height to the chord is equal to the radius. Therefore the chord becomes a diameter and the angle subtended by a diameter is 180 degrees.
      $endgroup$
      – Barry
      Jun 28 '12 at 13:12
















    $begingroup$
    Thanks that works. But can you explain me how v, can be 180 degress. Isn't it a triangle like on this image matb1htx.systime.dk/fileadmin/indhold/ISBNXXXXXXXXXXXXX/…
    $endgroup$
    – NikolajSvendsen
    Jun 28 '12 at 12:59






    $begingroup$
    Thanks that works. But can you explain me how v, can be 180 degress. Isn't it a triangle like on this image matb1htx.systime.dk/fileadmin/indhold/ISBNXXXXXXXXXXXXX/…
    $endgroup$
    – NikolajSvendsen
    Jun 28 '12 at 12:59














    $begingroup$
    In your example, the height to the chord is equal to the radius. Therefore the chord becomes a diameter and the angle subtended by a diameter is 180 degrees.
    $endgroup$
    – Barry
    Jun 28 '12 at 13:12




    $begingroup$
    In your example, the height to the chord is equal to the radius. Therefore the chord becomes a diameter and the angle subtended by a diameter is 180 degrees.
    $endgroup$
    – Barry
    Jun 28 '12 at 13:12











    0












    $begingroup$

    Yep, that is correct:

    $v=2arccos(1-frac{h}{r})$



    And this is how we get it:



    $cos(frac{v}{2}) = frac{r-h}{r}$

    $cos(frac{v}{2}) = 1-frac{h}{r}$

    $frac{v}{2} = arccos(1-frac{h}{r})$

    $v=2arccos(1-frac{h}{r})$



    See this homework help resource for more.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Yep, that is correct:

      $v=2arccos(1-frac{h}{r})$



      And this is how we get it:



      $cos(frac{v}{2}) = frac{r-h}{r}$

      $cos(frac{v}{2}) = 1-frac{h}{r}$

      $frac{v}{2} = arccos(1-frac{h}{r})$

      $v=2arccos(1-frac{h}{r})$



      See this homework help resource for more.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Yep, that is correct:

        $v=2arccos(1-frac{h}{r})$



        And this is how we get it:



        $cos(frac{v}{2}) = frac{r-h}{r}$

        $cos(frac{v}{2}) = 1-frac{h}{r}$

        $frac{v}{2} = arccos(1-frac{h}{r})$

        $v=2arccos(1-frac{h}{r})$



        See this homework help resource for more.






        share|cite|improve this answer











        $endgroup$



        Yep, that is correct:

        $v=2arccos(1-frac{h}{r})$



        And this is how we get it:



        $cos(frac{v}{2}) = frac{r-h}{r}$

        $cos(frac{v}{2}) = 1-frac{h}{r}$

        $frac{v}{2} = arccos(1-frac{h}{r})$

        $v=2arccos(1-frac{h}{r})$



        See this homework help resource for more.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jun 29 '12 at 12:49









        rschwieb

        108k12103252




        108k12103252










        answered Jun 28 '12 at 19:51









        KliKli

        1




        1






























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