Results 1 to 4 of 4

Math Help - Help with radians, arcs and areas please?

  1. #1
    Newbie
    Joined
    Jan 2013
    From
    uk
    Posts
    15

    Help with radians, arcs and areas please?

    A badge is made of a circle centre O, with radius 5cm and a chord AB of length 6cm creating a blue minor segment at the bottom. Give all your answers to 1 decimal place.

    Find:
    The angle AOB at the centre
    The length of the minor arc AB
    The area of the triangle AOB.
    The area of the yellow major segment.


    How do I do this? Please show the steps!! Thank you, I am so confused



    Here is the screen shot of the question (it has a diagram) Please help me. thanks - also i just typed in random numbers to get the grey answers on the bottom



    mymathsradians | Flickr - Photo Sharing!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,790
    Thanks
    1687
    Awards
    1

    Re: Help with radians, arcs and areas please?

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2013
    From
    uk
    Posts
    15

    Re: Help with radians, arcs and areas please?

    Cool. thanks i'll have a look now.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,963
    Thanks
    1631

    Re: Help with radians, arcs and areas please?

    One way to get the "angle at the center" is to use the cosine law: c^2= a^2+ b^2- 2ab cos(C) where a, b, and c are the lengths of the three sides of a triangle and C is the angle opposite the side "c". Here, the two sides a, and b are the radii of length 5 cm and c has length 6 cm.

    Once you know that angle, the length of the arc is easy- the arc length is just the angle, in radians, times the radius.

    You could get the area of the triangle, now that you know three sides by "Heron's formula" A= \sqrt{s(s-a)(s-b)(s- c)} where s is the "semi-perimeter", (a+b+c)/2.

    But since this is an isosceles triangle, it follows that the "height" bisects the opposite side. So you have two right triangles with hypotenuse of length 5 cm and one leg of length 3 cm.
    Use the Pyhagorean theorem to find the length of the other leg, the height of the original triangle. And then use A= (1/2)bh.

    The area of the entire circle is, of course, \pi r^2= 25\pi. The area of the "pie section", between the radii, is proportional to the part of 2\pi the angle, in radians, is, so its area is 25\pi\frac{\theta}{2\pi}= \frac{25}{2}\theta where \theta is the angle you found in the first part of the problem. Subtract the area of the triangle, that you just found, to find the area of the blue section, then subtract that from the area of the entire triangle, 25\pi to find the area of the yellow part.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Arcs
    Posted in the Geometry Forum
    Replies: 1
    Last Post: July 1st 2010, 05:24 AM
  2. Chords and Arcs
    Posted in the Geometry Forum
    Replies: 4
    Last Post: May 28th 2009, 04:17 AM
  3. Arcs and Angles
    Posted in the Geometry Forum
    Replies: 1
    Last Post: May 17th 2009, 09:36 AM
  4. Clock and arcs
    Posted in the Geometry Forum
    Replies: 1
    Last Post: April 30th 2008, 04:54 PM
  5. Conjectures For Arcs
    Posted in the Geometry Forum
    Replies: 1
    Last Post: January 12th 2008, 12:28 PM

Search Tags


/mathhelpforum @mathhelpforum