Results 1 to 3 of 3

Thread: finding areas perimeters and segments in circle

  1. #1
    Junior Member
    Joined
    May 2009
    Posts
    28

    finding areas perimeters and segments in circle

    Hi, i really need help on working this one out as it is very important for school, thanks in advance.

    A circle has centre o and radius 1cm, points A,B,C,D,E and F lie on the cirumference, so that A,B,C,D,E,F is a regular hexagone(6 equal sides)


    (a): Show that triangle OAB is equilateral, Find its area and the area of the hexagone. Find the area of the circle

    (b) : deduce the area of the minor segment defined on AB

    (c): Find perimeter of this segment
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello llkkjj24

    Welcome to Math Help Forum!
    Quote Originally Posted by llkkjj24 View Post
    Hi, i really need help on working this one out as it is very important for school, thanks in advance.

    A circle has centre o and radius 1cm, points A,B,C,D,E and F lie on the cirumference, so that A,B,C,D,E,F is a regular hexagone(6 equal sides)


    (a): Show that triangle OAB is equilateral, Find its area and the area of the hexagone. Find the area of the circle

    (b) : deduce the area of the minor segment defined on AB

    (c): Find perimeter of this segment
    (a) All angles at O are equal, because the hexagon is regular. Therefore $\displaystyle \angle AOB = 60^o$

    $\displaystyle AO = OB$, radii

    Therefore $\displaystyle \angle OAB = \angle ABO$ (isosceles triangle)

    Therefore $\displaystyle \angle OAB = 60^o$ (angle sum of triangle)

    Therefore $\displaystyle \triangle AOB$ is equilateral.

    Using $\displaystyle \triangle = \tfrac12 bc\sin A$, its area = $\displaystyle \tfrac12.1.1.\sin 60^o = 0.433\, cm^2$

    So area of hexagon $\displaystyle = 0.433 \times 6 = 2.598\, cm^2$.

    Area of circle $\displaystyle = \pi.1^2 = 3.142\, cm^2$

    (b) Therefore the area of the minor segment $\displaystyle AB = \tfrac16(3.142 - 2.598) = 0.088\, cm^2$

    (c) Arc $\displaystyle AB = \tfrac162\pi.1 = 1.047 \,cm$

    $\displaystyle AB = 1\, cm$

    So the perimeter of this segment $\displaystyle = 2.047 \,cm$

    Grandad
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    May 2009
    Posts
    28
    thank you very much for the help, very detailed
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Area of segments - circle help
    Posted in the Geometry Forum
    Replies: 2
    Last Post: Nov 21st 2009, 06:14 AM
  2. Find the perimeters and areas of the figure.
    Posted in the Geometry Forum
    Replies: 3
    Last Post: Oct 21st 2009, 06:52 PM
  3. Replies: 8
    Last Post: Sep 3rd 2009, 12:38 PM
  4. Areas and Perimeters?
    Posted in the Algebra Forum
    Replies: 10
    Last Post: May 31st 2009, 02:06 AM
  5. Areas and Perimeters.
    Posted in the Geometry Forum
    Replies: 1
    Last Post: Mar 29th 2009, 10:24 PM

Search tags for this page

Search Tags


/mathhelpforum @mathhelpforum