Results 1 to 5 of 5

Math Help - Find the blue area

  1. #1
    Member
    Joined
    Apr 2009
    From
    Jerusalem - Israel
    Posts
    108

    Find the blue area

    Find the blue area-area5.jpg
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,546
    Thanks
    539
    Hello, razemsoft21!

    Find the shaded area.

    Code:
                  * * *
              *     |r/2  *
            *      A|       *
           *        o        *
                   .|.
          *       .:|:.       *
          *      .::O::.      *
          *     .::*|*::.     *
               .::*:|:*::.
           *  .::*::|::*::.  *
            * ::*:::|:::*:: *
             Bo - - - - - oC
                    r

    \,O is the center of the circle.
    Draw AB and AC.

    Note that \Delta OBC is equilateral.
    Its height is: . \frac{\sqrt{3}}{2}r

    The height of \Delta ABC is: . \frac{r}{2} + \frac{\sqrt{3}}{2}r \:=\:\left(\frac{1+\sqrt{3}}{2}\right)r


    The area of \Delta ABC is: . \frac{1}{2}bh \;=\;\frac{1}{2}(r)\left(\frac{1+\sqrt{3}}{2}\righ  t)r \;=\;\dfrac{1+\sqrt{3}}{4}\,r^2

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Apr 2009
    From
    Jerusalem - Israel
    Posts
    108
    good job Soroban but I need the blue area
    and as you see it is a part of circle not a full
    circle to substact the area of triangle !!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,546
    Thanks
    539

    Ha! . . . That was stupid!
    I found the wrong area . . . *blush*


    Let's see, the shaded area is the area of the circle,
    . . minus the area of \Delta ABC, minus that missing segment.


    The area of the segment is the area of the 60^o sector
    . . minus equilateral triangle OBC.

    The area of the sector is: . \frac{1}{2}r^2(\frac{\pi}{3}) \:=\:\frac{\pi}{6}r^2

    The area of the equilateral triangle is: . \frac{\sqrt{3}}{4}r^2

    So the area of the segment is: . \frac{\pi}{6}r^2 - \frac{\sqrt{3}}{2}r^2 \:=\:\left(\frac{\pi}{6} - \frac{\sqrt{3}}{4}\right)r^2


    Therefore, the desired area is:

    . . \displaystyle \pi r^2 - \left(\frac{1+\sqrt{3}}{4}\right)r^2 - \left(\frac{\pi}{6} - \frac{\sqrt{3}}{4}\right)r^2 \;=\;\left(\frac{5\pi}{6} - \frac{1}{4}\right)r^2

    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Here is a slightly different method. In the attached picture (where the top of the white equilateral triangle is the centre of the circle), the blue and red regions together form 5/6ths of the area of the circle, so their combined area is \tfrac56r^2. From that we must subtract the red area. Each red triangle has area half the product of two sides times the sine of the included angle. In this case the sides are \tfrac12r and r, and the included angle is 150^\circ. So the area of each triangle is \tfrac14r^2\sin(150^\circ) = \tfrac18r^2. Therefore the blue area is \bigl(\tfrac{5\pi}6-\tfrac14\bigr)r^2, as Soroban already showed.

    Sometimes it's worth doing a problem in more than one way, just to check that you have the correct answer.
    Attached Thumbnails Attached Thumbnails Find the blue area-circle.jpg  
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Java Blue J
    Posted in the Math Software Forum
    Replies: 4
    Last Post: September 5th 2010, 11:28 AM
  2. Container red and blue
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 18th 2010, 09:55 PM
  3. Blue and Red
    Posted in the Algebra Forum
    Replies: 2
    Last Post: May 4th 2009, 06:38 AM
  4. Replies: 1
    Last Post: April 9th 2009, 12:20 PM
  5. Probability: Red or Blue
    Posted in the Statistics Forum
    Replies: 2
    Last Post: March 20th 2007, 07:19 AM

Search Tags


/mathhelpforum @mathhelpforum