Results 1 to 4 of 4

Math Help - area of triangle

  1. #1
    Junior Member
    Joined
    May 2008
    From
    California
    Posts
    36

    area of triangle

    I have no idea how to do this can someone teach me step by step please?!

    using integration, find the area of the triangle with vertices (0,0)(2,1)(-1,6)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Haven's Avatar
    Joined
    Jul 2009
    Posts
    197
    Thanks
    8
    Given a function f(x), \int_a^b f(x) dx gives us the area between the function and the x-axis, from a to b.

    Naturally, when we have two functions f(x) and g(x), where f(x) is always bigger than g(x), then \int_a^b f(x) - g(x) dx gives us the area between the two functions.

    So, we have 3 points (0,0), (2,1), (-1,6). We can find three lines from these points, and each line will cross two of the points. These will be the sides of the triangle. We can obtain these with simple algebra

    L_1(x) = \frac{1}{2}x crosses (0,0) and (2,1)

    L_2(x) = -6x crosses (0,0) and (-1,6)

    L_3(x) = \frac{-5}{3}x + \frac{13}{3} crosses (2,1) and (-1,6)

    Now if we draw out the triangle on the xy plane, we see that L_3 is above L_1 when 0\leq x \leq 2

    So the area between the two lines is given by \int_0^2 L_3(x) - L_1(x) dx.

    Similarly, L_3 is about L_2 when -1 \leq x \leq 0 and the area between the two lines is given by \int_{-1}^0 L_3(x) - L_2(x) dx.

    I'll leave you to solve these integrals. Now we see that by our setup, the sum of the two areas, is the area of the triangle. So the area of the triangle is \int_0^2 L_3(x) - L_1(x)dx + \int_{-1}^0 L_3(x) - L_2(x)dx
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2010
    Posts
    11
    Hi gearshifter,

    Drawing a rectangle from (-1,0) to (2,6) creates 3 triangles around yours. The surface of your triangle can be written as the surface of the rectangle minus the surface of the three triangles around your triangle:

    Which leads to:

    18-[3+1+7.5] = 6.5

    This can be done for all possible triangles!



    In your case the surfaces of the three triangles around the main traingle can be calcluated using integrals:

    -6x integrated from -1 to 6 = 3
    triangle 1: 3

    0.5x integrated from 0 to 2 = 1
    triangle 2: 1

    4,33-(5/3)x integrated from -1 to 2 = 10.5
    triangle 3: 3*6-10.5=7.5

    This method is not as elegant as one given by Haven, but hopefully better understandable.
    Last edited by Jeroentje; March 12th 2010 at 05:14 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2010
    Posts
    11
    If you have time left you can try the Gauss method:

    The surface of an figure can be calculated by multiplying the following factors per line. The summation over all lines leads to the surface.

    a) The length of the line
    b) The average x-coordinate
    c) The x value of the normal on the line

    First line ([0,0] [2,1])
    a1=5^0.5=2.236
    b1=1
    c1=cos(atan(2))=0.447

    second line ([2,1] [-1,6])
    a2=34^0.5=5.831
    b2=0.5
    c2=cos(atan(3/5))=0.857

    third line ([-1,6] [0,0])
    a3=37^0.5=6.083
    b3=-0.5
    c3=-cos(atan(1/6))=-0.986
    (NOTICE the normal vector has a negative x-value)

    surface = a1*b1*c1 + a2*b2*c2 + a3*b3*c3 = 6.5

    This method is far too complicated for this case, but can be usefull for more complicated geometries.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Area of Triangle #2
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 8th 2010, 08:19 PM
  2. Replies: 1
    Last Post: October 28th 2008, 07:02 PM
  3. Replies: 7
    Last Post: July 19th 2008, 06:53 AM
  4. The area of a triangle within a triangle
    Posted in the Geometry Forum
    Replies: 4
    Last Post: April 28th 2008, 03:27 AM
  5. Replies: 27
    Last Post: April 27th 2008, 10:36 AM

Search Tags


/mathhelpforum @mathhelpforum