Results 1 to 4 of 4

Math Help - quadrilateral with vertices

  1. #1
    Member
    Joined
    Jan 2008
    Posts
    173

    quadrilateral with vertices

    Determine the area of the quadrilateral whose vertices are (1,3), (6,5), (9,1) and
    (3,-3).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by sri340 View Post
    Determine the area of the quadrilateral whose vertices are (1,3), (6,5), (9,1) and
    (3,-3).
    Maybe you can use this method: http://www.mathhelpforum.com/math-he...ngle-help.html

    See post #4.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello sri340
    Quote Originally Posted by sri340 View Post
    Determine the area of the quadrilateral whose vertices are (1,3), (6,5), (9,1) and
    (3,-3).
    Look at the attached diagram. Calculate the areas of the four right-angled triangles and the area of the rectangle. Add together. Done.

    (If my arithmetic is correct, it's 35 sq. units.)

    Grandad
    Attached Thumbnails Attached Thumbnails quadrilateral with vertices-untitled.jpg  
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jan 2009
    Posts
    591
    Quote Originally Posted by sri340 View Post
    Determine the area of the quadrilateral whose vertices are (1,3), (6,5), (9,1) and (3,-3).
     \dfrac{1}{3} \text{ : } \dfrac{6}{5} \text{ : } \dfrac{9}{1} \text{ : } \dfrac{3}{-3} \text{ : } \dfrac{1}{3}

    (1) sum of all the [ a numerator multiplied by the next denominator]
     1 \cdot 5 + 6\cdot 1 + 9 \cdot -3 + 3 \cdot 3 = -7

    (2) sum of all the [ denominator multiplied by the next numerator ]
     3 \cdot 6 + 5 \cdot 9 + 1 \cdot 3 + -3 \cdot 1 = 63

    subtract the second sum from the first (-7 -63 = -70)
    and divide that absolute value by 2 ( 70/2 = 35 )

    This method will calculate the area of any polygon, when you know the coordinates of the points around the polygon. If you proceed in a clockwise direction around the perimeter (as was done above) you will have negative quantity; If you proceed in a counterclockwise direction the resulting quantity will be positive.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: October 9th 2010, 01:40 PM
  2. Vertices help
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: June 19th 2010, 12:26 AM
  3. Vertices
    Posted in the Geometry Forum
    Replies: 1
    Last Post: November 19th 2009, 11:05 AM
  4. Vertices of a parallelogram?
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: March 20th 2008, 06:12 AM
  5. Vertices help
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: August 19th 2007, 10:28 PM

Search Tags


/mathhelpforum @mathhelpforum