Facebook Page
Twitter
RSS
+ Reply to Thread
Results 1 to 5 of 5
  1. MHB Craftsman

    Status
    Offline
    Join Date
    Jul 2016
    Posts
    341
    Thanks
    460 times
    Thanked
    168 times
    Trophies
    7 Highscores
    Awards
    MHB Model User Award (2016)
    #1
    Last edited by greg1313; November 10th, 2016 at 08:25. Reason: Improve title

  2. MHB Journeyman
    MHB Math Helper

    Status
    Offline
    Join Date
    Jan 2012
    Posts
    611
    Thanks
    225 times
    Thanked
    717 times
    Thank/Post
    1.173
    #2
    I presume that the "right triangle" referred to is the one at the right of the parallelogram. It should be clear to you that the right triangle at the left is identical to the first one and has the same area. So in order that the area of either of those two right triangles be "exactly half" the area of the parallelogram, there must be nothing but those two right triangles. That is, there is no "middle section"- the "two" verticals must be one- that vertical goes from vertex D to vertex B. What does that look like?

  3. MHB Craftsman

    Status
    Offline
    Join Date
    Jul 2016
    Posts
    341
    Thanks
    460 times
    Thanked
    168 times
    Trophies
    7 Highscores
    Awards
    MHB Model User Award (2016)
    #3 Thread Author
    Quote Originally Posted by HallsofIvy View Post
    I presume that the "right triangle" referred to is the one at the right of the parallelogram. It should be clear to you that the right triangle at the left is identical to the first one and has the same area. So in order that the area of either of those two right triangles be "exactly half" the area of the parallelogram, there must be nothing but those two right triangles. That is, there is no "middle section"- the "two" verticals must be one- that vertical goes from vertex D to vertex B. What does that look like?
    Hello HallsofIvy,

    This should be that identical right angled triangle at the left



    Then the triangle which has half the area of the parallelogram would be,



    Correct ?

    Many THanks

  4. MHB Seeker
    MHB Global Moderator
    MHB Math Scholar
    I like Serena's Avatar
    Status
    Offline
    Join Date
    Mar 2012
    Location
    Netherlands
    Posts
    6,029
    Thanks
    4,066 times
    Thanked
    11,384 times
    Thank/Post
    1.888
    Awards
    MHB Model Helper Award (2016)  

MHB Best Ideas (2016)  

MHB LaTeX Award (2016)  

MHB Calculus Award (2014)  

MHB Discrete Mathematics Award (Jul-Dec 2013)
    #4
    Quote Originally Posted by mathlearn View Post
    This should be that identical right angled triangle at the left

    Then the triangle which has half the area of the parallelogram would be,

    Correct ?
    Hey mathlearn!

    I think the right triangle is fixed.
    We have something like this:
    \begin{tikzpicture}[shorten >=1pt, shorten <=1pt, font=\large]
    %preamble \usetikzlibrary{arrows}

    \def\x{6};

    \draw[ultra thick, blue]
    (0,0) coordinate (A)
    -- ({2*\x - 4},0) coordinate (B)
    -- ({2*\x - 4 + \x - 4},{abs(\x - 4)}) coordinate (C)
    -- ({\x - 4},{abs(\x - 4)}) coordinate (D)
    -- cycle;

    \node[left] at (A) {A};
    \node[right] at (B) {B};
    \node[right] at (C) {C};
    \node[left] at (D) {D};

    \draw[dashed] ({2*\x - 4},0) -- ({2*\x - 4},{abs(\x - 4)}) coordinate (E);
    \node[above,yshift=3mm] at (E) {E};

    \draw ({2*\x - 4},{abs(\x - 4)}) rectangle ({2*\x - 4 + 0.2},{abs(\x - 4) - 0.2});

    \draw ({2*\x - 4 - 0.2},{abs(\x - 4)/2 + 0.05}) -- ({2*\x - 4 + 0.2},{abs(\x - 4)/2 + 0.05});
    \draw ({2*\x - 4 - 0.2},{abs(\x - 4)/2 - 0.05}) -- ({2*\x - 4 + 0.2},{abs(\x - 4)/2 - 0.05});
    \draw ({\x*5/2 - 6 - 0.05},{abs(\x - 4) + 0.2}) -- ({\x*5/2 - 6 - 0.05},{abs(\x - 4) - 0.2});
    \draw ({\x*5/2 - 6 + 0.05},{abs(\x - 4) + 0.2}) -- ({\x*5/2 - 6 + 0.05},{abs(\x - 4) - 0.2});

    \draw[triangle 45-triangle 45] ({\x - 4},{abs(\x - 4) + 0.25}) -- node[above] {$x$} (({2*\x - 4},{abs(\x - 4) + 0.25});
    \draw[triangle 45-triangle 45] (({2*\x - 4},{abs(\x - 4) + 0.25}) -- node[above] {$x-4$} ({2*\x - 4 + \x - 4},{abs(\x - 4) + 0.25});

    \end{tikzpicture}

    And if we make $x$ smaller, we get:
    \begin{tikzpicture}[shorten >=1pt, shorten <=1pt, font=\large]
    %preamble \usetikzlibrary{arrows}

    \def\x{5};

    \draw[ultra thick, blue]
    (0,0) coordinate (A)
    -- ({2*\x - 4},0) coordinate (B)
    -- ({2*\x - 4 + \x - 4},{abs(\x - 4)}) coordinate (C)
    -- ({\x - 4},{abs(\x - 4)}) coordinate (D)
    -- cycle;

    \node[left] at (A) {A};
    \node[right] at (B) {B};
    \node[right] at (C) {C};
    \node[left] at (D) {D};

    \draw[dashed] ({2*\x - 4},0) -- ({2*\x - 4},{abs(\x - 4)}) coordinate (E);
    \node[above,yshift=6mm] at (E) {E};

    \draw ({2*\x - 4},{abs(\x - 4)}) rectangle ({2*\x - 4 + 0.2},{abs(\x - 4) - 0.2});

    \draw ({2*\x - 4 - 0.2},{abs(\x - 4)/2 + 0.05}) -- ({2*\x - 4 + 0.2},{abs(\x - 4)/2 + 0.05});
    \draw ({2*\x - 4 - 0.2},{abs(\x - 4)/2 - 0.05}) -- ({2*\x - 4 + 0.2},{abs(\x - 4)/2 - 0.05});
    \draw ({\x*5/2 - 6 - 0.05},{abs(\x - 4) + 0.2}) -- ({\x*5/2 - 6 - 0.05},{abs(\x - 4) - 0.2});
    \draw ({\x*5/2 - 6 + 0.05},{abs(\x - 4) + 0.2}) -- ({\x*5/2 - 6 + 0.05},{abs(\x - 4) - 0.2});

    \draw[triangle 45-triangle 45] ({\x - 4},{abs(\x - 4) + 0.25}) -- node[above] {$x$} (({2*\x - 4},{abs(\x - 4) + 0.25});
    \draw[triangle 45-triangle 45] (({2*\x - 4},{abs(\x - 4) + 0.25}) -- node[above] {$x-4$} ({2*\x - 4 + \x - 4},{abs(\x - 4) + 0.25});

    \end{tikzpicture}

    Hmm... let's make $x$ negative:
    \begin{tikzpicture}[shorten >=1pt, shorten <=1pt, font=\large]
    %preamble \usetikzlibrary{arrows}

    \def\x{-2};

    \draw[ultra thick, red]
    (0,0) coordinate (A)
    -- ({2*\x - 4},0) coordinate (B)
    -- ({2*\x - 4 + \x - 4},{abs(\x - 4)}) coordinate (C)
    -- ({\x - 4},{abs(\x - 4)}) coordinate (D)
    -- cycle;

    \node[right] at (A) {A};
    \node[left] at (B) {B};
    \node[left] at (C) {C};
    \node[right] at (D) {D};

    \draw[dashed] ({2*\x - 4},0) -- ({2*\x - 4},{abs(\x - 4)}) coordinate (E);
    \node[above,yshift=6mm] at (E) {E};

    \draw ({2*\x - 4},{abs(\x - 4)}) rectangle ({2*\x - 4 + 0.2},{abs(\x - 4) - 0.2});

    \draw ({2*\x - 4 - 0.2},{abs(\x - 4)/2 + 0.05}) -- ({2*\x - 4 + 0.2},{abs(\x - 4)/2 + 0.05});
    \draw ({2*\x - 4 - 0.2},{abs(\x - 4)/2 - 0.05}) -- ({2*\x - 4 + 0.2},{abs(\x - 4)/2 - 0.05});
    \draw ({\x*5/2 - 6 - 0.05},{abs(\x - 4) + 0.2}) -- ({\x*5/2 - 6 - 0.05},{abs(\x - 4) - 0.2});
    \draw ({\x*5/2 - 6 + 0.05},{abs(\x - 4) + 0.2}) -- ({\x*5/2 - 6 + 0.05},{abs(\x - 4) - 0.2});

    \draw[triangle 45-triangle 45] ({\x - 4},{abs(\x - 4) + 0.25}) -- node[above] {$x$} (({2*\x - 4},{abs(\x - 4) + 0.25});
    \draw[triangle 45-triangle 45] (({2*\x - 4},{abs(\x - 4) + 0.25}) -- node[above] {$x-4$} ({2*\x - 4 + \x - 4},{abs(\x - 4) + 0.25});

    \end{tikzpicture}

    I think we need $x=0$:
    \begin{tikzpicture}[shorten >=1pt, shorten <=1pt, font=\large]
    %preamble \usetikzlibrary{arrows}

    \def\x{0};

    \draw[ultra thick, red]
    (0,0) coordinate (A)
    -- ({2*\x - 4},0) coordinate (B)
    -- ({2*\x - 4 + \x - 4},{abs(\x - 4)}) coordinate (C)
    -- ({\x - 4},{abs(\x - 4)}) coordinate (D)
    -- cycle;

    \node[right] at (A) {A};
    \node[left] at (B) {B};
    \node[left] at (C) {C};
    \node[right] at (D) {D};

    \draw[dashed] ({2*\x - 4},0) -- ({2*\x - 4},{abs(\x - 4)}) coordinate (E);
    \node[above,yshift=6mm] at (E) {E};

    \draw ({2*\x - 4},{abs(\x - 4)}) rectangle ({2*\x - 4 - 0.2},{abs(\x - 4) - 0.2});

    \draw ({2*\x - 4 - 0.2},{abs(\x - 4)/2 + 0.05}) -- ({2*\x - 4 + 0.2},{abs(\x - 4)/2 + 0.05});
    \draw ({2*\x - 4 - 0.2},{abs(\x - 4)/2 - 0.05}) -- ({2*\x - 4 + 0.2},{abs(\x - 4)/2 - 0.05});
    \draw ({\x*5/2 - 6 - 0.05},{abs(\x - 4) + 0.2}) -- ({\x*5/2 - 6 - 0.05},{abs(\x - 4) - 0.2});
    \draw ({\x*5/2 - 6 + 0.05},{abs(\x - 4) + 0.2}) -- ({\x*5/2 - 6 + 0.05},{abs(\x - 4) - 0.2});

    \draw[triangle 45-triangle 45] ({\x - 4},{abs(\x - 4) + 0.25}) -- node[above,xshift=3mm] {$x$} (({2*\x - 4},{abs(\x - 4) + 0.25});
    \draw[triangle 45-triangle 45] (({2*\x - 4},{abs(\x - 4) + 0.25}) -- node[above] {$x-4$} ({2*\x - 4 + \x - 4},{abs(\x - 4) + 0.25});

    \end{tikzpicture}

    Now the right triangle takes up half of the parallellogram.
    Last edited by I like Serena; December 10th, 2016 at 20:10. Reason: Updated preamble

  5. MHB Journeyman
    MHB Math Helper

    Status
    Offline
    Join Date
    Jan 2012
    Posts
    611
    Thanks
    225 times
    Thanked
    717 times
    Thank/Post
    1.173
    #5
    Quote Originally Posted by mathlearn View Post
    Hello HallsofIvy,

    This should be that identical right angled triangle at the left



    Then the triangle which has half the area of the parallelogram would be,



    Correct ?

    Many THanks
    No, those are not right triangles. And it is always true that the two triangles you get by drawing a diagonal have area half the parallelogram.

    - -

    - - - Updated - - -

    Quote Originally Posted by I like Serena View Post
    Hey mathlearn!

    I think the right triangle is fixed.
    We have something like this:
    \begin{tikzpicture}[shorten >=1pt, shorten <=1pt, font=\large]

    \def\x{6};

    \draw[ultra thick, blue]
    (0,0) coordinate (A)
    -- ({2*\x - 4},0) coordinate (B)
    -- ({2*\x - 4 + \x - 4},{abs(\x - 4)}) coordinate (C)
    -- ({\x - 4},{abs(\x - 4)}) coordinate (D)
    -- cycle;

    \node[left] at (A) {A};
    \node[right] at (B) {B};
    \node[right] at (C) {C};
    \node[left] at (D) {D};

    \draw[dashed] ({2*\x - 4},0) -- ({2*\x - 4},{abs(\x - 4)}) coordinate (E);
    \node[above,yshift=3mm] at (E) {E};

    \draw ({2*\x - 4},{abs(\x - 4)}) rectangle ({2*\x - 4 + 0.2},{abs(\x - 4) - 0.2});

    \draw ({2*\x - 4 - 0.2},{abs(\x - 4)/2 + 0.05}) -- ({2*\x - 4 + 0.2},{abs(\x - 4)/2 + 0.05});
    \draw ({2*\x - 4 - 0.2},{abs(\x - 4)/2 - 0.05}) -- ({2*\x - 4 + 0.2},{abs(\x - 4)/2 - 0.05});
    \draw ({\x*5/2 - 6 - 0.05},{abs(\x - 4) + 0.2}) -- ({\x*5/2 - 6 - 0.05},{abs(\x - 4) - 0.2});
    \draw ({\x*5/2 - 6 + 0.05},{abs(\x - 4) + 0.2}) -- ({\x*5/2 - 6 + 0.05},{abs(\x - 4) - 0.2});

    \draw[triangle 45-triangle 45] ({\x - 4},{abs(\x - 4) + 0.25}) -- node[above] {$x$} (({2*\x - 4},{abs(\x - 4) + 0.25});
    \draw[triangle 45-triangle 45] (({2*\x - 4},{abs(\x - 4) + 0.25}) -- node[above] {$x-4$} ({2*\x - 4 + \x - 4},{abs(\x - 4) + 0.25});

    \end{tikzpicture}

    And if we make $x$ smaller, we get:
    \begin{tikzpicture}[shorten >=1pt, shorten <=1pt, font=\large]

    \def\x{5};

    \draw[ultra thick, blue]
    (0,0) coordinate (A)
    -- ({2*\x - 4},0) coordinate (B)
    -- ({2*\x - 4 + \x - 4},{abs(\x - 4)}) coordinate (C)
    -- ({\x - 4},{abs(\x - 4)}) coordinate (D)
    -- cycle;

    \node[left] at (A) {A};
    \node[right] at (B) {B};
    \node[right] at (C) {C};
    \node[left] at (D) {D};

    \draw[dashed] ({2*\x - 4},0) -- ({2*\x - 4},{abs(\x - 4)}) coordinate (E);
    \node[above,yshift=6mm] at (E) {E};

    \draw ({2*\x - 4},{abs(\x - 4)}) rectangle ({2*\x - 4 + 0.2},{abs(\x - 4) - 0.2});

    \draw ({2*\x - 4 - 0.2},{abs(\x - 4)/2 + 0.05}) -- ({2*\x - 4 + 0.2},{abs(\x - 4)/2 + 0.05});
    \draw ({2*\x - 4 - 0.2},{abs(\x - 4)/2 - 0.05}) -- ({2*\x - 4 + 0.2},{abs(\x - 4)/2 - 0.05});
    \draw ({\x*5/2 - 6 - 0.05},{abs(\x - 4) + 0.2}) -- ({\x*5/2 - 6 - 0.05},{abs(\x - 4) - 0.2});
    \draw ({\x*5/2 - 6 + 0.05},{abs(\x - 4) + 0.2}) -- ({\x*5/2 - 6 + 0.05},{abs(\x - 4) - 0.2});

    \draw[triangle 45-triangle 45] ({\x - 4},{abs(\x - 4) + 0.25}) -- node[above] {$x$} (({2*\x - 4},{abs(\x - 4) + 0.25});
    \draw[triangle 45-triangle 45] (({2*\x - 4},{abs(\x - 4) + 0.25}) -- node[above] {$x-4$} ({2*\x - 4 + \x - 4},{abs(\x - 4) + 0.25});

    \end{tikzpicture}

    Hmm... let's make $x$ negative:
    \begin{tikzpicture}[shorten >=1pt, shorten <=1pt, font=\large]

    \def\x{-2};

    \draw[ultra thick, red]
    (0,0) coordinate (A)
    -- ({2*\x - 4},0) coordinate (B)
    -- ({2*\x - 4 + \x - 4},{abs(\x - 4)}) coordinate (C)
    -- ({\x - 4},{abs(\x - 4)}) coordinate (D)
    -- cycle;

    \node[right] at (A) {A};
    \node[left] at (B) {B};
    \node[left] at (C) {C};
    \node[right] at (D) {D};

    \draw[dashed] ({2*\x - 4},0) -- ({2*\x - 4},{abs(\x - 4)}) coordinate (E);
    \node[above,yshift=6mm] at (E) {E};

    \draw ({2*\x - 4},{abs(\x - 4)}) rectangle ({2*\x - 4 + 0.2},{abs(\x - 4) - 0.2});

    \draw ({2*\x - 4 - 0.2},{abs(\x - 4)/2 + 0.05}) -- ({2*\x - 4 + 0.2},{abs(\x - 4)/2 + 0.05});
    \draw ({2*\x - 4 - 0.2},{abs(\x - 4)/2 - 0.05}) -- ({2*\x - 4 + 0.2},{abs(\x - 4)/2 - 0.05});
    \draw ({\x*5/2 - 6 - 0.05},{abs(\x - 4) + 0.2}) -- ({\x*5/2 - 6 - 0.05},{abs(\x - 4) - 0.2});
    \draw ({\x*5/2 - 6 + 0.05},{abs(\x - 4) + 0.2}) -- ({\x*5/2 - 6 + 0.05},{abs(\x - 4) - 0.2});

    \draw[triangle 45-triangle 45] ({\x - 4},{abs(\x - 4) + 0.25}) -- node[above] {$x$} (({2*\x - 4},{abs(\x - 4) + 0.25});
    \draw[triangle 45-triangle 45] (({2*\x - 4},{abs(\x - 4) + 0.25}) -- node[above] {$x-4$} ({2*\x - 4 + \x - 4},{abs(\x - 4) + 0.25});

    \end{tikzpicture}

    I think we need $x=0$:
    \begin{tikzpicture}[shorten >=1pt, shorten <=1pt, font=\large]

    \def\x{0};

    \draw[ultra thick, red]
    (0,0) coordinate (A)
    -- ({2*\x - 4},0) coordinate (B)
    -- ({2*\x - 4 + \x - 4},{abs(\x - 4)}) coordinate (C)
    -- ({\x - 4},{abs(\x - 4)}) coordinate (D)
    -- cycle;

    \node[right] at (A) {A};
    \node[left] at (B) {B};
    \node[left] at (C) {C};
    \node[right] at (D) {D};

    \draw[dashed] ({2*\x - 4},0) -- ({2*\x - 4},{abs(\x - 4)}) coordinate (E);
    \node[above,yshift=6mm] at (E) {E};

    \draw ({2*\x - 4},{abs(\x - 4)}) rectangle ({2*\x - 4 - 0.2},{abs(\x - 4) - 0.2});

    \draw ({2*\x - 4 - 0.2},{abs(\x - 4)/2 + 0.05}) -- ({2*\x - 4 + 0.2},{abs(\x - 4)/2 + 0.05});
    \draw ({2*\x - 4 - 0.2},{abs(\x - 4)/2 - 0.05}) -- ({2*\x - 4 + 0.2},{abs(\x - 4)/2 - 0.05});
    \draw ({\x*5/2 - 6 - 0.05},{abs(\x - 4) + 0.2}) -- ({\x*5/2 - 6 - 0.05},{abs(\x - 4) - 0.2});
    \draw ({\x*5/2 - 6 + 0.05},{abs(\x - 4) + 0.2}) -- ({\x*5/2 - 6 + 0.05},{abs(\x - 4) - 0.2});

    \draw[triangle 45-triangle 45] ({\x - 4},{abs(\x - 4) + 0.25}) -- node[above,xshift=3mm] {$x$} (({2*\x - 4},{abs(\x - 4) + 0.25});
    \draw[triangle 45-triangle 45] (({2*\x - 4},{abs(\x - 4) + 0.25}) -- node[above] {$x-4$} ({2*\x - 4 + \x - 4},{abs(\x - 4) + 0.25});

    \end{tikzpicture}

    Now the right triangle takes up half of the parallellogram.
    Yes, that was what I meant. Nice drawings!

Similar Threads

  1. Parallelogram Help?
    By Storminnorman in forum Geometry
    Replies: 1
    Last Post: January 18th, 2016, 22:56
  2. Area of a Parallelogram
    By jljarrett18 in forum Geometry
    Replies: 11
    Last Post: July 14th, 2014, 10:33
  3. Area of parallelogram
    By leprofece in forum Trigonometry
    Replies: 6
    Last Post: March 10th, 2014, 18:14
  4. Parallelogram Problem
    By Albert in forum Challenge Questions and Puzzles
    Replies: 1
    Last Post: July 9th, 2013, 20:40
  5. Replies: 2
    Last Post: March 20th, 2013, 16:54

Tags for this Thread

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •  
Math Help Boards