Thread: Draw sketch to represent the area of the right triangle is exactly half the area of a parallelogram

1. 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?

Originally Posted by HallsofIvy
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

3. Originally Posted by mathlearn
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.

4. Originally Posted by mathlearn
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 - - -

Originally Posted by I like Serena
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!