MHB Draw Sketch: Right Triangle Area = Half Parallelogram Area (Party)

[mathlearn]

View attachment 6172

Any thoughts on the sketch? (Party)

Many Thanks :)

 

[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?

 

[mathlearn]

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?

(Wave) Hello HallsofIvy,

This should be that identical right angled triangle at the left

View attachment 6176

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

View attachment 6177

Correct ? (Happy)

Many THanks (Party)

 

[I like Serena]

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 ? (Happy)

Hey mathlearn! (Smile)

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

at (A) {A};
\node

at (B) {B};
\node

at (C) {C};
\node

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

at (A) {A};
\node

at (B) {B};
\node

at (C) {C};
\node

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

at (A) {A};
\node

at (B) {B};
\node

at (C) {C};
\node

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

at (A) {A};
\node

at (B) {B};
\node

at (C) {C};
\node

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. (Happy)​

 

[HallsofIvy]

(Wave) 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 ? (Happy)

Many THanks (Party)

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 - - -

Hey mathlearn! (Smile)

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

at (A) {A};
\node

at (B) {B};
\node

at (C) {C};
\node

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

at (A) {A};
\node

at (B) {B};
\node

at (C) {C};
\node

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

at (A) {A};
\node

at (B) {B};
\node

at (C) {C};
\node

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

at (A) {A};
\node

at (B) {B};
\node

at (C) {C};
\node

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. (Happy)​

Yes, that was what I meant. Nice drawings!​