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

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Discussion Overview

The discussion revolves around the relationship between the areas of right triangles and a parallelogram, specifically exploring how the area of a right triangle can be half that of the parallelogram. Participants are sharing sketches and reasoning about the geometric configurations involved.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • Some participants suggest that the right triangles being discussed are those located at the sides of the parallelogram, and they emphasize that these triangles must be identical for their areas to be half of the parallelogram's area.
  • There is a proposal that for the area of either right triangle to be exactly half of the parallelogram, the configuration must exclude any middle section, implying a specific arrangement of the vertices.
  • One participant presents a series of sketches using TikZ to illustrate their understanding of the right triangle's area in relation to the parallelogram, suggesting that as the variable \(x\) changes, the area relationship is maintained.
  • Another participant asserts that the two triangles formed by drawing a diagonal of the parallelogram always have areas that are half of the parallelogram, challenging the earlier claims about the right triangles.

Areas of Agreement / Disagreement

Participants express differing views on the configuration of the triangles and their relationship to the parallelogram's area. There is no consensus on the correctness of the claims regarding the right triangles and their areas.

Contextual Notes

The discussion includes various assumptions about the geometric configurations and the definitions of the areas involved, which remain unresolved. The mathematical steps and reasoning behind the area calculations are not fully detailed.

mathlearn
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View attachment 6172

Any thoughts on the sketch? (Party)

Many Thanks :)
 

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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?
 
HallsofIvy said:
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)
 

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mathlearn said:
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)​
 
Last edited:
mathlearn said:
(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 - - -

I like Serena said:
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!​
 

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