- #1
mathlearn
- 331
- 0
Attachments
Last edited by a moderator:
Draw Sketch: Right Triangle Area = Half Parallelogram Area (Party)
- MHB
- Thread starter mathlearn
- Start date
In summary: Therefore, in order for the two right triangles to have exactly half the area of the parallelogram, the ratio of their sides must be equal to the ratio of their base and height. This means that there can be no "middle section" between the two right triangles, and the vertical line connecting vertices D and B must be one continuous line. Visually, this would look like a right triangle with a smaller right triangle inside on one side, and a vertical line connecting the two vertices on the other side. In summary, the two right triangles
- #2
HallsofIvy
Science Advisor
Homework Helper
- 42,988
- 981
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
mathlearn
- 331
- 0
HallsofIvy said:
(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)
Attachments
- #4
I like Serena
Homework Helper
MHB
- 16,336
- 258
mathlearn said:
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
\node
at (B) {B};
\node
\node
at (C) {C};
\node
\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
\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
\node
at (B) {B};
\node
\node
at (C) {C};
\node
\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
\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
\node
at (B) {B};
\node
\node
at (C) {C};
\node
\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
\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
\node
at (B) {B};
\node
\node
at (C) {C};
\node
\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)
\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:
- #5
HallsofIvy
Science Advisor
Homework Helper
- 42,988
- 981
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.mathlearn said:
- -
- - - Updated - - -
I like Serena said:
Yes, that was what I meant. Nice drawings!
FAQ: Draw Sketch: Right Triangle Area = Half Parallelogram Area (Party)
1) How do you draw a sketch to show that the area of a right triangle is half the area of a parallelogram?
To draw a sketch to show that the area of a right triangle is half the area of a parallelogram, start by drawing a parallelogram and labeling its base and height. Then, draw a right triangle with the same base and height as the parallelogram. Next, draw a line from the top vertex of the triangle to the midpoint of the other side of the parallelogram. This line will divide the parallelogram into two equal parts, with one part being the triangle. Therefore, the area of the right triangle is half of the area of the parallelogram.
2) What is the mathematical proof for the relationship between the area of a right triangle and a parallelogram?
The mathematical proof for the relationship between the area of a right triangle and a parallelogram is based on the formula for the area of a triangle, which is 1/2 x base x height, and the formula for the area of a parallelogram, which is base x height. By drawing a sketch and comparing the two shapes, it can be shown that the area of a right triangle is half the area of a parallelogram.
3) Why is it important to understand the relationship between the area of a right triangle and a parallelogram?
Understanding the relationship between the area of a right triangle and a parallelogram is important because it allows us to easily calculate the area of a triangle if we know the area of a parallelogram with the same base and height. This relationship is also useful in real-life situations, such as in construction or engineering, where right triangles and parallelograms are commonly used shapes.
4) Can the relationship between the area of a right triangle and a parallelogram be applied to other shapes?
Yes, the relationship between the area of a right triangle and a parallelogram can be applied to other shapes. This relationship is known as the "half base times height" formula and can be used to find the area of any shape that can be divided into a right triangle and a parallelogram with the same base and height.
5) How does the relationship between the area of a right triangle and a parallelogram relate to the concept of congruence?
The relationship between the area of a right triangle and a parallelogram is closely related to the concept of congruence. If two shapes are congruent, they have the same size and shape. Therefore, if a right triangle and a parallelogram are congruent, their areas will also be equal. In other words, if two shapes have the same base and height, their areas will be equal, regardless of their specific shape. This is why the "half base times height" formula can be applied to various shapes.