Area is the measure of the size of a shape, or region of a space, it can be seen as the two-dimensional analogue of the length of a curve.
Rectangle
\usetikzlibrary{arrows.meta}
\begin{document}
\begin{tikzpicture}
\fill[blue!20, opacity=0.5] (0,0) rectangle (5,2) node[midway, opacity=1] {A};
\draw[dashed,<->,>={Latex[length=4pt]}] (0,0) -- (5,0) node[midway, below, yshift=-4] {l};
\draw[dashed,<->,>={Latex[length=4pt]}] (5,0) -- (5,2) node[midway, right, xshift=2] {w};
\end{tikzpicture}
\end{document}where is length and is width.
Square
\usetikzlibrary{arrows.meta}
\begin{document}
\begin{tikzpicture}
\begin{axis}
\addplot+ [name path=A,domain=0:1,samples=2] {x};
\addplot+ [name path=B] table {
x y
0 2
0.5 -1
1 3
};
\addplot fill between [of=A and B];
\end{axis}
\end{tikzpicture}where is the length of a side.
Trapezium
(trapezoid)
\usetikzlibrary{arrows.meta}
\begin{document}
\begin{tikzpicture}[scale=1.5]
% Define the coordinates of the trapezium's vertices
\fill[blue!20, opacity=0.5] (0,0) -- (2,0) -- (4,2) -- (-1,2) -- cycle;
\draw[dashed,<->,>={Latex[length=4pt]}] (0,0) -- (0,2) node[left, midway] {h};
\draw[dashed,<->,>={Latex[length=4pt]}] (0,0) -- (2,0) node[left, midway] {b};
\draw[dashed,<->,>={Latex[length=4pt]}] (-1,2) -- (4,2) node[left, midway] {a};
\path (0,0) -- ++(2,2) node[midway] {A};
\end{tikzpicture}
\end{document}where and are the lengths of the bases (the parallel sides), and is the height.
Parallelogram
\usetikzlibrary{arrows.meta}
\begin{document}
\begin{tikzpicture}[scale=2]
% Define the coordinates of the trapezium's vertices
\fill[blue!20, opacity=0.5] (0,0) -- (2,0) -- (3,2) -- (1,2) -- cycle;
\draw[dashed,<->,>={Latex[length=4pt]}] (2,0) -- (2,2) node[left, midway] {h};
\draw[dashed,<->,>={Latex[length=4pt]}] (0,0) -- (2,0) node[left, midway] {b};
\draw[dashed,<->,>={Latex[length=4pt]}] (1,2) -- (3,2) node[left, midway] {b};
\path (0,0) -- ++(2.5,2) node[midway] {A};
\end{tikzpicture}
\end{document}where is the length of the base and is the height.
We can observe the following:
Corollary
The parallelogram is just a special case of the trapezium (where both bases are equal in size, so the average value is just the base itself).
Corollary
The rectangle is a special case of the parallelogram, where every angle is a right angle. (or viewed as a rectangle but sheared - since a shear transformation preserves area).
Corollary
The square is just a special case of the rectangle where .
Triangle
\usetikzlibrary{arrows.meta}
\begin{document}
\begin{tikzpicture}
\coordinate (o) at (0,0);
\coordinate (A) at (-1,0);
\coordinate (B) at (4,0);
\coordinate (H) at (0,5);
\fill[blue!20, opacity=0.5] (A) -- (H) -- (B) -- cycle node[pos=0.6, above=40, opacity=1] {A};
\draw[dashed,<->,>={Latex[length=4pt]}] (o) -- (H) node[right, midway] {h};
\draw[dashed,<->,>={Latex[length=4pt]}] (A) -- (B) node[below, midway] {b};
\end{tikzpicture}
\end{document}where is the base and is height.
Ellipse
\usetikzlibrary{arrows.meta}
\begin{document}
\begin{tikzpicture}[range=-2.2:2.2]
\fill[blue!20, opacity=0.5] (0,0) ellipse [x radius=4,y radius=2] node[midway, xshift=-8, opacity=1] {A};
\draw[dashed,<->,>={Latex[length=4pt]}] (0,0) -- (4,0) node[above, midway] {a};
\draw[dashed,<->,>={Latex[length=4pt]}] (0,0) -- (0,2) node[left, midway] {b};
\path (current bounding box.south west) +(-.1,-.1) (current bounding box.north east) +(.1,.1);
\end{tikzpicture}
\end{document}where and are the lengths of the semi-major and semi-minor axes, respectively.
Circle
\usetikzlibrary{arrows.meta}
\begin{document}
\begin{tikzpicture}[scale=0.8]
\fill[blue!20, opacity=0.5] (0,0) circle (3) node[midway, xshift=-8, opacity=1] {A};
\draw[dashed,<->,>={Latex[length=4pt]}] (0,0) -- (3*0.707,3*0.707) node[left, midway] {r};
\end{tikzpicture}
\end{document}where is the radius.
Corollary
The circle is just a special case of the ellipse where .
Parabola
Definition
Every parabola is the graph of a quadratic function .
The area of any parabolic segment (region enclosed parabola and a line intersecting it) is:
\usetikzlibrary{arrows.meta}
\begin{document}
\begin{tikzpicture}[domain=-3:3, scale=2]
\coordinate (a) at (-1.89, 0.43);
\coordinate (b) at (1.59, 1.48);
\coordinate (t) at (-0.15, 3.98);
\coordinate (_) at (-0.15, 0.96);
\begin{scope}
\clip (-3,0) rectangle (3, 4.2);
\draw[black!10, thick, smooth, range=0:2] plot (\x, {-(\x*\x) + 4});
\end{scope}
\draw[black!10, domain=-3:-1.89] plot (\x, {0.3*\x + 1});
\draw[black!10, domain=1.59:3] plot (\x, {0.3*\x + 1});
\draw[black!10, dashed, very thin, domain=-3:2] plot (\x, {0.3*\x + 4.025});
\begin{scope}
\clip (a) -- (b) -- (3, 4.2) -- (-3, 4.2) -- cycle;
\fill[blue!20, opacity=0.5] (a) -- plot (\x, {-(\x*\x) + 4}) -- (b) -- cycle;
\end{scope}
\draw[dashed,<->,>={Latex[length=4pt]}] (_) -- (t) node[right, midway] {h};
\draw[dashed,<->,>={Latex[length=4pt]}] (a) -- (b) node[below, midway, xshift=8] {b};
\draw[green!60, thick, dashed] (a) -- (t) -- (b) -- cycle;
\end{tikzpicture}
\end{document}(From Quadrature_of_the_Parabola )
where the base of the inscribed triangle is the chord of the parabola, the third vertex of the triangle is at the point where the tangent line is parallel to the chord.
Alternative interpretation
\usetikzlibrary{arrows.meta} \begin{document} \begin{tikzpicture}[domain=-2:2, scale=2] \begin{scope}[transparency group] \begin{scope}[blend mode=difference] \clip (-2, 0) (-1,0) — (1,0) — (1,2) — (-1, 2) — cycle; \fill[red!20!black!90, opacity=0.8] (-1,0) — (1,0) — (1,2) — (-1, 2) — cycle; \fill[blue!20, opacity=0.5, samples=50, range=0:2] plot (\x, {2*(\x*\x)}); \end{scope} \end{scope} \draw[very thin, color=gray] (-2,0) grid (2,2); \draw[dashed,←>,>={Latex[length=4pt]}] (-1,2) — (1,2) node[above, midway] {b}; \draw[dashed,←>,>={Latex[length=4pt]}] (0,2) — (0,0) node[right, midway] {h}; \end{tikzpicture} \end{document}
where $b$ is the length of the base and $h$ is the height. Let's define $b$ to be two unit vectors, this will make the calculations easier. $$ \newcommand{\i}{\hat{\imath}} b=2\,\i $$Then the area of the red square is just $$ A_{rec} = b\,h = 2\,\i\,h $$If we manage to get the area under the parabola inside the red square, we can subtract it from the red area and get the area inside the parabola. We know that the parabola must have a function of form $$f(x)=𝑘𝑥^2 + mx + C$$We are taking it to have its vertex at the origin, so we can ignore the $mx$ and $C$ terms (there is no horizontal or vertical translation), so we're left with $$ f(x) = kx^2 $$ from our observation of the graph and $A_{rec}$ we can see $$
\begin{align} h = k,\i^2 &\implies k = \frac{h}{\i^2} \ & \implies f(x) = \frac{h}{\i^2}x^2 \end{align}
If we take the integral of this from 0 up to $\i$ (a.k.a $1$ in our toy example) we will get the right half of the red area that doesn't intersect the parabola, so we simply double this value to get the full red area.
\require{physics} \begin{align} A_{red} = 2\int_{0}^{\i} k,x^2 ,dx &=2 \cdot \eval{\frac{k,x^3}{3}}0^{\i} \ &= 2 \cdot \left( \frac{h,\i^3}{3,\i^2} - 0 \right) \ &= \frac{2,\i ,h}{3} = \frac{1}{3}bh \ &= \frac{1}{3} A{rec} \end{align}
\begin{align} A_{par} &= A_{rec} - A_{red} \ &= A_{rec} - \frac{1}{3}A_{rec} \ &= \frac{2}{3}A_{rec} = \frac{2}{3}bh \end{align}