nyuuzyou/wikis huggingface dataset:
Area Sampling/Examples/Arbitrary Example 1

== Example of Area Sampling ==
In a study of farming, it may be appropriate to use a grid of squares $5 \, \mathrm {km} \times 5 \, \mathrm {km}$ in area.
Hence one would include in the sample each of the farms lying inside each of the selected squares.
== Sources ==
* 
* 
Category:Examples of Area Sampling
Area between Radii and Curve in Polar Coordinates

== Theorem ==
Let $C$ be a curve expressed in polar coordinates $\polar {r, \theta}$ as:
:$r = \map g \theta$
where $g$ is a real function.
Let $\theta = \theta_a$ and $\theta = \theta_b$ be the two rays from the pole at angles $\theta_a$ and $\theta_b$ to the polar axis respectively.
Then the area $\AA$ between $\theta_a$, $\theta_b$ and $C$ is given by:
:$\ds \AA = \int \limits_{\theta \mathop = \theta_a}^{\theta \mathop = \theta_b} \frac {\paren {\map g \theta}^2 \rd \theta} 2$
as long as $\paren {\map g \theta}^2$ is integrable.
== Proof ==
:
Consider the area of the brown triangle.
This would be:
:$a_\triangle = \dfrac 1 2 r^2 \map \sin {\delta \theta}$
We will be using nonstandard analysis, so let $\delta \theta = \varepsilon > 0$, an infinitesimal. 
Thus:
:$a_\triangle = \dfrac 1 2 r^2 \sin \varepsilon$
Using the Power Series Expansion for Sine Function: 
== Sources ==
* : $\S 13.16$
* 
* 
Category:Area Formulas
Area between Radii and Whorls of Archimedean Spiral

== Theorem ==
Let $S$ be the Archimedean spiral defined by the equation:
:$r = a \theta$
Let $\theta = \theta_1$ and $\theta = \theta_2$ be the two rays from the pole at angles $\theta_1$ and $\theta_b$ to the polar axis respectively.
Let $R$ be the figure enclosed by:
:$\theta_1$ and $\theta_2$
:the $n$th turn of $S$ and the $n+1$th turn of $S$
The area $\AA$ of $R$ is given by:
:$\AA = a^2 \pi \paren {\theta_2 - \theta_1} \paren {\theta_2 + \theta_1 + 2 \pi \paren {2 n + 1} }$
== Proof ==
The straight line boundaries of $R$ are given as $\theta_1$ and $\theta_2$.
The corners of $R$ are located where:
:$\theta = \theta_1 + 2 n \pi$
:$\theta = \theta_2 + 2 n \pi$
:$\theta = \theta_1 + 2 \paren {n + 1} \pi$
:$\theta = \theta_2 + 2 \paren {n + 1} \pi$
== Historical Note ==
Category:Archimedean Spiral
Area between Two Non-Intersecting Chords

== Theorem ==
Let $AB$ and $CD$ be two chords of a circle whose center is at $O$ and whose radius is $r$.
:
:
Let $\alpha$ and $\theta$ be respectively the measures in radians of the angles $\angle COD$ and $\angle AOB$.
Then the area $\AA$ between the two chords is given by:
:$\AA = \dfrac {r^2} 2 \paren {\theta - \sin \theta - \alpha + \sin \alpha}$
if $O$ is not included in the area, and:
:$\AA = r^2 \paren {\pi - \dfrac 1 2 \paren {\theta - \sin \theta + \alpha - \sin \alpha} }$
if $O$ is included in the area.
== Proof ==
Let $\SS_\alpha$ be the area of the segment whose base subtends $\alpha$.
Let $\SS_\theta$ be the area of the segment whose base subtends $\theta$.
=== Case $(1)$: Center included in Area ===
Let the center $O$ be included in the area.
The area between the two chords is given by:
:the area of the whole circle
minus:
:the areas of the segments $\SS_\alpha$ and $\SS_\theta$ .
Thus:
=== Case $(2)$: Center not included in Area ===
Let $\theta \ge \alpha$.
The area between the two chords is given by:
:the area of the segment $\SS_\theta$
minus:
:the area of the segment $\SS_\alpha$.
Thus:
Category:Circles
Area contained by Apotome and Binomial Straight Line Commensurable with Terms of Apotome and in same Ratio

== Theorem ==
=== Porism ===
== Proof ==
:
Let the rectangle $AB \cdot CD$ be contained by the apotome $AB$ and the binomial straight line $CD$.
Let $CE$ be the greater term of $CD$.
Let:
:$CE$ be commensurable in length with $AF$
:$ED$ be commensurable in length with $FB$
:$CE : ED = AF : FB$
Let the "side" of the rectangle $AB \cdot CD$ be $G$.
It is to be demonstrated that $G$ is rational.
Let $H$ be a rational straight line.
Let a rectangle equal to $H^2$ be applied to $CD$ which produces $KL$ as breadth.
By definition, $KL$ is an apotome.
Let the terms of $KL$ be $KM$ and $ML$.
Let $KM$ and $ML$ be commensurable with the terms $CE$ and $ED$ of the binomial straight line $CD$.
But from :
:$CE$ and $ED$ are commensurable with the terms $AF$ and $FB$ of the apotome $CD$.
Therefore:
:$AF : FB = KM : ML$
and so:
:$AF : KM = BF : LM$
and so by :
:$AB : KL = AF : KM$
But from :
:$AF$ is commensurable in length with $KM$.
Therefore from :
:$AB$ is commensurable in length with $KL$.
From :
:$AB : KL = CD \cdot AB : CD \cdot KL$
Therefore from :
:$CD \cdot AB$ is commensurable with $CD \cdot KL$.
But $CD \cdot KL = H^2$.
Therefore $CD \cdot AB$ is commensurable with $H^2$.
But $G^2 = CD \cdot AB$.
Therefore $G^2$ is commensurable with $H^2$.
But $H^2$ is rational.
Therefore $G^2$ is also rational.
Therefore $G$ is rational straight line.
But $G^2$ is the "side" of the rectangle $AB \cdot CD$.
Hence the result.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Area inside Astroid

== Theorem ==
The area inside an astroid $H$ constructed within a circle of radius $a$ is given by:
:$\AA = \dfrac {3 \pi a^2} 8$
== Proof ==
Let $H$ be embedded in a cartesian plane with its center at the origin and its cusps positioned on the axes.
:
By symmetry, it is sufficient to evaluate the area shaded yellow and to multiply it by $4$.
By Equation of Astroid:
:$\begin{cases}
x & = a \cos^3 \theta \\
y & = a \sin^3 \theta
\end{cases}$
Thus:
Simplifying the integrand:
Thus:
== Sources ==
* : $\S 11$: Special Plane Curves: Hypocycloid with Four Cusps: $11.10$
* : Chapter $\text {B}.21$: The Cycloid: Problem $6$
* : $\S 9$: Special Plane Curves: Hypocycloid with Four Cusps: $9.10.$
Category:Astroids
Area inside Cardioid

== Theorem ==
Consider the cardioid $C$ embedded in a polar plane given by its polar equation:
:$r = 2 a \paren {1 + \cos \theta}$
The area inside $C$ is $6 \pi a^2$.
== Proof ==
Let $\AA$ denote the area inside $C$.
The boundary of $C$ is traced out where $-\pi \le \theta \le \pi$.
Thus:
== Sources ==
* : $\S 11$: Special Plane Curves: Cardioid: $11.13$
* : $\S 9$: Special Plane Curves: Cardioid: $9.13.$
Category:Cardioids
Category:Area Formulas
Area of Annulus

== Theorem ==
Let $A$ be an annulus whose inner radius is $r$ and whose outer radius is $R$.
The area of $A$ is given by:
:$\map \Area A = \pi \paren {R^2 - r^2}$
== Proof ==
:
The area of $A$ is seen to be:
:the area of the outer circle with the area of the inner circle removed.
From Area of Circle:
:the area of the outer circle is $\pi R^2$
:the area of the inner circle is $\pi r^2$
The result follows.
== Sources ==
* 
* 
* 
* 
* 
Category:Annuli
Area of Annulus as Area of Rectangle

== Theorem ==
Let $A$ be an annulus whose inner radius is $r$ and whose outer radius is $R$.
The area of $A$ is given by:
:$\map \Area A = 2 \pi \paren {r + \dfrac w 2} \times w$
where $w$ denotes the width of $A$.
That is, it is the area of the rectangle contained by:
:the width of $A$
:the circle midway in radius between the inner radius and outer radius of $A$.
== Proof ==
:
The result follows from Area of Rectangle.
== Sources ==
* 
* 
Category:Annuli
Area of Circle

== Theorem ==
The area $A$ of a circle is given by:
:$A = \pi r^2$
where $r$ is the radius of the circle.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Kepler's Proof ==
== Proof 6 ==
== Proof 7 ==
== Examples ==
== Also see ==
* Perimeter of Circle
== Sources ==
* : $\S 4$: Geometric Formulas: Circle of Radius $r$: $4.11$
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$
* 
* 
* : $\S 7$: Geometric Formulas: Circle of Radius $r$: $7.11.$
* 
* : Appendix $1$: Areas and volumes
* 
* : Appendix $1$: Areas and volumes
Category:Area of Circle
Category:Circles
Category:Area Formulas
Area of Circle/Examples/Radius 2

== Example of Area of Circle ==
Let $C$ be a circle whose radius is $2$.
Then the area of $C$ is $4 \pi$.
== Proof ==
From Area of Circle:
:$A = \pi r^2$
Hence when $r = 2$ we have:
:$A = \pi \times 2^2 = 4 \pi$
== Sources ==
* 
* 
Category:Examples of Area of Circle
Area of Circle/Kepler's Proof

== Theorem ==
== Proof ==
Let the circle of radius $r$ be divided into many sectors:
:
If they are made small enough, they can be approximated to triangles whose heights are all $r$.
Let the bases of these triangles be denoted:
:$b_1, b_2, b_3, \ldots$
From Area of Triangle in Terms of Side and Altitude, their areas are:
:$\dfrac {r b_1} 2, \dfrac {r b_2} 2, \dfrac {r b_3} 2, \ldots$
The area $\AA$ of the circle is given by the sum of the areas of each of these triangles:
But $b_1 + b_2 + b_3 + \cdots$ is the length of the circumference of the circle.
From Perimeter of Circle:
:$b_1 + b_2 + b_3 + \cdots = 2 \pi r$
Hence:
It needs to be noted that this proof is intuitive and non-rigorous.
== Historical Note ==
== Sources ==
* : Chapter $\text {A}.10$: Kepler ()
Category:Area of Circle
Area of Circle/Proof 1

== Theorem ==
== Proof ==
From Equation of Circle:
:$x^2 + y^2 = r^2$
Thus $y = \pm \sqrt {r^2 - x^2}$.
It follows that from the geometric interpretation of the definite integral:
Let $x = r \sin \theta$ (note that we can do this because $-r \le x \le r$).
Thus $\theta = \map \arcsin {\dfrac x r}$ and $\rd x = r \cos \theta \rd \theta$.
== Sources ==
* : Chapter $\text I$: $\S 1$. Area of a Circle
Category:Area of Circle
Area of Circle/Proof 3

== Theorem ==
== Proof ==
:
Construct a circle with radius $r$ and circumference $c$, whose area is denoted by $C$.
Construct a triangle with height $r$ and base $c$, whose area is denoted by $T$.
=== Lemma $1$ ===
=== Lemma $2$ ===
=== Lemma $3$ ===
=== Final Proof ===
From Lemma $2$:
:$T \ge C$
From Lemma $3$:
:$T \le C$
Therefore:
:$T \mathop = C$
and so from Lemma $1$:
:$C \mathop = T \mathop = \pi r^2$
== Historical Note ==
== Sources ==
* : Chapter $\text I$: $\S 1$. Area of a Circle
* : Chapter $\text {A}.5$: Archimedes (ca. B.C.)
* 
* 
Category:Area of Circle
Area of Circle/Proof 3/Lemma 1

== Lemma for Area of Circle: Proof 3 ==
:
Construct a circle with radius $r$ and circumference $c$, whose area is denoted by $C$.
Construct a triangle with height $r$ and base $c$, whose area is denoted by $T$.
Then:
:$T = \pi r^2$
== Proof ==
Category:Area of Circle
Area of Circle/Proof 3/Lemma 2

== Lemma for Area of Circle: Proof 3 ==
:
Construct a circle with radius $r$ and circumference $c$, whose area is denoted by $C$.
Construct a triangle with height $r$ and base $c$, whose area is denoted by $T$.
Then:
:$T \ge C$
== Proof ==
:
 $T < C$.
It should be possible to construct a regular polygon with area $P$, where $T < P < C$.
For any given regular polygon:
:$P = \dfrac {h q} 2$
where:
:$q$ is the perimeter of the regular polygon
:$h$ is the height of any given triangular part of it
:$P$ is the area.
 
Onthe one hand:
:$P > T \implies \dfrac {h q} 2 > \dfrac {r c} 2$
On the other hand:
:$0 < h < r \land 0 < q < c \implies \dfrac {h q} 2 < \dfrac {r c} 2$
Hence a contradiction is obtained.
Hence:
:$\neg T < C$
and so:
:$T \ge C$
Category:Area of Circle
Area of Circle/Proof 3/Lemma 3

== Lemma for Area of Circle: Proof 3 ==
:
Construct a circle with radius $r$ and circumference $c$, whose area is denoted by $C$.
Construct a triangle with height $r$ and base $c$, whose area is denoted by $T$.
Then:
:$T \le C$
== Proof ==
:
 $T > C$.
It should be possible to construct a regular polygon with area $P$, where $C < P < T$.
From Area of Polygon by Inradius and Perimeter:
:$P = \dfrac {h q} 2$
where:
:$q$ is the perimeter of the regular polygon
:$h$ is the inradius of the regular polygon
:$P$ is the area.
as each triangle has:
:base $B = \dfrac q n$
:area $A = \dfrac {h q} {2 n}$
And with $n$ triangles we get:
:$P = \dfrac {h q} 2$
On one hand:
:$P < T \implies \dfrac {h q} 2 < \dfrac {r c} 2$
On the other hand:
:$0 < h = r \land 0 < c < q \implies \dfrac {h q} 2 > \dfrac {r c} 2$
Hence a contradiction is obtained.
Hence:
:$\neg T > C$.
and so:
:$T \le C$.
Category:Area of Circle
Area of Circle/Proof 6

== Theorem ==
== Proof ==
From Equation of Circle:
:$x^2 + y^2 = r^2$
Let $A$ be the area of the circle whose equation is given by $x^2 + y^2 = r^2$.
We have that:
:$y = \pm \sqrt {r^2 - x^2}$
For the upper half of the circle:
:$y = +\sqrt {r^2 - x^2}$
Thus for the right hand half of the upper half of the circle:
Hence the result.
Category:Area of Circle
Area of Circle/Proof 7

== Theorem ==
== Proof ==
By the method of exhaustion:
:
== Construction ==
For step $1$ of the construction, construct a circle $C$ and a diameter of $C$, thereby dividing the circumference of $C$ into $2$ arcs.
Let $A$ be the area of $C$ and $r$ be the radius of $C$.
For $n > 1$, step $n$ of the construction consists of constructing a regular $2^n$-gon $P_n$ inscribed in $C$.
For each arc $EB$ of $C$ which has not been subdivided, construct the midpoint $G$ of arc $EB$ and join segments $\overline{EG}$ and $\overline{GB}$.
=== Lemma: The area of $P_n$ converges to $A$ as $n \to \infty$ ===
Let $E_n = A - P_n$ be the area outside of $P_n$ and inside of $C$.
As this area is divided into $2^n$ disjoint, congruent sectors, let $e_n = \dfrac {E_n} {2^n}$ be the area of one such sector.
Consider a sector with endpoints $E$ and $B$, with $G$ the midpoint of the bounding arc.
Construct a line $\ell$ through $G$ parallel to $\overline{EB}$.
Construct perpendiculars to $\ell$ through $E$ and $B$ and let them meet $\ell$ at $H$ and $I$ respectively.
Then quadrilateral $EHIB$ is a rectangle, as it has four right angles by construction.
Therefore, its area is twice that of $\triangle EGB$. 
:
We have:
It can be shown by induction that $E_n$ is bounded above by the geometric sequence $\paren {\dfrac 1 2}^n E_1$ which converges to $0$.
Since each term of $E_n$ is an area, it is bounded below by the constant sequence $0$, which converges to $0$.
By the Squeeze Theorem, $E_n$ converges to $0$.
By Combination Theorem for Sequences, $P_n$ converges to $A$. 
Observe that $P_n$ is the union of $2^n$ congruent isosceles triangles whose bases are the edges of $P_n$, whose legs are radii of $C$, and which have a common vertex at the center of $C$.
Thus if $T$ is the area of one of these triangles, the area of $P_n$ is $2^n T$. 
For each such triangle, the angle at the vertex is $\dfrac {2 \pi} {2^n}$.
By Area of Isosceles Triangle, the area of each triangle is $\dfrac 1 2 r^2 \map \sin {\dfrac {2 \pi} {2^n} }$.
Thus, the area of $P_n$ is $2^{n - 1} r^2 \map \sin {\dfrac {2 \pi} {2^n} }$.
By the Lemma, the area of $C$ is then:
:$\ds \lim_{n \mathop \to \infty} 2^{n - 1} r^2 \map \sin {\dfrac \pi {2^{n - 1} } }$
Let $m = \dfrac \pi {2^{n-1} }$.
Because $m \to 0$ as $n \to \infty$, this leads to:
== Historical Note ==
== Sources ==
* : $\S 2$: Continuum Property: $\S 2.1$: Achilles and the tortoise
Category:Area of Circle
Area of Ellipse

== Theorem ==
Let $K$ be an ellipse whose major axis is of length $2 a$ and whose minor axis is of length $2 b$.
The area $\AA$ of $K$ is given by:
:$\AA = \pi a b$
== Proof 1 ==
== Proof 2 ==
== Historical Note ==
== Sources ==
* : $\S 4$: Geometric Formulas: Ellipse of Semi-major Axis $a$ and Semi-minor Axis $b$: $4.22$
* : Chapter $\text {A}.5$: Archimedes (ca. B.C.)
* 
* 
* : $\S 7$: Geometric Formulas: Ellipse of Semi-major Axis $a$ and Semi-minor Axis $b$: $7.22.$
Category:Area of Ellipse
Category:Ellipses
Category:Area Formulas
Area of Ellipse/Proof 1

== Theorem ==
== Proof ==
Let $K$ be an ellipse aligned in a cartesian plane in reduced form.
Then from Equation of Ellipse in Reduced Form:
:$\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$
Thus:
:$y = \pm b \sqrt {1 - \dfrac {x^2} {a^2} }$
From the geometric interpretation of the definite integral:
Let $x = a \sin \theta$ (note that we can do this because $-a \le x \le a$).
Thus:
:$\theta = \map \arcsin {\dfrac x a}$
and:
:$\d x = a \cos \theta \rd \theta$
Then:
Category:Area of Ellipse
Area of Ellipse/Proof 2

== Theorem ==
== Proof ==
Let $K$ be an ellipse aligned in a cartesian plane in reduced form.
Then from Equation of Ellipse in Reduced Form:
:$\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$
Thus:
:$y = \pm \dfrac b a \sqrt {a^2 - x^2}$
Consider a circle of radius $a$ whose center is at the origin.
From Equation of Circle center Origin, its equation is given by:
:$x^2 + y^2 = a^2$
and so:
:$y = \pm \sqrt {a^2 - x^2}$
The formulas show that each ordinate of the ellipse is $\dfrac b a$ the ordinate of the circle.
Since the same thing is true of the vertical chords:
== Historical Note ==
== Sources ==
* : Chapter $\text {A}.14$: Cavalieri ()
Category:Area of Ellipse
Area of Equilateral Triangle

== Theorem ==
Let $T$ be an equilateral triangle.
Let the length of one side of $T$ be $s$.
Let $\AA$ be the area of $T$.
Then:
:$\AA = \dfrac {s^2 \sqrt 3} 4$
== Proof ==
:
From Area of Triangle in Terms of Two Sides and Angle:
:$\AA = \dfrac {s^2} 2 \sin 60 \degrees$
From Sine of $60 \degrees$:
:$\sin 60 \degrees = \dfrac {\sqrt 3} 2$
The result follows.
Category:Areas of Triangles
Category:Equilateral Triangles
Area of Isosceles Triangle

== Theorem ==
Let $\triangle ABC$ be an isosceles triangle whose apex is $A$.
Let $\theta$ be the angle of the apex $A$.
Let $r$ be the length of a leg of $\triangle ABC$.
Then the area $\AA$ of $\triangle ABC$ is given by:
:$\AA = \dfrac 1 2 r^2 \sin \theta$
== Proof 1 ==
== Proof 2 ==
Category:Areas of Triangles
Category:Isosceles Triangles
Category:Area of Isosceles Triangle
Area of Isosceles Triangle in terms of Sides

== Theorem ==
Let $\triangle ABC$ be an isosceles triangle whose apex is $A$.
Let $r$ be the length of a leg of $\triangle ABC$.
Let $b$ be the length of the base of $\triangle ABC$.
Then the area $\AA$ of $\triangle ABC$ is given by:
:$\AA = \dfrac b 4 \sqrt {4 r^2 - b^2}$
== Proof ==
:
Let $h$ be the height of $\triangle ABC$.
Category:Areas of Triangles
Category:Isosceles Triangles
Area of Lobe of Lemniscate of Bernoulli

== Theorem ==
Consider the lemniscate of Bernoulli $M$ embedded in a Cartesian plane such that its foci are at $\tuple {a, 0}$ and $\tuple {-a, 0}$ respectively.
Let $O$ denote the origin.
The area of one lobe of $M$ is $a^2$.
== Proof ==
By the definition of the lemniscate of Bernoulli, we have that the polar equation of $M$ is:
:$r^2 = 2 a^2 \cos 2 \theta$
Let $\AA$ denote the area of one lobe of $M$.
The boundary of the right hand lobe of $M$ is traced out where $-\dfrac \pi 2 \le 2 \theta \le \dfrac \pi 2$.
Thus:
== Sources ==
* : $\S 11$: Special Plane Curves: Lemniscate: $11.4$
* : $\S 9$: Special Plane Curves: Lemniscate: $9.4.$
Category:Lemniscate of Bernoulli
Area of Loop of Folium of Descartes

== Theorem ==
Consider the folium of Descartes $F$, given in parametric form as:
:$\begin {cases} x = \dfrac {3 a t} {1 + t^3} \\ y = \dfrac {3 a t^2} {1 + t^3} \end {cases}$
The area $\AA$ of the loop of $F$ is given as:
:$\AA = \dfrac {3 a^2} 2$
== Proof ==
From Behaviour of Parametric Equations for Folium of Descartes according to Parameter we have that the loop is traversed for $0 \le t < +\infty$.
We convert the parametric equation to polar form:
Then we have:
Then:
which leads us to:
Hence the result.
== Sources ==
* : $\S 11$: Special Plane Curves: Folium of Descartes: $11.26$
* : $\S 9$: Special Plane Curves: Folium of Descartes: $9.26.$
* 
Category:Folium of Descartes
Area of Parallelogram

== Theorem ==
The area of a parallelogram equals the product of one of its bases and the associated altitude.
== Proof ==
There are three cases to be analysed: the square, the rectangle and the general parallelogram.
=== Square ===
=== Rectangle ===
=== Parallelogram ===
== Sources ==
* 
* 
Category:Area of Parallelogram
Category:Parallelograms
Category:Areas of Quadrilaterals
Category:Area Formulas
Area of Parallelogram/Parallelogram

== Theorem ==
Let $ABCD$ be a parallelogram whose adjacent sides are of length $a$ and $b$ enclosing an angle $\theta$.
The area of $ABCD$ equals the product of one of its bases and the associated altitude:
where:
:$b$ is the side of $ABCD$ which has been chosen to be the base
:$h$ is the altitude of $ABCD$ from $b$.
== Proof ==
:
Let $ABCD$ be the parallelogram whose area is being sought.
Let $F$ be the foot of the altitude from $C$
Also construct the point $E$ such that $DE$ is the altitude from $D$ (see figure above).
Extend $AB$ to $F$.
Then:
Thus:
:$\triangle AED \cong \triangle BFC \implies \map \Area {AED} = \map \Area {BFC}$
So:
== Sources ==
* : $\S 4$: Geometric Formulas: Parallelogram of Altitude $h$ and Base $b$: $4.3$
* : $\S 7$: Geometric Formulas: Parallelogram of Altitude $h$ and Base $b$: $7.3.$
* 
* : Appendix $1$: Areas and volumes
* : Appendix $1$: Areas and volumes
Category:Area of Parallelogram
Area of Parallelogram/Rectangle

== Theorem ==
The area of a rectangle equals the product of one of its bases and the associated altitude.
== Proof ==
Let $ABCD$ be a rectangle.
:
Then construct the square with side length:
:$\map \Area {AB + BI}$
where $BI = BC$, as shown in the figure above.
Note that $\square CDEF$ and $\square BCHI$ are squares.
Thus:
:$\square ABCD \cong \square CHGF$
Since congruent shapes have the same area:
:$\map \Area {ABCD} = \map \Area {CHGF}$ (where $\map \Area {FXYZ}$ denotes the area of the plane figure $FXYZ$).
Let $AB = a$ and $BI = b$.
Then the area of the square $AIGE$ is equal to:
== Sources ==
* : Chapter $\text I$: $\S 1$. Area of a Circle
* : $\S 4$: Geometric Formulas: Rectangle of Length $b$ and Width $a$: $4.1$
* 
* 
* : $\S 7$: Geometric Formulas: Rectangle of Length $b$ and Width $a$: $7.1.$
* : Appendix $1$: Areas and volumes
* : Appendix $1$: Areas and volumes
Category:Area of Parallelogram
Area of Parallelogram/Square

== Theorem ==
The area of a square equals the product of one of its bases and the associated altitude.
== Proof ==
:
From Area of Square:
:$\paren {ABCD} = a^2$
where $a$ is the length of one of the sides of the square.
The altitude of a square is the same as its base.
Hence the result.
Category:Area of Parallelogram
Area of Parallelogram from Determinant

== Theorem ==
Let $OABC$ be a parallelogram in the Cartesian plane whose vertices are located at:
The area of $OABC$ is given by:
:$\map \Area {OABC} = \begin {vmatrix} a & b \\ c & d \end {vmatrix}$
where $\begin {vmatrix} a & b \\ c & d \end {vmatrix}$ denotes the determinant of order $2$.
== Proof ==
Arrange for the parallelogram to be situated entirely in the first quadrant.
:
First need we establish that $OABC$ is actually a parallelogram in the first place.
Indeed:
Thus:
:$OA = CB$
:$OC = AB$
and it follows from Opposite Sides Equal implies Parallelogram that $OABC$ is indeed a parallelogram.
Now we calculate the area of $OABC$ as equal to:
:the area occupied by the large rectangle in the diagram above
less:
:the $4$ triangles
:the $2$ small rectangles.
Thus:
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace
Category:Area of Parallelogram‎
Area of Parallelogram in Complex Plane

== Theorem ==
Let $z_1$ and $z_2$ be complex numbers expressed as vectors.
Let $ABCD$ be the parallelogram formed by letting $AD = z_1$ and $AB = z_2$.
Then the area $\AA$ of $ABCD$ is given by:
:$\AA = z_1 \times z_2$
where $z_1 \times z_2$ denotes the cross product of $z_1$ and $z_2$.
== Proof ==
:
From Area of Parallelogram:
:$\AA = \text{base} \times \text{height}$
In this context:
:$\text {base} = \cmod {z_2}$
and:
:$\text {height} = \cmod {z_1} \sin \theta$
The result follows by definition of complex cross product.
== Sources ==
* : $1$: Complex Numbers: Dot and Cross Product: $4.$
* : $1$: Complex Numbers: Solved Problems: Dot and Cross Product: $41$
Category:Area of Parallelogram
Category:Complex Cross Product
Category:Geometry of Complex Plane