nyuuzyou/wikis huggingface dataset:
Area of Quadrilateral in Determinant Form

== Theorem ==
Let $A = \tuple {x_1, y_1}$, $B = \tuple {x_2, y_2}$, $C = \tuple {x_3, y_3}$ and $D = \tuple {x_4, y_4}$ be points in the Cartesian plane.
Let $A$, $B$, $C$ and $D$ form the vertices of a quadrilateral.
The area $\AA$ of $\Box ABCD$ is given by:
:$\AA = \dfrac 1 2 \paren {\size {\paren {\begin{vmatrix}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1 \\
\end{vmatrix} } } + \size {\paren {\begin{vmatrix}
x_1 & y_1 & 1 \\
x_4 & y_4 & 1 \\
x_3 & y_3 & 1 \\
\end{vmatrix} } } }$
== Proof ==
$\Box ABCD$ can be divided into $2$ triangles: $\triangle ABC$ and $\triangle ADC$.
Hence $\AA$ is the sum of the areas of $\triangle ABC$ and $\triangle ADC$.
From Area of Triangle in Determinant Form:
Hence the result.
== Examples ==
Category:Areas of Quadrilaterals
Category:Area of Quadrilateral in Determinant Form
Area of Quadrilateral in Determinant Form/Examples/Vertices at (2, -1), (4, 3), (-1, 2), (-3, -2)

== Example of Area of Quadrilateral in Determinant Form ==
Let $Q$ be a quadrilateral embedded in the cartesian plane with vertices at $\tuple {2, -1}$, $\tuple {4, 3}$, $\tuple {-1, 2}$ and $\tuple {-3, -2}$.
The area of $Q$ is given by:
:$\map \Area Q = 18$
== Proof ==
From Area of Quadrilateral in Determinant Form:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: The Dot and Cross Product: $115$
Category:Area of Quadrilateral in Determinant Form
Area of Quadrilateral with Given Sides is Greatest when Quadrilateral is Cyclic

== Theorem ==
Let $Q$ be a quadrilateral whose sides are $a$, $b$, $c$ and $d$.
Let $\AA$ be the area of $Q$.
Then:
:$Q$ is a cyclic quadrilateral
:
:$\AA$ is the greatest area possible for one with sides $a$, $b$, $c$ and $d$.
== Proof ==
From Bretschneider's Formula:
:$\AA = \sqrt {\paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d} - a b c d \map {\cos^2} {\dfrac {\alpha + \gamma} 2} }$
where $s$ is the semiperimeter of $Q$.
Hence $\AA$ is greatest exactly when:
:$\map {\cos^2} {\dfrac {\alpha + \gamma} 2} = 0$
That is, when:
:$\map \cos {\dfrac {\alpha + \gamma} 2} = 0$
When this happens:
:$\AA = \sqrt {\paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d} }$
which is Brahmagupta's Formula.
That is, $\AA$ is the area of a cyclic quadrilateral with sides $a$, $b$, $c$ and $d$.
This happens exactly when:
:$\dfrac {\alpha + \gamma} 2 = \dfrac \pi 2$
That is, when:
:$\alpha + \gamma = \pi = 180 \degrees$
Hence the result from Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles.
Category:Cyclic Quadrilaterals
Area of Regular Hexagon

== Theorem ==
Let $H$ be a regular hexagon.
Let the length of one side of $H$ be $s$.
Let $\AA$ be the area of $H$.
Then:
:$\AA = \dfrac {3 \sqrt 3} 2 s^2$
== Proof 1 ==
== Proof 2 ==
Category:Area of Regular Hexagon
Category:Hexagons
Area of Regular Hexagon/Proof 1

== Theorem ==
== Proof ==
From Regular Hexagon is composed of Equilateral Triangles, it follows that a regular hexagon can be dissected into six congruent equilateral triangles:
:
Let $\AA_T$ be the area of the bottom triangle. 
Then by Area of Equilateral Triangle:
: $ \AA_T = \dfrac{\sqrt 3} 4 s^2 $
As $H$ consists of six congruent triangles, it follows that:
: $ \AA = 6\AA_T = \dfrac{6\sqrt 3} 4 s^2 = \dfrac{3\sqrt3} 2 s^2 $
Category:Area of Regular Hexagon
Area of Regular Polygon

== Theorem ==
Let $P$ be a regular $n$-sided polygon whose side length is $b$.
Then the area of $P$ is given by:
:$\Box P = \dfrac 1 4 n b^2 \cot \dfrac \pi n$
where $\cot$ denotes cotangent.
== Proof ==
:
Let $H$ be the center of the regular $n$-sided polygon $P$.
Let one of its sides be $AB$.
Consider the triangle $\triangle ABH$.
As $P$ is regular and $H$ is the center, $AH = BH$ and so $\triangle ABH$ is isosceles.
Thus $b = AB$ is the base of $\triangle ABH$.
Let $h = GH$ be its altitude.
See the diagram.
Then:
The full polygon $P$ is made up of $n$ such triangles, each of which has the same area as $\triangle ABH$.
Hence:
:$\Box P = \dfrac 1 4 n b^2 \cot \dfrac \pi n$
== Sources ==
* : $\S 4$: Geometric Formulas: Regular Polygon of $n$ Sides Each of Length $b$: $4.9$
* : $\S 7$: Geometric Formulas: Regular Polygon of $n$ Sides Each of Length $b$: $7.9.$
Category:Regular Polygons
Category:Area Formulas
Area of Regular Polygon by Circumradius

== Theorem ==
Let $P$ be a regular $n$-gon.
Let the circumradius of $P$ be $r$.
Then the area $\AA$ of $P$ is given by:
:$\AA = \dfrac 1 2 n r^2 \sin \dfrac {2 \pi} n$
== Proof ==
:
From Regular Polygon is composed of Isosceles Triangles, let $\triangle OAB$ be one of the $n$ isosceles triangles that compose $P$.
Then $\AA$ is equal to $n$ times the area of $\triangle OAB$.
Let $d$ be the length of one side of $P$.
Then $d$ is the length of the base of $\triangle OAB$.
Let $h$ be the altitude of $\triangle OAB$.
The angle $\angle AOB$ is equal to $\dfrac {2 \pi} n$.
Then:
:$(1): \quad h = r \cos \dfrac \pi n$
:$(2): \quad d = 2 r \sin \dfrac \pi n$
So:
== Sources ==
* : $\S 4$: Geometric Formulas: Regular Polygon of $n$ Sides Inscribed in Circle of Radius $r$: $4.17$
* : $\S 7$: Geometric Formulas: Regular Polygon of $n$ Sides Inscribed in Circle of Radius $r$: $7.17.$
Category:Regular Polygons
Category:Circumradii
Category:Area Formulas
Area of Regular Polygon by Inradius

== Theorem ==
Let $P$ be a regular $n$-gon.
Let $C$ be an incircle of $P$.
Let the radius of $C$ be $r$.
Then the area $\AA$ of $P$ is given by:
:$\AA = n r^2 \tan \dfrac \pi n$
== Proof ==
:
From Regular Polygon is composed of Isosceles Triangles, let $\triangle OAB$ be one of the $n$ isosceles triangles that compose $P$.
Then $\AA$ is equal to $n$ times the area of $\triangle OAB$.
Also, $r$ is the length of the altitude of $\triangle OAB$.
Let $d$ be the length of one side of $P$.
Then $d$ is the length of the base of $\triangle OAB$.
The angle $\angle AOB$ is equal to $\dfrac {2 \pi} n$.
Then $d = 2 r \tan \dfrac \pi n$
So:
== Sources ==
* : $\S 4$: Geometric Formulas: Regular Polygon of $n$ Sides Circumscribing a Circle of Radius $r$: $4.19$
* : $\S 7$: Geometric Formulas: Regular Polygon of $n$ Sides Circumscribing a Circle of Radius $r$: $7.19.$
Category:Regular Polygons
Category:Area Formulas
Area of Right Parabolic Segment

== Theorem ==
Let $ABC$ be a right parabolic segment where:
:$AB$ is the defining chord $\LL$ of $ABC$
:$C$ is the vertex of the defining parabola $\PP$ of $ABC$.
:
The area $\AA$ of $ABC$ is given by:
:$\AA = \dfrac {2 a b} 3$
where:
:$a$ is the length of the line segment $CF$, where $F$ is the point at which the axis of $\PP$ intersects $AB$
:$b$ is the length of the line segment $AB$.
== Proof ==
Construct the triangle $\triangle ABC$:
:
From Quadrature of Parabola:
:$\AA = \dfrac 4 3 \triangle ABC$
From Area of Triangle in Terms of Side and Altitude, the area of $\triangle ABC$ equals $\dfrac {a b} 2$.
The result follows.
== Sources ==
* : $\S 4$: Geometric Formulas: Segment of a Parabola: $4.24$
* : $\S 7$: Geometric Formulas: Segment of a Parabola: $7.24.$
* 
Category:Right Parabolic Segments
Category:Area Formulas
Area of Sector

== Theorem ==
Let $\CC = ABC$ be a circle whose center is $A$ and with radii $AB$ and $AC$.
Let $BAC$ be the sector of $\CC$ whose angle between $AB$ and $AC$ is $\theta$.
:
Then the area $\AA$ of sector $BAC$ is given by:
:$\AA = \dfrac {r^2 \theta} 2$
where:
:$r = AB$ is the length of the radius of the circle
:$\theta$ is measured in radians.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 4$: Geometric Formulas: Sector of Circle of Radius $r$: $4.13$
* 
* 
* : $\S 7$: Geometric Formulas: Sector of Circle of Radius $r$: $7.13.$
* 
Category:Area of Sector
Category:Sectors of Circles
Category:Area Formulas
Area of Sector/Proof 1

== Theorem ==
== Proof ==
Let $\CC$ be a circle of radius $r$ whose center $A$ is at the origin and with radii $AB$ and $AC$.
Let $B$ and $C$ be arbitrary points on the circumference of $\CC$.
Let $C$ be positioned at $\tuple {0, r}$ so that $AC$ coincides with the $y$-axis.
Let $BAC$ be the sector of $\CC$ whose angle between $AB$ and $AC$ is $\theta$.
:
We have that $\CC$ is centered at the origin.
From Equation of Circle center Origin, the equation of $\CC$ is given by:
:$r^2 = x^2 + y^2$
Let $B$ have co-ordinates:
:$\tuple {x_0, y_0} = \tuple {x_0, \sqrt {r^2 - {x_0}^2} }$
Triangle $\triangle ADB$ is a right triangle whose leg $DB$ is opposite $\theta$, and whose hypotenuse is $AB$.
Hence:
Consider the line $AB$, which by definition passes through $A = \tuple {0, 0}$ and $B = \tuple {x_0, \sqrt {r^2 - {x_0}^2} }$.
Let:
:$\Delta y$ be the change in $y$ between $A$ and $B$
:$\Delta x$ be the change in $x$
:$m$ be the slope of $AB$.
Let $K$ be the region between $\CC$, $AC$ and $AB$, from $0$ to $x_0$, orange in the diagram above.
This is the sector of $\CC$ whose area we are to find.
Let $\AA$ be the area of $K$.
Then:
Category:Area of Sector
Area of Sector/Proof 2

== Theorem ==
== Proof ==
From Area of Circle, the area of $\CC$ is $\pi r^2$.
From Measurement of Full Angle, the angle within $\CC$ is $2 \pi$.
The fraction of the area of $\CC$ within the sector $BAC$ is therefore $\pi r^2 \times \dfrac \theta {2 \pi}$.
Hence the result.
Category:Area of Sector
Area of Segment of Circle

== Theorem ==
Let $C$ be a circle of radius $r$.
Let $S$ be a segment of $C$ such that its base subtends an angle of $\theta$ at the center of the circle.
Then the area $\AA$ of $S$ is given by:
:$\AA = \dfrac 1 2 r^2 \paren {\theta - \sin \theta}$
where $\theta$ is measured in radians.
== Proof ==
:
Let $BDCE$ be the segment $S$.
Let $b$ be the length of the base of $S$.
Let $BACE$ be the sector of $C$ whose angle is $\theta$.
The $\AA$ is equal to the area of $BACE$ minus the area of the isosceles triangle $\triangle ABC$ whose base is $b$.
Let $h$ be the altitude of $\triangle ABC$.
From Area of Sector, the area of sector $BACE$ is $\dfrac 1 2 r^2 \theta$.
From Area of Isosceles Triangle, the area of $\triangle ABC$ is $\dfrac 1 2 r^2 \sin \theta$.
Thus:
== Sources ==
* : $\S 4$: Geometric Formulas: Geometric Formulas: Segment of Circle of Radius $r$: $4.21$
* 
* 
* : $\S 7$: Geometric Formulas: Segment of Circle of Radius $r$: $7.21.$
Category:Circles
Category:Area Formulas
Area of Smallest Rectangle accommodating Re-Entrant Knight's Tour

== Theorem ==
The area of the smallest rectangular chessboard on which a re-entrant knight's tour is possible is $30$ squares.
This can be configured either as a $5 \times 6$ chessboard or a $3 \times 10$ chessboard.
== Proof ==
== Sources ==
* : $30$
* : $30$
Category:30
Category:Knight's Tours
Area of Spherical Lune

== Theorem ==
Let $S$ be a sphere of radius $r$.
Let $\CC_1$ and $\CC_2$ be two great circles on $S$ such that the spherical angle between $\CC_1$ and $\CC_2$ is $\theta$.
Let $\LL$ be the lune described by $\CC_1$ and $\CC_2$.
Then the area of $\LL$ is given by:
:$\map \Area \LL = 2 r^2 \theta$
or, if $\theta$ is measured in degrees:
:$\map \Area \LL = \dfrac {4 \pi r^2 \theta} {360}$
== Proof ==
== Sources ==
* 
* 
Category:Lunes (Spherical Geometry)
Area of Spherical Polygon

== Theorem ==
Let $P$ be a spherical polygon with $n$ sides on the surface of a sphere $S$.
The area $\AA$ of $P$ is given by:
:$\AA = \dfrac {\pi r^2 E} {180}$
where:
:$r$ is the radius of $S$
:$E$ is the spherical excess of $P$.
== Proof ==
== Sources ==
* 
* 
Category:Spherical Polygons
Category:Area Formulas
Area of Spherical Triangle

== Theorem ==
Let $T$ be a spherical triangle on the surface of a sphere $S$.
The area $\AA$ of $T$ is given by:
:$\AA = \dfrac {\pi r^2 E} {180}$
where:
:$r$ is the radius of $S$
:$E$ is the spherical excess of $T$.
== Proof ==
== Sources ==
* : $\S 4$: Geometric Formulas: Spherical Triangle of Angles $A, B, C$ on Sphere of Radius $r$: $4.44$
* 
* 
* : $\S 7$: Geometric Formulas: Spherical Triangle of Angles $A, B, C$ on Sphere of Radius $r$: $7.44.$
Category:Spherical Triangles
Category:Area Formulas
Area of Square

== Theorem ==
A square has an area of $L^2$ where $L$ is the length of a side of the square.
Thus we have that the area is a function of the length of the side:
:$\forall L \in \R_{\ge 0}: \map \Area L = L^2$
where it is noted that the domain of $L$ is the set of non-negative real numbers.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $2 \text B$: The Meaning of the Term Function of One Independent Variable: Example $2.2$
Category:Areas of Quadrilaterals
Category:Area of Square
Area of Square/Proof 1

== Theorem ==
== Proof ==
=== Integer Side Length ===
In the case where $L = 1$, the statement follows from the definition of area.
If $L \in \N, L > 1$, then we can divide the square into smaller squares, each of side length one.
Since there will be $L$ squares of side length one on each side, it follows that there will be $L \cdot L = L^2$ squares of side length one.
Thus, the area of the square of side length $L$ is $L^2 \cdot 1 = L^2$.
=== Rational Side Length ===
Let $A$ be the area of the square $S$ with side length $L$.
If $L$ is a rational number, then:
:$\exists p, q \in \N: L = \dfrac p q$
Create a square of side length $p$.
From the integer side length case, its area equals $p^2$.
Divide the sides into $q$ equal parts.
Thus the square of side length $p$ is divided into $q^2$ small squares.
As they all have side length $\dfrac p q$, each of them equals $S$.
It follows arithmetically:
:$A = \dfrac {p^2} {q^2} = \paren {\dfrac p q}^2 = L^2$
=== Irrational Side Length ===
Let $L$ be an irrational number.
Then from Rationals are Everywhere Dense in Topological Space of Reals we know that within an arbitrarily small distance $\epsilon$ from $L$, we can find a rational number less than $L$ and a rational number greater than $L$.
In formal terms, we have:
:$\forall \epsilon > 0: \exists A, B \in \Q_+: A < L < B: \left|{A - L}\right| < \epsilon, \left|{B - L}\right| < \epsilon$
Thus:
: $\displaystyle \lim_{\epsilon \to 0^+} A = L$
: $\displaystyle \lim_{\epsilon \to 0^+} B = L$
Since a square of side length $B$ can contain a square of side length $L$, which can in turn contain a square of side length $A$, then:
: $\operatorname {area} \Box B \ge \operatorname {area} \Box L \ge \operatorname {area}\Box A$
By the result for rational numbers:
: $\operatorname {area}\Box B = B^2$
: $\operatorname {area}\Box A = A^2$
We also note that:
: $\displaystyle \lim_{B \to L} B^2 = L^2 = \lim_{A \to L} A^2$
Thus:
: $\displaystyle \lim_{B \to L} \operatorname {area} \Box B = \lim_{B \to L} B^2 = L^2$
: $\displaystyle \lim_{A \to L} \operatorname {area} \Box A = \lim_{A \to L} A^2 = L^2$
Finally:
: $L^2 \ge \operatorname {area}\Box L \ge L^2$
so:
: $\operatorname {area}\Box L = L^2$
Category:Area of Square
Area of Square/Proof 2

== Theorem ==
== Proof ==
Let $\Box ABCD$ be a square whose side $AB$ is of length $L$.
Let $\Box EFGH$ be a square whose side $EF$ is of length $1$.
:
From the Axioms of Area, the area of $\Box EFGH$ is $1$.
By definition, $AB : EF = L : 1$.
From Similar Polygons are composed of Similar Triangles, the ratio of the areas of $\Box ABCD$ to $\Box EFGH$ is the duplicate ratio of the ratio of $AB$ to $EF$.
Thus by definition of duplicate ratio:
: $\Box ABCD : \Box EFGH = \left({AB : EF}\right)^2$
That is:
: $\dfrac {\Box ABCD} {\Box EFGH} = \left({\dfrac L 1}\right)^2 = L^2$
That is, the area of $\Box ABCD$ has $L^2$ as many units as $\Box EFGH$.
Hence the result.
Category:Area of Square
Area of Square/Proof 3

== Theorem ==
== Proof ==
Let a square have a side length $a \in \R$.
This square is equivalent to the area under the graph of $\map f x = a$ from $0$ to $a$.
Thus from the geometric interpretation of the definite integral, the area of the square will be the integral:
:$\ds A = \int_0^a a \rd l$
Thus:
=== Warning ===
This proof is circular.
The use of the definite integral to represent area is based on the fact that the area of a rectangle is the product of the rectangle's width and height.
That fact is in turn derived from this one.
However, this demonstration neatly parallels the integration based proofs of the areas of other figures, for example Area of Circle.
Category:Area of Square
Area of Square on Greater Segment of Straight Line cut in Extreme and Mean Ratio

== Theorem ==
== Proof ==
:
Let the line segment $AB$ be cut in extreme and mean ratio at the point $C$.
Let $AC$ be the greater segment.
Let the straight line $AD$ be produced in a straight line with $CA$.
Let $AD = \dfrac {AB} 2$.
It is to be demonstrated that:
:$CD^2 = 5 \cdot AD^2$
Let the squares $AE$ and $DF$ be drawn on $AB$ and $DC$.
Let the figure in $DF$ be drawn.
Let $FC$ be produced to $G$.
We have that $AB$ has been cut in extreme and mean ratio at $C$.
Therefore from:
:
and:
:
it follows that:
:$AB \cdot BC = AC^2$
From the construction:
:$CE = AB \cdot BC$
:$FH = AC^2$
Therefore:
:$CE = FH$
We have that:
:$BA = 2 \cdot AD$
while:
:$BA = KA$
:$AD = AH$
Therefore:
:$KA = 2 \cdot AH$
But from 
:$KA : AH = CK : CH$
Therefore:
:$CK = 2 \cdot CH$
But:
:$LH + HC = 2 \cdot CH$
Therefore:
:$KC = LH + HC$
But:
:$CE = HF$
Therefore $AE$ equals the gnomon $MNO$.
We have that:
:$BA = 2 \cdot AD$
Therefore:
:$BA^2 = 4 \cdot A^2$
That is:
:$AE = 4 \cdot DH$
But:
:$AE = MNO$
Therefore:
:$MNO = 4 \cdot AP$
Therefore:
:$DF = 5 \cdot AP$
But:
:$DF$ is the square on $DC$
and:
:$AP$ is the square on $DA$.
Hence the result.
== Sources ==
* : Book $\text{XIII}$. Propositions
Category:Golden Mean
Area of Square on Lesser Segment of Straight Line cut in Extreme and Mean Ratio

== Theorem ==
== Proof ==
:
Let the line $AB$ be cut in extreme and mean ratio at the point $C$.
Let $AC$ be the greater segment.
Let $AC$ be bisected at $D$.
It is to be demonstrated that $BD^2 = 5 \cdot DC^2$.
Let the squares $AE$ be described on $AB$.
Let the figure be drawn as above.
We have that:
: $AC = 2 \cdot DC$
Therefore:
: $AC^2 = 4 \cdot DC^2$
That is:
:$RS = 4 \cdot FG$
We have by definition of extreme and mean ratio that:
:$AB \cdot BC = AC^2$
and:
:$CE = AB \cdot BC$
Therefore:
:$CE = RS$
But:
:$RS = 4 \cdot FG$
Therefore:
:$CE = 4 \cdot FG$
We have that:
:$AD = DC$
and so:
:$HK = KF$
Hence:
:$GF = HL$
Therefore:
:$GK = KL$
that is:
:$MN = NE$
Hence:
:$MF = CG$
Therefore:
:$CG = FE$
and so:
:$CG + CN = FE + CN$
That is, the gnomon $OPQ$ equals the rectangle $CE$.
But $CE = 4 \cdot GF$
Therefore:
:$OPQ = 4 \cdot GF$
Therefore:
:$OPQ + FG = 5 \cdot GF$
But:
:$OPQ + FG = DN$
while:
:$DN = DB^2$
and:
:$GF = DC^2$
Therefore:
:$BD^2 = 5 \cdot DC^2$
== Sources ==
* : Book $\text{XIII}$. Propositions
Category:Golden Mean
Area of Squares on Whole and Lesser Segment of Straight Line cut in Extreme and Mean Ratio

== Theorem ==
== Proof ==
:
Let the line $AB$ be cut in extreme and mean ratio at the point $C$.
Let $AC$ be the greater segment.
It is to be demonstrated that:
:$AB^2 + BC^2 = 3 \cdot CA^2$
Let the square $ADEB$ be described on $AB$.
Let the figure be drawn as above.
We have that $AB$ be cut in extreme and mean ratio at $C$ such that $AC > CB$.
Therefore from:
:
and:
:
it follows that:
:$AB \cdot BC = AC^2$
We have that:
:$AK = AB \cdot BC$
and:
:$HG = AC^2$
Therefore:
:$AK = HG$
We have that:
:$AF = FE$
So:
:$AF + CK = FE + CK$
Therefore:
:$AK = CE$
and so:
:$AK + CE = 2 \cdot CE$
But:
:$AK + CE$ are the gnomon $LMN$ and the square $CK$.
Therefore:
:$LMN + CK = 2 \cdot AK$
But we have that:
:$AK = HG$
Therefore:
:$LMN + CK + HG = 3 \cdot HG$
But:
:$LMN + CK + HG = AE + CK$
while:
:$HG = AC^2$
Therefore:
:$AB^2 + BC^2 = 3 \cdot CA^2$
== Sources ==
* : Book $\text{XIII}$. Propositions
Category:Golden Mean
Area of Surface of Revolution

== Theorem ==
Let $\SS$ be a surface of revolution such that:
:$\SS$ is embedded in a Cartesian $3$-space
:the axis of revolution of $\SS$ is aligned with the $x$-axis
:the curve $\CC$ being rotated to generate $\SS$ is the plane curve $y = \map f x$
:$\CC$ has endpoints at $x = a$ and $x = b$.
Then the area of $\SS$ is given by:
:$\map \AA \SS = \ds \int_a^b 2 \pi y \sqrt {1 + \paren {\dfrac {\d y} {\d x} }^2} \rd x$
== Proof ==
== Sources ==
* 
* 
Category:Surfaces of Revolution
Category:Area Formulas
Area of Surface of Revolution from Astroid

== Theorem ==
Let $H$ be the astroid constructed within a circle of radius $a$.
The surface of revolution formed by rotating $H$ around the $x$-axis:
:
evaluates to:
:$\SS = \dfrac {12 \pi a^2} 5$
== Proof ==
By symmetry, it is sufficient to calculate the surface of revolution of $H$ for $0 \le x \le a$.
From Area of Surface of Revolution, this surface of revolution is given by:
:$\ds \SS = 2 \int_0^{\pi / 2} 2 \pi y \, \sqrt {\paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2} \rd \theta$
From Equation of Astroid:
:$\begin{cases}
x & = a \cos^3 \theta \\
y & = a \sin^3 \theta
\end{cases}$
so:
Hence:
Thus:
== Sources ==
* : Chapter $\text {B}.21$: The Cycloid: Problem $8$
Category:Astroids
Category:Surfaces of Revolution
Area of Trapezium

== Theorem ==
:
Let $ABCD$ be a trapezium:
:whose parallel sides are of lengths $a$ and $b$
and
:whose height is $h$.
Then the area of $ABCD$ is given by:
:$\Box ABCD = \dfrac {h \paren {a + b} } 2$
== Proof ==
:
Extend line $AB$ to $E$ by length $a$.
Extend line $DC$ to $F$ by length $b$.
Then $BEFC$ is another trapezium whose parallel sides are of lengths $a$ and $b$ and whose height is $h$.
Also, $AEFD$ is a parallelogram which comprises the two trapezia $ABCD$ and $BEFC$.
So $\Box ABCD + \Box BEFC = \Box AEFD$ and $\Box ABCD = \Box BEFC$.
$AEFD$ is of altitude $h$ with sides of length $a + b$.
Thus from Area of Parallelogram the area of $AEFD$ is given by:
:$\Box AEFD = h \paren {a + b}$
It follows that $\Box ABCD = \dfrac {h \paren {a + b} } 2$
== Usage Differences ==
== Sources ==
* : $\S 4$: Geometric Formulas: Trapezoid of Altitude $h$ and Parallel Sides $a$ and $b$: $4.7$
* 
* 
* : $\S 7$: Geometric Formulas: Trapezoid of Altitude $h$ and Parallel Sides $a$ and $b$: $7.7.$
* 
* : Appendix $1$: Areas and volumes
* : Appendix $1$: Areas and volumes
Category:Areas of Quadrilaterals
Category:Trapezia
Area of Triangle

== Theorem ==
This page gathers a variety of formulas for the area of a triangle.
=== In Terms of Side and Altitude ===
=== In Terms of Two Sides and Angle ===
=== In Terms of Inradius ===
=== In Terms of Circumradius ===
=== In Terms of Exradius ===
=== In Terms of Inradius and Exradii ===
=== Heron's Formula ===
Category:Areas of Triangles
Area of Triangle Inscribed in Parabola

== Theorem ==
Let $T$ be the parabola which is
the locus of points $\tuple {x, y}$ satisfying $y = x^2$.
Let $A$ and $B$ be two arbitrary points on $T$ with coordinates:
:$A = \tuple {u, u^2}$
:$B = \tuple {v, v^2}$, with $u > v$
Let $C$ be a third point on $T$ whose $x$-coordinate is the average of those of $A$ and $B$.
The area of $\triangle ABC$ is:
:$\map \AA {\triangle ABC} = \dfrac 1 8 \paren {u - v}^3$
== Proof 1 ==
== Proof 2 ==
Category:Area of Triangle Inscribed in Parabola
Category:Triangles
Category:Parabolas
Category:Areas of Triangles