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Difference between revisions of "Area"

m (→‎Also true, but more complex than above)
(→‎Other formulas K = f(a,b,c) equivalent to Heron's)
 
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<math> K =\tfrac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)} </math>
 
<math> K =\tfrac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)} </math>
  
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These are especially useful when <math>(a,b,c) = (\sqrt{X_1}, \sqrt{X_1}, \sqrt{X_3})</math>, for <math>X_i \in \mathbb{Z}</math>:
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These are especially useful when <math>(a,b,c) = (\sqrt{X_1}, \sqrt{X_2}, \sqrt{X_3})</math>, for <math>X_i \in \mathbb{Z}</math>:
  
 
<math> K =\tfrac{1}{4}\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)} = \tfrac{1}{4} \sqrt{(\sum_i X_i)^2 - 2 \sum_i {X_i^2} } </math>
 
<math> K =\tfrac{1}{4}\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)} = \tfrac{1}{4} \sqrt{(\sum_i X_i)^2 - 2 \sum_i {X_i^2} } </math>

Latest revision as of 13:01, 23 October 2024

In mathematics, area refers to the size of the region that a two-dimensional figure occupies. The size of a region in higher dimensions is referred to as volume.

It is often possible to find the area of a region bounded by parts of circles and line segments through elementary means. One can find the area of even more complex regions via the use of calculus.

Rectangles are the most basic figures whose area we can study. It makes sense that the area of a rectangle with length $l$ and width $w$ is simply $l\cdot w$.

Once we know the area of a rectangle, we can easily find the area of a triangle by just noting that if our triangle has base $b$ and height $h$, then the rectangle with length $b$ and width $h$ has exactly twice as much area as the original triangle. Thus, the area of a triangle is

$A=\frac 12 bh.$

We can now find the area of any polygon by breaking it up into triangles.

Introductory Videos

https://youtu.be/51K3uCzntWs?t=842 \\ https://youtu.be/j3QSD5eDpzU

Notation

The letters $A$ and $K$ are frequently used to stand for area. When there are multiple regions under consideration, subscripts are often employed: $A_1, K_2,\ldots$ might be used to denote the areas of particular regions, or $A_{ABC}, K_{BCD},\ldots$. For example, $K_{ABCDEF}$ would mean the area of hexagon $ABCDEF$.

An alternative notation is to use square brackets around the name of the region to denote its area, e.g. $[ABC]$ for the area of triangle $\triangle ABC$.

Area of a Regular Polygon

The area of any regular polygon can be found as follows:

Inscribe the figure, with $n$ sides of length $s$, in a circle and draw a line from two adjacent vertices to the circumcenter. This creates a triangle that is $\frac{1}{n},$ of the total area (consider the regular octagon below as an example).

Regularoctagon.PNG

Drawing the apothem creates two right triangles, each with an angle of $\frac{180}{n}^{\circ}$ at the top vertex. If the polygon has side length $s$, the height of the triangle can be found using trigonometry to be of length $\frac s2 \cot \frac{180}{n}^{\circ}$.

The area of each triangle is $\frac12$ the base times the height, which can also be expressed as $\frac{s^2}{4} \cot\frac{180}{n}^{\circ}$ and the area of the entire polygon is $\frac{n\cdot s^2}{4} \cot\frac{180}{n}^{\circ}$.

Area of a Triangle

There are many ways to find the area $[ABC]$ of a triangle. In all of these formulae, ${K}$ will be used to indicate area.

  • $K=\frac{bh}{2}$ where $b$ is a base and $h$ is the altitude of the triangle to that base.
  • Heron's formula: $K=\sqrt{s(s-a)(s-b)(s-c)}$, where $a, b$ and $c$ are the lengths of the sides and $s$ is the semi-perimeter $s=\frac{a+b+c}{2}$.
  • $K=rs$, where $r$ is the radius of the incircle and s is the semi-perimeter.
  • $K=\frac{ab\sin{\theta}}{2}$ where $a$ and $b$ are adjacent sides of the triangle and $\theta$ is the measure of the angle between them.
  • $K=\frac{abc}{4R}$, where $a,b,c$ are the lengths of the sides of the triangle and $R$ is the circumradius.
  • $\frac{1}{K}=4\sqrt{H(H-h_a^{-1})(H-h_b^{-1})(H-h_c^{-1})}$, where $H=\frac{(h_a^{-1}+h_b^{-1}+h_c^{-1})}{2}$ and the triangle has altitudes $h_a$, $h_b$, $h_c$.

Other formulas $K = f(a,b,c)$ equivalent to Heron's

$K =\tfrac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}$

These are especially useful when $(a,b,c) = (\sqrt{X_1}, \sqrt{X_2}, \sqrt{X_3})$, for $X_i \in \mathbb{Z}$:

$K =\tfrac{1}{4}\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)} = \tfrac{1}{4} \sqrt{(\sum_i X_i)^2 - 2 \sum_i {X_i^2} }$

$K = \frac1{2}\sqrt{a^2 c^2 - \left(\frac{a^2 + c^2 - b^2}{2}\right)^2}$

Also true, but more complex than above

$K =\tfrac{1}{4} \sqrt{ 2(a^2 b^2+a^2c^2+b^2c^2)-(a^4+b^4+c^4)} = \tfrac{1}{4}\sqrt{2 \sum_{i \neq j} {X_i X_j} - \sum_i X_i^2 }$

$K = \tfrac{1}{4}\sqrt{4(a^2b^2+a^2c^2+b^2c^2)-(a^2+b^2+c^2)^2} = \tfrac{1}{4} \sqrt{4 \sum_{i \neq j} {X_i X_j} - {(\sum_i X_i})^2 }$

Area of a Quadrilateral

To find the area of most quadrilaterals, you must divide the quadrilateral up into smaller triangles and find the area of each triangle. However, some quadrilaterals have special formulas to find their areas. Again, $K$ is the area.

  • Kite - $K=\frac{d_1\cdot d_2}{2}$ where the $d$s represent the lengths of the diagonals of the kite.
  • Parallelogram - ${K=bh}$, where $b$ is the base and $h$ is the height to that base.
  • Trapezoid - $K=\frac{b_1+b_2}{2}\cdot h$, where the $b$s are the parallel sides and $h$ is the distance between those bases.
  • Rhombus - a special case of a kite and parallelogram, so either formula may be used here.
  • Rectangle - ${K=lw}$, where $l$ is the length of the rectangle and $w$ is the width. (This is a special case of the formula for a parallelogram where the height and a side happen to coincide.)
  • Square - $K=s^2$, where $s$ is the length of a side.
  • Any quadrilateral - $K=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\left(\dfrac{B+D}{2}\right)}$, where $s$ is the semiperimeter, $a$, $b$, $c$, and $d$ are the side lengths, and $B$ and $D$ are the measures of angles $B$ and $D$, respectively.
  • Cyclic quadrilateral - $K=\sqrt{(s-a)(s-b)(s-c)(s-d)}$ where $s$ is the semiperimeter and $a$, $b$, $c$, and $d$ are the side lengths. (This is a special case of the formula for the area of any quadrilateral; $\cos^2\left(\dfrac{B+D}{2}\right)=0$.)

See Also

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