Perimeter of a polygon:

The perimeter of a Polygon is the sum of the lengths of all sides of the polygon.

Perimeter of a triangle:

-For a scalene Triangle of sides $a$ , $b$ and $c$ , the perimeter $P$ is:

$P=a+b+c$

-For Isoceles Triangle of Equal side $s$ and base $b$ :

$P=b+2s$

-For an Equilateral Triangle of side $s$ :

$P=3s$

-Perimeter of Quadrilaterals:

-For a quadrilateral of sides $a$ , $b$ , $c$ and $d$ :

$P=a+b+c+d$

-For a Rectangle of Length $L$ and width $w$

$P=2L+2w$

$P=2(L+w)$

-For a Square(or Rhombus) of side $s$ :

$P=4s$

-For a Parallelogram of parallel equal sides $b$ and parallel equal sides $s$ :

$P=2b+2s$

$P=2(b+s)$

Areas of Circles and Polygons:

A triangular region consists of a triangle and its interior while a circular region is one of a circle and its interior.

The area of a triangle can be calculated using methods depending on information we have:

For a scalene triangle:

-Taking any altitude (as height, h) and its opposite side (as base, b).

The Area is:

$A=\frac{b\times h}{2}$

We can also write:

$A=\frac{1}{2}bh$

For an isosceles triangle, given the three sides: a,a and b

We can calculate $h$

$h^2=a^2-\left(\frac{b}{2}\right)^{2}$

Finally:

$h=\frac{1}{2}\sqrt{4a^2-b^2}$

And the area of the triangle:

$A=\frac{1}{4}b\sqrt{4a^2-b^2}$

If we use the Heron formula, we find the same formula

$A=\sqrt{s(s-a)(s-b)(s-c}$

In our situation $b=a$ and $c=b$

That means $s=\frac{1}{2}(2a+b)$

$s-a=\frac{1}{2}(2a+b-2a)=\frac{1}{2}(b)$

$s-b=s-a=\frac{1}{2}(2a+b-2a)=\frac{1}{2}(b)$

$s-c=s-b=\frac{1}{2}(2a+b-2b)=\frac{1}{2}(2a-b)$

We get the following area:

$A=\sqrt{(\frac{2a+b}{2})(\frac{b}{2})(\frac{b}{2})(\frac{2a-b}{2})}$

$A=\sqrt{(\frac{b^2}{4})(\frac{4a^2-b^2}{4}})$

Finally:

$A=\frac{1}{4}b\sqrt{4a^2-b^2}$

Special isosceles and right triangle:

In this situation we have :

$h=\frac{1}{2}b$

Without going far, we can notice that each of the right triangle legs can be the base (a) or the height (a):

$A=\frac{1}{2}a\times a$

$A=\frac{1}{2}a^2$

Using b:

$A=\frac{b^2}{4}$

This can be found from the previous formula as well.

Equilateral Triangle

We have the nomrmal formula:

$A=\frac{1}{2}bh$

$h^2=a^2-\frac{a^2}{4}$

$h^2=\frac{1}{4}(4a^2-a^2)$

$h^2=\frac{1}{4}(3a^2)$

$h=\frac{\sqrt{3}a}{2}$

Now for the area we have $b=a$

$A=\frac{1}{2}(a)(\frac{\sqrt{3}a}{2})$

finally:

$A=\frac{\sqrt{3}}{4}a^2$

Area Triangle using angles:

We can see that:

$Area=\frac{1}{2}ch$

$\sin(A)=\frac{h}{b}$

So: $h=b\sin(A)$

The Area:

$Area=\frac{1}{2}bc\sin(A)$

We can find all three relationships:

$Area=\frac{1}{2}bc\sin(A)$

$Area=\frac{1}{2}ac\sin(B)$

$Area=\frac{1}{2}ab\sin(C)$

Circumference of a circle

The Circumference $C$ of a circle is as follows:

$C=2\pi r$

where $r$ is the radius of the circle.

Since the diameter $D=2r$ , we can write:

$C=D\times\pi$

Areas and circumferences of circular figures can be calculated without replacing $\pi$ with its value. That means in terms of $\pi$ .

Length of an arc:

The length of an arc of a circle is related to the central angle $\alpha$ that intercepts it.

If $\alpha$ is in $radians$ , we have:

$m\stackrel\frown{AB}=\alpha r$

For an angle in $degrees$ , we use the proportion to the circumference:

The length of $m\stackrel\frown{AB}=\frac{\alpha^{\circ}}{360}2\pi r$

This is the same as:

$m\stackrel\frown{AB}=\frac{\alpha^{\circ}}{180}\pi r$

Area of a Circle of radius r:

If $r$ is the radius of a circle, the $Area$ is:

$Area=\pi r^{2}$

This formula is widely used in geometry and other fields.

Area of a Sector AOB of a circle:

If $r$ is the radius of a circle, $\alpha$ the central angle (or intercepted arc) in $radians$ , the $Area$ of $Sector \; AOB$ is:

$Area\; AOB=\frac{\alpha}{2\pi}\pi r^2$

We simplify and get:

$Area\; AOB=\frac{\alpha}{2}r^2$

If $\alpha$ the central angle (or intercepted arc) is in $degrees$ :

$Area\; AOB=\frac{\alpha^{\circ}}{360}\pi r^2$

This is a simple conversion of $\alpha$ radians above. $\alpha (rad)=\frac{\alpha^{\circ}\pi}{180}$

Area of a segment of a circle

This can be approached 2 ways:

-Area of the sector minus area of isosceles triangle $AOB$

-By deriving the formula.

The central angle’s bisector bisects the chord AB at D.

We get

$\sin\frac{\alpha}{2}=\frac{AD}{OA}$

But $OA=r$

$AD=r\sin\frac{\alpha}{2}$

OD is the altitude:

$OD=h$

$h^2=r^2-(AD)^2$

$h^2=r^2-r^2\sin^{2}\frac{\alpha}{2}$

$h=\sqrt{r^2-r^2\sin^{2}\frac{\alpha}{2}}$

$h=r\sqrt{1-\sin^{2}\frac{\alpha}{2}}$

But we know that:

$1-\sin^{2}\frac{\alpha}{2}=\cos^{2}\frac{\alpha}{2}$

That returns:

$h=r\cos\frac{\alpha}{2}$

The Area of $\triangle AOB$ is:

$Area\;\triangle AOB=AD \cdot h$

We plug in:

$Area\;\triangle AOB=r\sin\frac{\alpha}{2}\cdot r\cos\frac{\alpha}{2}$

Finally:

$Area\;\triangle AOB=\frac{1}{2}r^2\sin\alpha$

For sector AOB:

$Area\;Sector\; AOB=\frac{\alpha}{2}r^2$

We take the difference to get the area of the segment:

$Area\; Segment=\frac{1}{2}\alpha r^2-\frac{1}{2}r^2\sin\alpha$

$Area\; Segment=\frac{1}{2}r^2 \left(\alpha-\sin\alpha \right)$

Al-Kashi, Heron,Bretschneider’s, Brahmagupta’s and Coolidge formulas

The Bretscheiner’s formula helps find the area of a non-cyclic quadrilateral, that cannot be inscribed in a circle,

using only side lengths and possibly one angle measure or one diagonal length.

After the Bretschneider’s formula, we’ll simplify the quadrilateral to make it cyclic. We’ll get the Brahmagupta’s formula.

With the cyclic quadrilateral the product of the diagonals $e$ and $f$ is $ef=ac+bd$ and opposite angles are supplementary.

Let $K$ be area.

From the figure, we notice $K=A_{ABD}+A_{DCB}$

$K=\frac{1}{2}ad \sin \alpha+ \frac{1}{2}bc \sin \phi$

$2K=ad \sin \alpha+ bc \sin \phi$

Squaring both sides:

$4K^{2}=(ad)^{2} \sin^{2} \alpha+ (bc)^{2} \sin^2 \phi+ 2adbc \sin \alpha \sin \phi$

$4K^{2}=(ad)^{2} \sin^{2} \alpha+ (bc)^{2} \sin^2 \phi+ 2adbc \sin \alpha \sin \phi$ $(1)$

Now let’s check 2 values of diagonal $e$ :

$e^{2}=a^{2}+d^{2}-2ad \cos \alpha$

$e^{2}=b^{2}+c^{2}-2bc \cos \phi$

$a^{2}+d^{2}-2ad \cos \alpha=b^{2}+c^{2}-2bc \cos \phi$

$a^{2}+d^{2}-b^{2}-c^{2}=2ad \cos \alpha-2bc \cos \phi$

$\frac{a^{2}+d^{2}-b^{2}-c^{2}}{2}=ad \cos \alpha-bc \cos \phi$

Taking the square:

$\frac{(a^{2}+d^{2}-b^{2}-c^{2})^{2}}{4}=(ad)^{2} \cos^{2} \alpha+(bc)^{2} \cos^{2} \phi-2abcd \cdot \cos \phi \cdot \cos \alpha$ $(2)$

Adding $(1)$ and $(2)$ :

$4K^{2}+\frac{(a^{2}+d^{2}-b^{2}-c^{2})^{2}}{4}=(ad)^{2}(\sin^{2} \alpha+\cos^{2} \alpha)+(bc)^{2}(\sin^{2} \phi+\cos^{2} \phi)+ 2adbc \cdot \sin \alpha \sin \phi -2abcd \cdot \cos \alpha \cos \phi$

$4K^{2}+\frac{(a^{2}+d^{2}-b^{2}-c^{2})^{2}}{4}=(ad)^{2}+ (bc)^{2}- 2adbc \cdot \cos (\alpha + \phi)$

$4K^{2}+\frac{(a^{2}+d^{2}-b^{2}-c^{2})^{2}}{4}=(ad +bc)^{2}-2adbc- 2adbc \cdot \cos (\alpha + \phi)$

$4K^{2}+\frac{(a^{2}+d^{2}-b^{2}-c^{2})^{2}}{4}=(ad +bc)^{2}- 2adbc \cdot( \cos (\alpha + \phi)+1)$

$4K^{2}+\frac{(a^{2}+d^{2}-b^{2}-c^{2})^{2}}{4}=(ad +bc)^{2}- 4adbc(\frac{\cos (\alpha + \phi)+1}{2})$ Injected 2 in the last term.

$4K^{2}+\frac{(a^{2}+d^{2}-b^{2}-c^{2})^{2}}{4}=(ad +bc)^{2}- 4adbc \cdot \cos^{2} (\frac{\alpha + \phi}{2})$

To the same denominator and multiplying both sides by 4:

$16K^{2}=4(ad +bc)^{2}-(a^{2}+d^{2}-b^{2}-c^{2})^{2}- 16adbc \cdot \cos^{2} (\frac{\alpha + \phi}{2})$

$16K^{2}=[4(ad +bc)^{2}]-[(a^{2}+d^{2}-b^{2}-c^{2})^{2}]- 16adbc \cdot \cos^{2} (\frac{\alpha + \phi}{2})$

$16K^{2}=(2ad +2bc+a^{2}+d^{2}-b^{2}-c^{2})(2ad +2bc-a^{2}-d^{2}+b^{2}+c^{2})- 16adbc \cdot\cos^{2} (\frac{\alpha + \phi}{2})$

$16K^{2}=(a^{2}+2ad+d^{2}-(b^{2}-2bc+c^{2}))(-a^{2}+2ad-d^{2}+(b^{2}+2bc+c^{2}))- 16adbc \cdot \cos^{2} (\frac{\alpha + \phi}{2})$

$16K^{2}=((a+d)^{2}-(b-c)^{2})((b+c)^{2}-(a-d)^{2})- 16adbc \cdot\cos^{2} (\frac{\alpha + \phi}{2})$

$16K^{2}=(a+d+b-c)(a+d-b+c)(b+c+a-d)(b+c-a+d)- 16adbc \cdot\cos^{2} (\frac{\alpha + \phi}{2})$

$K^{2}=\frac{b+c+d-a}{2}\cdot \frac{a+c+d-b}{2}\cdot \frac{a+b+d-c}{2}\cdot \frac{a+b+c-d}{2}- adbc\cdot \cos^{2} (\frac{\alpha + \phi}{2})$

$\boxed {K=\sqrt{\frac{b+c+d-a}{2}\cdot \frac{a+c+d-b}{2}\cdot \frac{a+b+d-c}{2}\cdot \frac{a+b+c-d}{2}- adbc\cdot \cos^{2} (\frac{\alpha + \phi}{2})}}$

If we call the semi-perimeter $s=\frac{a+b+c+d}{2}$

We have the final Bretscheiner’s formula:

$\boxed {K=\sqrt{(s-a)(s-b)(s-c)(s-d)-adbc\cdot \cos^{2} (\frac{\alpha + \phi}{2})}}$

Brahmagupta’s formula

The steps remain the same but this time since the quadrilateral is $CYCLIC$ we know that the two opposite angles are supplementary.

$\cos \frac{\pi}{2}=0$

We get from the Bretscheiner’s formula

$\boxed {K=\sqrt{(s-a)(s-b)(s-c)(s-d)}}$

Heron’s formula

The fourth side goes away and we get a triangle. From Brahmagupta’s formula:

$\boxed {K=\sqrt{s(s-a)(s-b)(s-c)}}$

In depth explanation:

From Al-Kashi to Bretshneider via Heron and Brahmagupta, we are going to go through some of their work:

Al-kashi is simply the Law of Cosines:

Law of cosines:

From what we know:
$\cos{A}=\frac{AD}{c} \Rightarrow AD=c\cdot \cos{A}$
We have two right triangles: $\triangle{ADB}$ and $\triangle{CDB}$
Using the Pythagorean Theorem:
${h}^2={c}^2-{AD}^2$
Also:
${h}^2={a}^2-{DC}^2$
${a}^2-{DC}^2={c}^2-{AD}^2$
But :
$\overline{AC}=\overline{AD}+\overline{DC} \Rightarrow \overline{DC}=\overline{AC}-\overline{AD}$
This means:

$<span class="ql-right-eqno"> </span><span class="ql-left-eqno"> </span><img src="https://www.mouctar.org/wp-content/ql-cache/quicklatex.com-847a7e12dc54dc19dc4aa9f48cdf9a85_l3.png" height="38" width="129" class="ql-img-displayed-equation quicklatex-auto-format" alt="\begin{align*}DC&=AC-AD\\&=b-AD \end{align*}" title="Rendered by QuickLaTeX.com"/>$

${a}^2-{b-AD}^2={c}^2-{AD}^2$
${a}^2-({b}^2-2 \cdot b \cdot AD+{AD}^2) ={c}^2-{AD}^2$
${a}^2-{b}^2+2 \cdot b \cdot AD-{AD}^2={c}^2-{AD}^2$
${a}^2-{b}^2+2 \cdot b \cdot (AD)={c}^2$
We know from above: $AD=c\cdot\cos{A}$
${a}^2-{b}^2+2 \cdot b \cdot c\cdot\cos{A}={c}^2$
Finally we have the $Law \;of\; cosines$ :

${a}^2={b}^2+{c}^2-2 \cdot b \cdot c\cdot\cos{A}$

${b}^2={a}^2+{c}^2-2 \cdot a \cdot c\cdot\cos{B}$

${c}^2={a}^2+{b}^2-2 \cdot a \cdot b\cdot\cos{C}$

Heron Formula:

This one is widely used. We use it when we have only sides and we want to find the Area of the triangle

Let $S$ be area and $s=\frac{a+b+c}{2}$ the semi-perimeter.

$S=\frac{1}{2}bc \sin A$ .

We should remember this because we’ll need it in the following paragraphs.

$\sin A=\frac{2S}{bc}$

$a^2=b^2+c^2-2bc \cos A \Rightarrow \cos A=\frac{b^2+c^2-a^2}{2bc}$

But we know that:

$\sin^{2} A+ \cos^{2} A=1$

$(\frac{2S}{bc})^{2}+(\frac{b^2+c^2-a^2}{2bc})^{2}=1$

$\frac{4S^{2}}{b^{2}c^{2}}+\frac{(b^2+c^2-a^2)^{2}}{4b^{2}c^{2}}=1$

$\frac{4\times 4S^{2}}{4 \times b^{2}c^{2}}+\frac{(b^2+c^2-a^2)^{2}}{4b^{2}c^{2}}=\frac{4b^{2}c^{2}}{4b^{2}c^{2}}$

$16S^{2}=4b^{2}c^{2}-(b^2+c^2-a^2)^{2}$

This is a difference of squares, we can factor:

$16S^{2}=(2bc+b^2+c^2-a^2)(2bc-b^2-c^2+a^2)$

$16S^{2}=[(b+c)^2-a^2][a^2-(b-c)^{2}]$

$16S^{2}=(b+c-a)(b+c+a)(a+b-c)(a+c-b)$

$16S^{2}=(a+b+c)(b+c-a)(a+c-b)(a+b-c)$

$S^{2}=\frac{(a+b+c)}{2}\frac{(b+c-a)}{2}\frac{(a+c-b)}{2}\frac{(a+b-c)}{2}$

$S=\sqrt{s(s-a)(s-b)(s-c)}$

This is the Heron formula:

$\boxed{S=\sqrt{s(s-a)(s-b)(s-c)}}$

The Bretschneider formula:

We will demonstrate the following forms of the Area $K$ per BRETSCHNEIDER

$K=\frac{1}{2}pq \sin \theta$

$K=\frac{1}{4}(b^{2}+d^{2}-a^{2}-c^{2}) \tan \theta$

$K=\frac{1}{4}\sqrt{4p^{2}q^{2}-(b^{2}+d^{2}-a^{2}-b^{2})^{2}}$

We split the diagonal $p$ into $(p-y)$ and $y$

The same way for $q$ : $(q-x)$ and $x$

We express the area of the four triangles:

$A_1=\frac{1}{2}xy \sin \theta$

$A_2=\frac{1}{2}x(p-y) \sin \theta$

$A_3=\frac{1}{2}(p-y)(q-x) \sin \theta$

$A_4=\frac{1}{2}(q-x)y \sin \theta$

Now we add all four areas:

$A_{ABCD}=A_1+A_2+A_3+A_4$

$A_{ABCD}=(\frac{1}{2}xy+\frac{1}{2}px-\frac{1}{2}xy+\frac{1}{2}pq-\frac{1}{2}px-\frac{1}{2}qy+\frac{1}{2}xy+\frac{1}{2}qy-\frac{1}{2}xy) \sin \theta$

After simplication:

$A_{ABCD}=\frac{1}{2}pq \sin \theta$

The first form:

$\boxed{K=\frac{1}{2}pq \sin \theta}$

Now the second form:

From the figure:

$c^{2}=x^{2}+y^{2}-2xy \cos \theta$ $(1)$

$a^{2}=(p-y)^{2}+(q-x)^{2}-2(p-y)(q-x)\cos \theta$

$a^{2}=p^{2}-2py+y^{2}+q^{2}-2qx+x^{2}-(pq-px-qy+xy)\cdot 2\cos \theta$

We expand and replace terms in $(1)$ by $c^{2}$

$a^{2}=p^{2}+c^{2}+q^{2}-2py-2qx-2pq\cos \theta+2px \cos \theta+ 2qy \cos \theta$

Now we move to the $b^{2}$

$b^{2}=(p-y)^{2}+x^{2}+2(p-y)x \cos \theta$

$b^{2}=p^{2}-2py+y^{2}+x^{2}+2px \cos \theta-2xy \cos \theta$

We expand and replace terms in $(1)$ by $c^{2}$

$b^{2}=p^{2}-2py+c^{2}+2px \cos \theta$

Now we move to the $d^{2}$

$d^{2}=(q-x)^{2}+y^{2}+2(q-x)y \cos \theta$

$d^{2}=q^{2}-2qx+x^{2}+y^{2}+2qy\cos \theta-2xy \cos \theta$

$d^{2}=q^{2}-2qx+c^{2}+2qy \cos \theta$

Let’s add as follows:

$b^{2}+d^{2}-a^{2}-c^{2}=$

$=p^{2}-2py+c^{2}+2px \cos \theta+q^{2}-2qx+c^{2}+2qy \cos \theta-p^{2}-c^{2}-q^{2}+2py+2qx+2pq\cos \theta$

$-2px \cos \theta- 2qy \cos \theta-c^{2}$

After simplification we get:

$b^{2}+d^{2}-a^{2}-c^{2}=2pq \cos \theta \Rightarrow pq=\frac{b^{2}+d^{2}-a^{2}-c^{2}}{2\cos \theta}$

But:

$K=\frac{1}{2}pq \sin \theta$

We plug in $pq$

$K=\frac{1}{4}(b^{2}+d^{2}-a^{2}-b^{2}) \tan \theta$

Second Form:

$\boxed{K=\frac{1}{4}(b^{2}+d^{2}-a^{2}-c^{2}) \tan \theta}$

Third form:

From the sequences above:

$K=\frac{1}{2}pq \sin \theta$

But: $\sin \theta=\sqrt{1- \cos^{2} \theta}$

$K=\frac{1}{2}pq\sqrt{1-(\frac{b^{2}+d^{2}-a^{2}-c^{2}}{2pq})^{2}}$

$K=\frac{1}{2}pq\sqrt{1-\frac{(b^{2}+d^{2}-a^{2}-c^{2})^{2}}{4p^{2}q^{2}}}$

$K=\frac{1}{2}\sqrt{p^{2}q^{2}-p^{2}q^{2}\frac{(b^{2}+d^{2}-a^{2}-c^{2})^{2}}{4p^{2}q^{2}}}$

$K=\frac{1}{4}\sqrt{4p^{2}q^{2}-(b^{2}+d^{2}-a^{2}-c^{2})^{2}}$

This is the actual BRETSCHNEIDER formula:

$\boxed{K=\frac{1}{4}\sqrt{4p^{2}q^{2}-(b^{2}+d^{2}-a^{2}-c^{2})^{2}}}$

According to wiki, the following work is credited to COOLIDGE in 1939:

Let $K$ be area.

From the figure, we notice $K=A_{ABD}+A_{DCB}$

$K=\frac{1}{2}ad \sin \alpha+ \frac{1}{2}bc \sin \phi$

$2K=ad \sin \alpha+ bc \sin \phi$

Squaring both sides:

$4K^{2}=(ad)^{2} \sin^{2} \alpha+ (bc)^{2} \sin^2 \phi+ 2adbc \sin \alpha \sin \phi$

$4K^{2}=(ad)^{2} \sin^{2} \alpha+ (bc)^{2} \sin^2 \phi+ 2adbc \sin \alpha \sin \phi$ $(1)$

Now let’s check 2 values of diagonal $e$ :

$e^{2}=a^{2}+d^{2}-2ad \cos \alpha$

$e^{2}=b^{2}+c^{2}-2bc \cos \phi$

$a^{2}+d^{2}-2ad \cos \alpha=b^{2}+c^{2}-2bc \cos \phi$

$a^{2}+d^{2}-b^{2}-c^{2}=2ad \cos \alpha-2bc \cos \phi$

$\frac{a^{2}+d^{2}-b^{2}-c^{2}}{2}=ad \cos \alpha-bc \cos \phi$

Taking the square:

$\frac{(a^{2}+d^{2}-b^{2}-c^{2})^{2}}{4}=(ad)^{2} \cos^{2} \alpha+(bc)^{2} \cos^{2} \phi-2abcd \cdot \cos \phi \cdot \cos \alpha$ $(2)$

Adding $(1)$ and $(2)$ :

$4K^{2}+\frac{(a^{2}+d^{2}-b^{2}-c^{2})^{2}}{4}=$

$=(ad)^{2}(\sin^{2} \alpha+\cos^{2} \alpha)+(bc)^{2}(\sin^{2} \phi+\cos^{2} \phi)+ 2adbc \cdot \sin \alpha \sin \phi -2abcd \cdot \cos \alpha \cos \phi$