অন্যান্য ত্রিভুজ ক্ষেত্রফল সূত্র

Geometric inequalities / how to bash with area formulas

WARNING: To avoid the danger of ugly algebra please keep this article away from bashers

First some definitions:

Let $\triangle ABC$ be the triangle.

The area of a triangle is $[ABC]$.

The sides opposite to $A$, $B$ and $C$ are called $a$, $b$ and $c$ respectively.

$\alpha = \angle CAB$, $\beta = \angle ABC$ and $\gamma = \angle BCA$.

$s$ is half the perimeter, so $\frac{a+b+c}{2}$.

$R$ and $r$ are the radii of the circumcircle and incircle respectively.

$r_A$, $r_B$ and $r_C$ are the radii of the excircles opposing $A$, $B$ and $C$ respectively.

$h_A$, $h_B$ and $h_C$ are the altitudes of the triangle through $A$, $B$ and $C$ respectively.

This is the most important area formula:

$[ABC] = \frac{ah_A}{2 } = \frac{bh_B}{2 } = \frac{ch_C}{2 }$

Look here for the proof

Now we are going to prove some other area formulas:

Now try to prove the following equations yourself!

$[ABC] = \frac{abc}{4R}$

$[ABC] = \sqrt{(s-a)(s-b)(s-c)s}$ (Heron’s formula)

$[ABC] = (s-a)r_A$

Now we can express the radii of the incircle, circumcircle and excircles in terms of $a$, $b$ and $c$. We also know the heights of the triangle. So are we done?

No, there is a very important last bit of information left: $a$, $b$ and $c$ are sides of a triangle. They are not just some random reals, they satisfy the triangle inequalities: $b+c>a$, $a+c>b$ and $a+b>c$. What are we going to do with this?

We make the substitutions $a=y+z$, $b=x+z$ and $c=x+y$. This is called the Ravi substitution. $x=(b+c-a)/2$, $y=(a+c-b)/2$ and $z=(a+b-c)/2$. So $x$, $y$ and $z$ will be positive reals if and only if the triangle inequality is satisfied. So we got rid of the nasty condition and didn’t lose any information!

Now you only have to finish the problem using your inequality skills (or expand everything and bash it with AM-GM).

$A=\frac{bh}{2}$ is not the only area formula for triangles. There are many others, some of them involving trigonometry. Listed below are some useful formulas. Note that $a,b,c$ (lowercase) always indicate the side lengths of the triangle.

$A=\sqrt{s(s-a)(s-b)(s-c)}$ in which $s=\frac{a+b+c}{2}$. This is also known as Heron's formula.

$A=\dfrac{ab \sin C}{2}$ in which $a$ and $b$ are two adjacent sides and $c$ is the angle at their common vertex. This formula is used in finding the area of a parallelogram using just sides and angles.

$A=\dfrac {ab \sqrt{1-\cos^2 C}}{2}$ This follows directly from the above formula, because $$\begin{align*} \cos^2 \alpha + \sin^2 \alpha &=1 \\ \sin^2 \alpha &= 1-\cos^2 \alpha \\ \sin \alpha &= \sqrt{1-\cos^2 \alpha}.\end{align*}$$ This formula is used in proving Heron's formula, detailed above.

$A=rs$ in which $r$ is the inradius of the triangle and $s$ is the semiperimeter--$\frac{a+b+c}{2}$

$A=\frac{abc}{4R}$ in which $R$ is the circumradius.

$A= \sqrt{rr_1r_2r_3}$. in which $r$ is the inradius and $r_1, r_2, r_3$ are the exradii of the triangle.

There are also other less useful formulas out there, some of which can be found on the Wikipedia page on the area of a triangle.