HERON'S FORMULA

Area of a Triangle

Triangle

  The plane closed figure, with three sides and three angles is called as a triangle. 

Types of triangles:
Based  on sides -  a) Equilateral b) Isosceles c) Scalene
Based  on angles - a) Acute angled triangle b) Right- angled triangle c) Obtuse angled triangle

Area of a triangle

$Area=\frac{1}{2}\times base\times height$

 
In case of equilateral and isosceles triangles, if the length of the sides of triangles are given then,
we use Pythagoras theorem in order to find the height of a triangle.

 

Area of an equilateral triangle

Consider an equilateral $\Delta ABC$, with each side as $a$ units. Let AO be perpendicular bisector of BC. In order to derive the formula for the area of equilateral triangle, we need to find height AO.

Equilateral triangle ABC

Using Pythagoras theorem,
$A{C}^2=O{A}^2+O{C}^2$
$O{A}^2=A{C}^2-O{C}^2$
Substitute $AC=a,OC=\frac{a}{2}$ to find OA
$O{A}^2=a^2-\frac{a^2}{4}$
$OA=\frac{\surd 3a}{2}$  
We know the area of triangle is
$A=\frac{1}{2}\times base\times height$,
$A=\frac{1}{2}\times a\times \frac{\surd 3a}{2}$

$\therefore AreaofEquilateraltriangle=\frac{\surd 3a^2}{4}$  

Area of an isosceles triangle

Consider an isosceles $\Delta ABC$ with equal sides as $a$ units and base as $b$ unit.
 

Isosceles triangle ABC

The height of the triangle can be found by Pythagoras’ Theorem :
$C{D}^2=A{C}^2-A{D}^2$ 
$\Rightarrow h=a^2-\frac{b^2}{4}=\frac{4a^2-b^2}{4}$
$\Rightarrow h=\frac{1}{2}\sqrt{4a^2-b^2}$
Area of triangle is $A=\frac{1}{2}bh$
$\therefore A=\frac{1}{2}\times b\times \frac{1}{2}\sqrt{4a^2-b^2}$

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

Area of a triangle - By Heron's formula

 Area of a $\Delta ABC$, given sides a, b, c  by Heron’s formula (Also known as Hero’s Formula) : 
 

Triangle ABC

Find semi perimeter (s ) $=\frac{a+b+c}{2}$

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

This formula is helpful to find area of a scalene triangle, given the lengths of all its sides.

Area of any polygon - By Heron's formula

Area of a quadrilateral whose sides and one diagonal are given, can be calculated by dividing the quadrilateral into two triangles and using the Heron’s formula.

Example :A park, in the shape of a quadrilateral ABCD, has $\angle C={90}^{\circ },$ AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?

$\Rightarrow$We draw the figure according to the information given.
 

Quadrilateral ABCD

The figure can be split into 2 triangles $\Delta BCDand\Delta ABD$
From $\Delta BCD$, we can find BD (Using Pythagoras’ Theorem)
$B{D}^2={12}^2+5^2=169$
$BD=13cm$
Semi-perimeter for $\Delta BCD{S}_1=\frac{12+5+13}{2}=15$
Semi-perimeter $\Delta ABD{S}_2=\frac{9+8+13}{2}=15$
Using Heron's formula we find $A_{1}and{A}_2$
$A_1=\sqrt{15(15-12)(15-5)(15-13)}=\sqrt{15\times 3\times 10\times 2}$
 $A_1=\sqrt{900}=30c{m}^2$
Similarly we find $A_2$ to be $35.49c{m}^2.$
The area of the quadrilateral $ABCD=A_1+A_2=65.49c{m}^2$