Heron’s Formula Class 9 Notes: Chapter 12

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=12×base×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 ΔABC, 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, 
AC2=OA2+OC2
OA2=AC2OC2
Substitute AC=a,OC=a2 to find OA

OA2=a2a24
OA=3a2  
We know the area of triangle is
A=12×base×height,
A=12×a×3a2

Area of Equilateral triangle=3a24  

Area of an isosceles triangle

Consider an isosceles ΔABC 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 :
CD2=AC2AD2 
h=a2b24=4a2b24
h=124a2b2
Area of triangle is A=12bh
A=12×b×124a2b2

A=14×b×4a2b2

Area of a triangle – By Heron’s formula

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

Triangle ABC

Find semi perimeter (s ) =a+b+c2

Area=s(sa)(sb)(sc)

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 C=90, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?

We draw the figure according to the information given.
 

Quadrilateral ABCD

The figure can be split into 2 triangles ΔBCD and ΔABD
From ΔBCD, we can find BD (Using Pythagoras’ Theorem)
BD2=122+52=169
BD=13cm
Semi-perimeter for ΔBCD S1=12+5+132=15
Semi-perimeter ΔABD S2=9+8+132=15
Using Heron’s formula we find A1 and A2 
A1=15(1512)(155)(1513)=15×3×10×2
 A1=900=30cm2
Similarly we find A2 to be 35.49cm2. 
The area of the quadrilateral ABCD=A1+A2=65.49 cm2

1 Comment

  1. superb byju’s

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