Introduction to Heron’s Formula
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=(1/2)×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 unit. Let AO be the perpendicular bisector of BC. In order to derive the formula for the area of an equilateral triangle, we need to find height AO.
![Class 9th Maths ncert solution](http://free-education.in/wp-content/uploads/2020/08/image-298.png)
Using Pythagoras theorem,
AC2=OA2+OC2
OA2=AC2−OC2
Substitute AC=a,OC= a/2 to find OA
OA2=a2−a2/4
OA=√3a/2
We know the area of the triangle is
A=(1/2)×base×height,
A=(1/2)×a×(√3a/2)
∴Area of Equilateral triangle=√3a2/4
Area of an isosceles triangle
Consider an isosceles ΔABC with equal sides as a units and base as b units.
![Class 9th Maths ncert solution](http://free-education.in/wp-content/uploads/2020/08/image-299.png)
Isosceles triangle ABC
The height of the triangle can be found by Pythagoras’ Theorem :
CD2=AC2−AD2
⇒h2=a2− (b2/4) = (4a2–b2)/4
⇒h=(1/2) √(4a2–b2)
Area of triangle is A=(1/2)bh
∴A=(1/2)×b×(1/2)√(4a2–b2)
∴A=(1/4)×b×√(4a2–b2)
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) :
![Class 9th Maths ncert solution](http://free-education.in/wp-content/uploads/2020/08/image-300.png)
Triangle ABC
Find semi perimeter (s ) = (a+b+c)/2
Area=√[s(s-a)(s-b)(s-c)]
This formula is helpful to find the area of a scalene triangle, given the lengths of all its sides.
Area of any polygon – By Heron’s formula
To find the area of a quadrilateral, when one of its diagonal value and the sides are given, the area can be calculated by splitting the given quadrilateral into two triangles and use the Heron’s formula.
Example :A park, in the shape of a quadrilateral ABCD, has ∠C=90∘, AB = 9 cm, BC = 12 cm, CD = 5 cm and AD = 8 cm. How much area does it occupy?
⇒We draw the figure according to the information given.
![Class 9th Maths ncert solution](http://free-education.in/wp-content/uploads/2020/08/image-301.png)
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+13)/2 = 15
Semi-perimeter ΔABD S2= (9+8+13)/2 = 15
Using Heron’s formula we find A1 and A2
A1= √[15(15-12)(15-5)(15-13)]
A1= √(15×3×10×2 )
A1=√900 = 30 cm2
Similarly, we find A2 to be 35.49 cm2.
The area of the quadrilateral ABCD=A1+A2=65.49 cm2