Heron’s Formula for Class 9 Maths Chapter 12

  • June 26, 2020
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Heron’s Formula for Class 9 Maths Chapter 12

Are you looking for Heron’s Formula or important points that are required to understand Heron’s for class 9 maths Chapter 12? You are the right place to get all information about Heron’s Class 9 maths chapters 12. Heron’s Formula plays a vital role in preparing you for the class 9 exam as well as higher studies. Heron’s Formula is very helpful for better scores in the exam.  Check Heron’s Formula according to class 9:

\[\ Area\;of\;Triangle = \sqrt{s(s-a)(s-b)(s-c)} \] \[\ Perimeter\;of\;Triangle = a+b+c \] \[\ Semi\;perimeter\;of\;Triangle = \frac{a+b+c}{2} \] Where, a, b, c are Side of Triangle

Proof Heron’s Formula

From the Heron’s Formula, you can find the area of any type of triangles such as the Equilateral triangle, isosceles, Scalene, right-angle triangle, Acute Angle triangle, Obtuse Angle triangle, and more. I think you should well understand the Heron’s Formula. Today I am going to proof Heron’s formulas for class 9.

in the ∆XYZ , h is a height of ∆XYZ and a, b, c is the Sides of ∆XYZ. The Perimeter of ∆XYZ = a+b+c, Semi-Perimeter of ∆XYZ, S = \( \frac{a+b+c}{2} \)

Area of ∆XYZ  = \( \frac{1}{2} Base \times height \)

=> \( \frac{1}{2} c\times h\)

We know that \( xp \perp yz \), it means \( \angle xpz = \angle xpy =90 \;degree \)

According to Pythagoras Theorem,

In the ∆XPZ, \( (b)^2 = h^2+ (c-d)^2 \) ————-(1)

In the ∆XPY, \( (a)^2 = h^2+ d^2 \) ————-(2)

Simplify equation (2) – (1)

\( a^2 – b^2 = h^2 + d^2 -(h^2 + (c-d)^2) \)

=> \( a^2 – b^2 = h^2 + d^2 -h^2 – (c-d)^2) \)

=> \( a^2 – b^2 = d^2 – (c-d)^2) \)

=> \( a^2 – b^2 = d^2 – (c^2+ d^2 -2cd) \)

=> \( a^2 – b^2 = d^2 – c^2- d^2 + 2cd) \)

=> \( a^2 – b^2 =  – c^2 + 2cd) \)

=> \( a^2 – b^2 + c^2 = 2cd) \)

=> \( \frac{a^2 – b^2 + c^2} {2c} = d \)

Put the Value of d in the Equation (2)

=> \( (a)^2 = h^2+ d^2 \)

=> \( (a)^2 = h^2+  (\frac{a^2 – b^2 + c^2} {2c})^2\)

=> \( (a)^2 -(\frac{a^2 – b^2 + c^2} {2c})^2 = h^2  \)

=> \( \sqrt {(a)^2 -(\frac{a^2 – b^2 + c^2} {2c})^2} = h  \)

Put the Value of h in the Area of ∆XYZ.

=> \( \frac{1}{2} \times c\times \sqrt {(a)^2 -(\frac{a^2 – b^2 + c^2} {2c})^2}\)

=> \( \frac{1}{2} \times c\times \sqrt {(a)^2 -\frac{(a^2 – b^2 + c^2)^2} {4c^2}}\)

=> \(  \sqrt {\frac{(a)^2\times c^2}{4} -\frac{c^2 \times (a^2 – b^2 + c^2)^2} {4 \times 4c^2}}\) [\( c^2 \) is Comman in the nominator and denominator then remove it]

=> \(  \sqrt {\frac{(a)^2\times c^2}{4} -\frac{(a^2 – b^2 + c^2)^2} {16}}\)

=> \(  \sqrt {\frac{(4 a^2 c^2) – (a^2 – b^2 + c^2)^2}{16}}\)

=> \(  \sqrt {\frac{(2ac)^2 – (a^2 – b^2 + c^2)^2}{16}}\) [Use Formula \( (a^2-b^2 = (a+b)(a-b) \)]

=> \(  \sqrt {\frac{(2ac + a^2 – b^2 + c^2) (2ac – a^2 + b^2 – c^2)}{16}}\)

=> \(  \sqrt {\frac{((a+c)^2 – b^2) (b^2 – (a-c)^2 )}{16}}\) [Use Formula \( (a^2-b^2 = (a+b)(a-b) \)]

=> \(  \sqrt {\frac{(a+c+b)(a+c-b) (b+a-c) (b-a+c)}{16}}\)

we can Write

=> \(  \sqrt { \frac{a+c+b}{2} \times \frac{a+c-b}{2} \times \frac{b+a-c}{2} \times \frac{b-a+c}{2}} \)

=> \(  \sqrt { \frac{a+c+b}{2} \times \frac{a+c+b-2b}{2} \times \frac{b+a+c-2c}{2} \times \frac{b-2a+c+a}{2}} \)

we can write

=> \(  \sqrt { \frac{a+c+b}{2}  (\frac{a+c+b}{2}- \frac{2b}{2})(\frac{a+c+b}{2}- \frac{2c}{2})(\frac{a+c+b}{2}- \frac{2a}{2})}   \)

Put the Value of \( \frac{a+b+c}{2} = s \)

=> \(  \sqrt { s  (s – b)(s – c)(s – a)}   \)

Proof Area of ∆XYZ = \(  \sqrt { s  (s – b)(s – c)(s – a)}   \)

Class 9 Maths Formulas By Chapters

Summary of Heron’s

We have shared a very important Formula for Heron’s which helps to score in the class 9 exams. If you have any questions and doubts related to Heron’s please let me know through comment or mail as well as social media. When you understand the Formula behind each Heron’s topics then it would be easier to solve the most complex problems related to Heron’s too.