Ex.12.1 Q5 Heron’s Formula Solution - NCERT Maths Class 9


Sides of a triangle are in the ratio of \(12:17:25\) and its perimeter is \(540\; \rm{cm}\). Find its area.

Text Solution

What is known?

Ratio of sides of the triangle and its perimeter.

What is unknown?

Area of the triangle.


By using Heron’s formula we can calculate the area of triangle.

The formula given by Heron about the area of a triangle


Where \(a, b\) and \(c\) are the sides of the triangle, and

\[\begin{align}s &= \text{Semi-perimeter}\\& = \begin{Bmatrix} \text{Half the Perimeter } \\ \text{ of the triangle} \;\end{Bmatrix}  \\&=\frac{(a+b+c)}{2}\end{align}\]


Suppose the sides are \(12x\;\rm {cm}\), \(17x\;\rm {cm}\) and \(25x\rm\; {cm}\).

Perimeter of the triangle \(=\) \(540\; \rm{cm}\)

\[\begin{align}{12 x+17 x+25 x=540} \\ {54 x=540} \\ {x=\frac{540}{54}} \\ {x=10 \mathrm{cm}}\end{align}\]

Therefore sides of triangle:

\[\begin{align}&12 x=12 \times 10=120 \;\mathrm{cm},\\ & 17 x=17 \times 10=170 \;\mathrm{cm}, \\ & 25 x=25 \times 10=250 \;\mathrm{cm}\end{align}\]

\(a = 120\; {\rm{cm}} ,\\ b = 170\;{\rm{cm}},\\ c = 250\;{\rm{cm}}\)

\[\begin{align}s &= \text{Half the Perimeter}\\ s&=\frac{540}{2}\\&=270 \;\mathrm{cm}\end{align}\]

By using Heron’s formula,

Area of a triangle

\(\begin{align}&={\sqrt{s(s-a)(s-b)(s-c)}} \\&=\!\sqrt{270(270\!-\!120)\!(270\!-\!170)\!(270\!-\!250)}\\ &=\sqrt{270 \times 150 \times 100 \times 20} \\ &=\sqrt{81000000} \\ &=9000 \mathrm{cm}^{2}\\\end{align}\)

Area of a triangle \(=\)\(9000 \;\rm{cm^2}.\) 

Learn from the best math teachers and top your exams

  • Live one on one classroom and doubt clearing
  • Practice worksheets in and after class for conceptual clarity
  • Personalized curriculum to keep up with school