Ex.12.1 Q4 Heron’s Formula Solution - NCERT Maths Class 9

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Find the area of a triangle two sides of which are \(18\;\rm{cm} \) and \(10\;\rm{cm}\) and the perimeter is \(42\;\rm{cm}\).

Text Solution

What is known?

Two sides of the triangle and its perimeter.

What is unknown?

Area of the triangle and one of its side.


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

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


The sides of triangle given: \(a =18 \, {\rm{cm}} , b = 10 \, {\rm{cm}}\)

Perimeter of the triangle

\[\begin{align} &= a + b + c\\ {42}&={18+10+c} \\ {42}&={28+c} \\ {c}&={42-28} \\ {c}&={14}\end{align}\]

Semi Perimeter

\[\begin{align}s & =\frac{(a+b+c)}{2}\\ &=\frac{42}{2} \\ &=21 \mathrm{cm}\end{align}\]

By using Heron’s formula,

Area of a triangle

\(\begin{align}&={\sqrt{s(s-a)(s-b)(s-c)}} \\ &=\sqrt{  21(21-18)  (21-10) (21-14)  } \\ &=\sqrt{21 \times 3 \times 11 \times 7} \\ &=21 \sqrt{11} \mathrm{cm}^{2}\end{align}\)

Area of a triangle \(=21\sqrt{11}\mathrm{cm}^{2}\)

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