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

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Question

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.

Reasoning:

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

The formula given by Heron about the area of a triangle

\(=\sqrt{s(s-a)(s-b)(s-c)}\)

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

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

Steps:

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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