June 14, 2024

# Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42cm

Class 9  Exercise 12.1  Question 4

find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is  42cm .

Solution:  Let $a,b$

and $c$ be the sides of the triangle  ABC.

Here we have perimeter of the triangle    $a+b+c=42&space;\,&space;cm$, $a=18\,&space;cm$ and $b=10\,&space;cm$.

First we will find the 3rd side of the triangle  by using perimeter,  then we will use  Heron’s formula  to find the area of the  given triangle.

The Heron’s formula  for finding the area of a triangle  given by

 Area of a triangle =    $\mathbf{\sqrt{s(s-a)(s-b)(s-c)}}$

where $a$, $b$ and $c$ are the sides of the triangle and s= semi-perimeter i.e. half of the perimeter of the triangle=$\frac{a+b+c}{2}$ .

It is given that  $a+b+c=42$. Putting the value of $a$ and $b$ , we get

18+10+$c$=42 $\Rightarrow&space;28+c=42&space;\Rightarrow&space;c=14&space;\,&space;cm$

Now semi-perimeter $s=\frac{a+b+c}{2}=\frac{42}{2}=21$

So  $s-a=21-18=3&space;\,&space;cm$

$s-b=21-10=11\,&space;cm$

$s-c=21-14=7\,&space;cm$

Therefore , area of the triangle = $\sqrt{s(s-a)(s-b)(s-c)}$

=$\sqrt{21&space;\times&space;3\times&space;11\times&space;7}=\sqrt{21\times&space;21\times&space;11}$

= $21\sqrt{11}\,&space;cm^{2}$

Area of triangle =    $21\sqrt{11}\,&space;cm^{2}$    .

Conclusion: The area of the triangle  two sides of which are 18 cm and 10 cm and the perimeter is
42 cm  is  $21\sqrt{11}\,&space;cm^{2}$    .

#### Bina singh

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