July 25, 2024

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

by Bina singhSeptember 15, 2021September 19, 2021

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.

to find area of a triangle

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

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 $\Rightarrow 28+c=42 \Rightarrow c=14 \, cm$

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

So $s-a=21-18=3 \, cm$

$s-b=21-10=11\, cm$

$s-c=21-14=7\, cm$

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

= $\sqrt{21 \times 3\times 11\times 7}=\sqrt{21\times 21\times 11}$

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

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

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

Bina singh

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