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

Q4.    Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42cm.

Views

Given the perimeter of the triangle is  $42cm$ and the sides length $a= 18cm$ and $b= 10cm$

So, $a+b+c = 42cm$

Or, $c = 42 - 18-10 = 14cm$

So, the semi perimeter of the triangle will be:

$s = \frac{P}{2} = \frac{42cm}{2} = 21cm$

Therefore, the area given by the Heron's Formula will be,

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

$= \sqrt{21(21-18)(21-10)(21-14)}$

$= \sqrt{(7\times3 )(3)(11)(7)}$

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

Hence, the area of the triangle is $21\sqrt{11}\ cm^2.$

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