How do you use Heron's formula to determine the area of a triangle with sides of that are 35, 28, and 41 units in length?

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Answer 1 · 2018-04-17T12:51:46

$483$ units squared

Heron's formula states that,

$A=sqrt(s(s-a)(s-b)(s-c))$

$s$ is the semiperimeter of the triangle, given by $s=(a+b+c)/2$ .

$a,b,c$ are the sides of the triangle

Let $a=35,b=28,c=41$

$:.s=(35+28+41)/2=104/2=52$

So, the area of this triangle will be:

$A=sqrt(52(52-35)(52-28)(52-41))$

$=sqrt(52*17*24*11)$

$=sqrt(233376)$

$~~483$

So, the area of the triangle will be $483$ units squared.