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?

Question

1642 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-17T12:51:46

$483$ $483$ units squared

Heron's formula states that,

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

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

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

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

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

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.