How do you find the area given $a=3.05, b=0.75, c=2.45$ ?

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Answer 1 · 2017-03-04T17:06:31

Area of triangle is $0.613$

Given the three sides $a$ , $b$ and $c$ of a triangle,

its area is given by $Delta=sqrt(s(s-a)(s-b)(s-c))$ ,

where $s=1/2(a+b+c)$ .

Here as $a=3.05$ , $b=0.75$ and $c=2.45$ and therefore

$s=1/2(3.05+0.75+2.45)=1/2xx6.25=3.125$

and $Delta=sqrt(3.125(3.125-3.05)(3.125-0.75)(3.125-2.45))$

= $sqrt(3.125xx0.075xx2.375xx0.675)$

= $=0.3757324=0.613$

How do you find the area given a=3.05, b=0.75, c=2.45a=3.05, b=0.75, c=2.45?