What is the area of a triangle with sides of length 2, 4, and 5?

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Answer 1 · 2018-08-04T17:04:36

Area is $3.8$ square units

If three sides of a triangle are $a,b$ and $c$ , then

its area is $sqrt(s(s-a)(s-b)(s-c))$ , where $s=(a+b+c)/2$

Here sides are $2,4$ and $5$ and hence $s=(2+4+5)/2=11/2$

and area of triangle is

$sqrt(11/2(11/2-2)(11/2-4)(11/2-5))$

= $sqrt(11/2xx7/2xx3/2xx1/2)$

= $1/4sqrt231=15.19868/4=3.79967$ or say $3.8$