How do you find the area of triangle ABC given a=2, b=3, c=4?

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Answer 1 · 2017-07-23T09:42:40

The area is $= 2.9 u^{2}$ $=2.9u^2$

We apply Heron's formula

$a r e a = \sqrt{s (s - a) (s - b) (s - c)}$ $area =sqrt(s(s-a)(s-b)(s-c))$

Where

$s = \frac{a + b + c}{2}$ $s=(a+b+c)/2$

Here,

$a = 2$ $a=2$

$b = 3$ $b=3$

$c = 4$ $c=4$

Therefore,

$s = \frac{2 + 3 + 4}{2} = \frac{9}{2} = 4.5$ $s=(2+3+4)/2=9/2=4.5$

So,

$a r e a = \sqrt{4.5 \cdot (4.5 - 2) \cdot (4.5 - 3) \cdot (4.5 - 4)}$ $area = sqrt (4.5*(4.5-2) * (4.5-3) * (4.5-4) )$

$= \sqrt{8.3475}$ $=sqrt8.3475$

$= 2.9 u^{2}$ $=2.9u^2$

Where $u$ $u$ , represents the units in this case.