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 $3.8$ $3.8$ square units

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

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

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

and area of triangle is

$\sqrt{\frac{11}{2} (\frac{11}{2} - 2) (\frac{11}{2} - 4) (\frac{11}{2} - 5)}$ $sqrt(11/2(11/2-2)(11/2-4)(11/2-5))$

= $\sqrt{\frac{11}{2} \times \frac{7}{2} \times \frac{3}{2} \times \frac{1}{2}}$ $sqrt(11/2xx7/2xx3/2xx1/2)$

= $\frac{1}{4} \sqrt{231} = \frac{15.19868}{4} = 3.79967$ $1/4sqrt231=15.19868/4=3.79967$ or say $3.8$ $3.8$