How do you use Heron's formula to determine the area of a triangle with sides of that are 14, 16, and 17 units in length?

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Answer 1 · 2016-04-10T17:44:05

$= 104.324$ $=104.324$ square units

Heron's formula for area of triangle is:

$A = \sqrt{s (s - a) (s - b) (s - c)}$ $A = sqrt(s(s-a)(s-b)(s-c)$ , where $s$ $s$ is the semi-pertimeter.

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

Here, $a = 14$ $a=14$ , $b = 16$ $b=16$ and $c = 17$ $c=17$ .

First find $s$ $s$ :

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

$= \frac{14 + 16 + 17}{2} = \frac{47}{2}$ $=(14+16+17)/2=47/2$

Now to calculate the area:

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

$= \sqrt{\frac{47}{2} (\frac{47}{2} - 14) (\frac{47}{2} - 16) (\frac{47}{2} - 17)}$ $= sqrt(47/2(47/2-14)(47/2-16)(47/2-17)$

$= \sqrt{\frac{47}{2} (\frac{47 - 28}{2}) (\frac{47 - 32}{2}) (\frac{47 - 34}{2})}$ $= sqrt(47/2((47-28)/2)((47-32)/2)((47-34)/2)$

$= \sqrt{\frac{47}{2} (\frac{19}{2}) (\frac{15}{2}) (\frac{13}{2})}$ $= sqrt(47/2(19/2)(15/2)(13/2)$

$= \sqrt{\frac{47 \times 19 \times 15 \times 13}{16}}$ $=sqrt((47xx19xx15xx13)/16)$

$= \frac{1}{4} \sqrt{47 \times 19 \times 15 \times 13}$ $=1/4sqrt(47xx19xx15xx13)$

$= \frac{1}{4} \sqrt{174135}$ $=1/4sqrt(174135)$

$= \frac{1}{4} \times 417.294$ $=1/4xx417.294$

$= 104.324$ $=104.324$