How do you use Heron's formula to find the area of a triangle with sides of lengths $14$ $14$ , $9$ $9$ , and $13$ $13$ ?

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Answer 1 · 2018-05-27T11:13:46

$Area = 18 \sqrt{10} square units$ $color(blue)("Area"=18sqrt(10)" square units")$

$A r e a = \sqrt{s (s - a) (s - b) (s - c)}$ $"Area=sqrt(s(s-a)(s-b)(s-c))$

Where $a, b and c$ $a, b and c$ are the lengths of the triangles sides.

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

Let: $a = 14$ $a=14$ , $b = 13$ $b=13$ . $c = 9$ $c=9$

Then:

$s = \frac{14 + 13 + 9}{2} =$ $s=(14+13+9)/2=$ 18

$Area = \sqrt{18 (18 - 14) (18 - 13) (18 - 9)}$ $"Area"=sqrt(18(18-14)(18-13)(18-9))$

$\ \ \ \ \ \ \ \ \ =sqrt(18(4)(5)(9)$

$\ \ \ \ \ \ \ \ \ =sqrt(3240)=18sqrt(10)$ square units

How do you use Heron's formula to find the area of a triangle with sides of lengths 14 1414 , 9 99 , and 13 1313 ?