How do you use Heron's formula to find the area of a triangle with sides of lengths $11$ $11$ , $14$ $14$ , and $13$ $13$ ?

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Answer 1 · 2016-02-24T17:35:37

$A r e a = 67.53 u n i t s^{2}$ $Area = 67.53 units^2$

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First we would find S which is the sum of the 3 sides divided by 2.

$S = \frac{11 + 14 + 13}{2}$ $S = (11 + 14 + 13)/2$ = $\frac{38}{2}$ $38/2$ = 19

Then use Heron's Equation to calculate the area.

$A r e a = \sqrt{S (S - A) (S - B) (S - C)}$ $Area = sqrt(S(S-A)(S-B)(S-C))$

$A r e a = \sqrt{19 (19 - 11) (19 - 14) (19 - 13)}$ $Area = sqrt(19(19-11)(19-14)(19-13))$

$A r e a = \sqrt{19 (8) (5) (6)}$ $Area = sqrt(19(8)(5)(6))$

$A r e a = \sqrt{4, 560}$ $Area = sqrt(4,560)$

$A r e a = 67.53 u n i t s^{2}$ $Area = 67.53 units^2$

How do you use Heron's formula to find the area of a triangle with sides of lengths 1111 , 1414 , and 1313 ?