How do you use Heron's formula to find the area of a triangle with sides of lengths $4$ $4$ , $4$ $4$ , and $5$ $5$ ?

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How do you use Heron's formula to find the area of a triangle with sides of lengths $4$ $4$ , $4$ $4$ , and $5$ $5$ ?

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Answer 1 · 2016-01-24T22:36:53

$A = \frac{5 \sqrt{39}}{4} \approx 7.806$ $A=(5sqrt39)/4approx7.806$

Explanation:

Heron's formula states that for a triangle with sides $a, b, c$ $a,b,c$ and a semiperimeter $s = \frac{a + b + c}{2}$ $s=(a+b+c)/2$ , the area of the triangle is

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

Here, we know that

$s = \frac{4 + 4 + 5}{2} = \frac{13}{2}$ $s=(4+4+5)/2=13/2$

which gives an area of

$A = \sqrt{\frac{13}{2} (\frac{13}{2} - 4) (\frac{13}{2} - 4) (\frac{13}{2} - 5)}$ $A=sqrt(13/2(13/2-4)(13/2-4)(13/2-5))$

$A = \sqrt{\frac{13}{2} (\frac{5}{2}) (\frac{5}{2}) (\frac{3}{2})}$ $A=sqrt(13/2(5/2)(5/2)(3/2))$

$A = \sqrt{\frac{39 \times 5^{2}}{4^{2}}}$ $A=sqrt((39xx5^2)/4^2)$

$A = \frac{5 \sqrt{39}}{4} \approx 7.806$ $A=(5sqrt39)/4approx7.806$

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How do you use Heron's formula to find the area of a triangle with sides of lengths $4$ $4$ , $4$ $4$ , and $5$ $5$ ?

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How do you use Heron's formula to find the area of a triangle with sides of lengths $4$ $4$ , $4$ $4$ , and $5$ $5$ ?