How do you use Heron's formula to find the area of a triangle with sides of lengths $3$ $3$ , $3$ $3$ , and $4$ $4$ ?

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How do you use Heron's formula to find the area of a triangle with sides of lengths $3$ $3$ , $3$ $3$ , and $4$ $4$ ?

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Answer 1 · 2016-01-30T12:57:27

$A r e a = 4.47213$ $Area=4.47213$ square units

Explanation:

Heron's formula for finding area of the triangle is given by
$A r e a = \sqrt{s (s - a) (s - b) (s - c)}$ $Area=sqrt(s(s-a)(s-b)(s-c))$

Where $s$ $s$ is the semi perimeter and is defined as
$s = \frac{a + b + c}{2}$ $s=(a+b+c)/2$

and $a, b, c$ $a, b, c$ are the lengths of the three sides of the triangle.

Here let $a = 3, b = 3$ $a=3, b=3$ and $c = 4$ $c=4$

$\Rightarrow s = \frac{3 + 3 + 4}{2} = \frac{10}{2} = 5$ $implies s=(3+3+4)/2=10/2=5$

$\Rightarrow s = 5$ $implies s=5$

$\Rightarrow s - a = 5 - 3 = 2, s - b = 5 - 3 = 2 and s - c = 5 - 4 = 1$ $implies s-a=5-3=2, s-b=5-3=2 and s-c=5-4=1$
$\Rightarrow s - a = 2, s - b = 2 and s - c = 1$ $implies s-a=2, s-b=2 and s-c=1$

$\Rightarrow A r e a = \sqrt{5 \cdot 2 \cdot 2 \cdot 1} = \sqrt{20} = 4.47213$ $implies Area=sqrt(5*2*2*1)=sqrt20=4.47213$ square units

$\Rightarrow A r e a = 4.47213$ $implies Area=4.47213$ square units

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