How do you use Heron's formula to determine the area of a triangle with sides of that are 4, 6, and 3 units in length?

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How do you use Heron's formula to determine the area of a triangle with sides of that are 4, 6, and 3 units in length?

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Answer 1 · 2016-04-27T10:32:43

${Area}_{△} \approx 5.3$ $"Area"_triangle~~5.3$ sq.units
(see below for use of Heron's formula)

Explanation:

Heron's formula tells us how to calculate the area of a triangle given the lengths of it's three sides.

If (for the general case) the lengths of the three sides are $a, b, and c$ $a, b, and c$ and the semi-perimeter is $s = \frac{a + b + c}{2}$ $s=(a+b+c)/2$

Then
$XXX {Area}_{△} = \sqrt{s (s - a) (s - b) (s - c)}$ $color(white)("XXX")"Area"_triangle=sqrt(s(s-a)(s-b)(s-c))$

For the given triangle with sides $4, 6, and 3$ $4, 6, and 3$
$XXX s = \frac{13}{2}$ $color(white)("XXX")s=13/2$
and
$XXX {Area}_{△} = \sqrt{(\frac{13}{2}) (\frac{13}{2} - 4) (\frac{13}{2} - 6) (\frac{13}{2} - 3)}$ $color(white)("XXX")"Area"_triangle = sqrt((13/2)(13/2-4)(13/2-6)(13/2-3))$

$XXXXXXX = \sqrt{\frac{13}{2} \times \frac{5}{2} \times \frac{1}{2} \times \frac{7}{2}}$ $color(white)("XXXXXXX")=sqrt(13/2xx5/2xx1/2xx7/2)$

$XXXXXXX = \frac{\sqrt{455}}{4}$ $color(white)("XXXXXXX")=sqrt(455)/4$

$XXXXXXX \approx 5.3$ $color(white)("XXXXXXX")~~5.3$ sq.units

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How do you use Heron's formula to determine the area of a triangle with sides of that are 4, 6, and 3 units in length?

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