# How do you use Heron's formula to find the area of a triangle with sides of lengths 6 , 6 , and 7 ?

Apr 24, 2018

$17.06 \setminus {\text{units}}^{2}$

#### Explanation:

Heron's formula states for any triangle, the area is given by:

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

• $s$ is the semiperimeter of the triangle, given by $s = \frac{a + b + c}{2}$.

• $a , b , c$ are the sides of the triangle

Here, $s = \frac{6 + 6 + 7}{2} = \frac{19}{2} = 9.5$ units.

And so, the area is:

$A = \sqrt{9.5 \left(9.5 - 6\right) \left(9.5 - 6\right) \left(9.5 - 7\right)}$

$= \sqrt{9.5 \cdot 3.5 \cdot 3.5 \cdot 2.5}$

$= \sqrt{290.9375}$

$\approx 17.06$

Apr 30, 2018

Archimedes' Theorem is a modern form of Heron's Formula. We set $a = c = 6 , b = 7$ in

$16 {A}^{2} = 4 {a}^{2} {b}^{2} - {\left({c}^{2} - {a}^{2} - {b}^{2}\right)}^{2} = 4655$

$A = \frac{7 \sqrt{95}}{4}$

#### Explanation:

Archimedes' Theorem is usually superior to Heron's Formula.

Given a triangle with sides $a , b , c$ and area $A$, we have

$16 {A}^{2} = 4 {a}^{2} {b}^{2} - {\left({c}^{2} - {a}^{2} - {b}^{2}\right)}^{2}$

$= {\left({a}^{2} + {b}^{2} + {c}^{2}\right)}^{2} - 2 \left({a}^{4} + {b}^{4} + {c}^{4}\right)$

$= \left(a + b + c\right) \left(- a + b + c\right) \left(a - b + c\right) \left(a + b - c\right)$

No semiperimeter, no multiple square roots when given coordinates.

We're free to choose whichever form is convenient and assign $a , b , c$ however we see fit. For an isosceles triangle, $a = c$ gives nice cancellation in the first form. Let's set $a = c = 6 , b = 7$ in

$16 {A}^{2} = 4 \left({6}^{2}\right) \left({7}^{2}\right) - {7}^{4} = 4655$

$A = \setminus \sqrt{\frac{4744}{16}} = \frac{7 \sqrt{95}}{4}$