# How do you use Heron's formula to determine the area of a triangle with sides of that are 25, 28, and 22 units in length?

Mar 10, 2016

Area of triangle is $262.72$ units

#### Explanation:

Heron's formula gives the area of a triangle with sides $a$, $b$ and $c$ as $\sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$ where $s = \frac{1}{2} \left(a + b + c\right)$.

Hence to determine the area of a triangle with sides of $25 , 28$ and $22$ units, first we find $s$, which is given by $s = \frac{1}{2} \left(25 + 28 + 22\right) = 37.5$.

Hence area of triangle is

$\sqrt{37.5 \times \left(37.5 - 25\right) \times \left(37.5 - 28\right) \times \left(37.5 - 22\right)}$ or

$\sqrt{37.5 \times 12.5 \times 9.5 \times 15.5}$ or $\sqrt{69 , 023.4375}$ or $262.72$ units

May 1, 2018

$16 {A}^{2} = \left(25 + 28 + 22\right) \left(- 25 + 28 + 22\right) \left(25 - 28 + 22\right) \left(25 + 28 - 22\right)$$= \left(75\right) \left(25\right) \left(19\right) \left(31\right)$

$A = \setminus \sqrt{\frac{3 \left(25\right) \left(25\right) \left(19\right) \left(31\right)}{16}} = \frac{25}{4} \sqrt{1767}$

#### Explanation:

Heron's formula is usually among the poorest choices. Here are some modern forms:

For a triangle with sides $a , b , c$ and area $A$,

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

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

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

$= 16 s \left(s - a\right) \left(s - b\right) \left(s - c\right)$ where $s = \frac{1}{2} \left(a + b + c\right)$

The last is of course Heron.

Usually given integer sides the factored form is fast and leads to the least messy exact answer:

$16 {A}^{2} = \left(25 + 28 + 22\right) \left(- 25 + 28 + 22\right) \left(25 - 28 + 22\right) \left(25 + 28 - 22\right)$

$= \left(75\right) \left(25\right) \left(19\right) \left(31\right)$

$A = \setminus \sqrt{\frac{3 \left(25\right) \left(25\right) \left(19\right) \left(31\right)}{16}} = \frac{25}{4} \setminus \sqrt{3 \left(19\right) \left(31\right)} = \frac{25}{4} \sqrt{1767}$

If we're given 2d coordinates, the Shoelace Theorem is the quickest way to the area. If we're given 3 or more dimensional coordinates, the first two forms, which rely only on squared lengths, are best.

The first form is great for deriving the special cases as well:

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

Equilateral triangle : $a = b = c$

$16 {A}^{2} = 4 {a}^{4} - {a}^{4} = 3 {a}^{4} \quad$ or $\quad A = \setminus \frac{\sqrt{3}}{4} {a}^{2}$

Isosceles triangle: $a = c ,$ base $b .$

$16 {A}^{2} = 4 {a}^{2} {b}^{2} - {b}^{4} = {b}^{2} \left(4 {a}^{2} - {b}^{2}\right) \mathmr{and} A = \frac{b}{4} \setminus \sqrt{4 {a}^{2} - {b}^{2}}$

Right Triangle ${c}^{2} = {a}^{2} + {b}^{2}$

$16 {A}^{2} = 4 {a}^{2} {b}^{2} - {\left({c}^{2} - {a}^{2} - {b}^{2}\right)}^{2} = 4 {a}^{2} {b}^{2} - 0$ or $A = \frac{1}{2} a b$

Cool. I'm not sure why they don't teach it in school.