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How do you use Heron's formula to find the area of a triangle with sides of lengths 15 , 16 , and 14 ?

Mar 8, 2018

$\text{Area} \approx 96.56$

Explanation:

Heron's formula says that:
$\text{Area} = \sqrt{S \left(S - A\right) \left(S - B\right) \left(S - C\right)}$ given $S = \frac{A + B + C}{2}$

For this triangle, $S = \frac{14 + 15 + 16}{2} = 22.5$

$\text{Area} = \sqrt{22.5 \left(22.25 - 14\right) \left(22.5 - 15\right) \left(22.5 - 16\right)} = \sqrt{\frac{149175}{16}} = \frac{15 \sqrt{663}}{4} \approx 96.56$

Apr 30, 2018

Know more than the other students! Archimedes' Theorem is a modern form of Heron's Formula. We set $a = 15 , b = 16 , c = 14$ in

$16 {A}^{2} = 4 {a}^{2} {b}^{2} - {\left({c}^{2} - {a}^{2} - {b}^{2}\right)}^{2} = 149175$

$A = \frac{15 \sqrt{663}}{4}$

Explanation:

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

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

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

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

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

No semiperimeter, no fractions till the end, 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.