Heron’s formula, derived in Question 11.2.13, states that the area of a triangle with side lengths $a$, $b$, and $c$ is \[\sqrt{s(s-a)(s-b)(s-c)}\] where $s = \frac{1}{2}(a + b + c)$. Find the area of the triangle with side lengths $6$, $25$, and $29$. Show answer
$a = 6$, $b = 25$, and $c = 29$. \begin{align*} s &= \frac{1}{2}(a + b + c)\\ &= \frac{1}{2}(6 + 25 + 29)\\ &= \frac{1}{2}60\\ &= 30. \end{align*} \begin{align*} \sqrt{s(s-a)(s-b)(s-c)} &= \sqrt{30(30 - 6)(30 - 25)(30 - 29)}\\ &= \sqrt{30\times 24 \times 5 \times 1}\\ &= \sqrt{30 \times 120}\\ &= \sqrt{3600}\\ &= 60 \end{align*}