Areas of Triangles

• $\text{Area} = \dfrac12\,b\,h$, where $h$ is the perpendicular to the chosen base.
• $\text{Area} = \dfrac12\,ab\sin C$, where $C$ is the angle between sides $a$ and $b$.
• In a right triangle either leg is the height for the other, and the altitude to the hypotenuse is $\dfrac{\text{leg}_1\times\text{leg}_2}{\text{hypotenuse}}$.

When the foot of an altitude is unknown, write the altitude's square twice (once from each side) and equate.

• Semi-perimeter $s = \dfrac{a + b + c}{2}$.
• $\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}$.
• Test first: if $a^2 + b^2 = c^2$, the area is $\dfrac12 ab$.
• Isosceles with equal sides $p$ and base $q$: the height is $\sqrt{p^2 - \dfrac{q^2}{4}}$.

Common triples: $3, 4, 5$; $5, 12, 13$; $8, 15, 17$; $7, 24, 25$; $9, 40, 41$; $11, 60, 61$; $20, 21, 29$.

For legs $a, b$ and hypotenuse $c$:

• $a + b = \text{perimeter} - c$ and $a^2 + b^2 = c^2$.
• $2ab = (a + b)^2 - c^2$, so the area is $\dfrac{(a+b)^2 - c^2}{4}$.
• Isosceles right triangle with leg $L$: perimeter $L(2 + \sqrt2)$, area $\dfrac{L^2}{2}$.
• With the hypotenuse fixed, the area is largest when the triangle is isosceles: $\dfrac{c^2}{4}$.
• The incircle of a right triangle has radius $\dfrac{a + b - c}{2}$.

For side $a$:

• Height (which is also the median and the angle bisector) $= \dfrac{\sqrt3}{2}a$.
• Area $= \dfrac{\sqrt3}{4}a^2$.
• From the height $h$: area $= \dfrac{h^2}{\sqrt3}$.

Areas of equilateral triangles are proportional to the squares of their sides, so $n^2$ small ones tile a big one of $n$ times the side.

• Similar figures: $\dfrac{\text{Area}_1}{\text{Area}_2} = \left(\dfrac{\text{side}_1}{\text{side}_2}\right)^2$.
• The midpoint triangle has $\dfrac14$ of the area; the trapezium left over has $\dfrac34$.
• A percentage change of $p\%$ in every side changes the area by the factor $\left(1 + \dfrac{p}{100}\right)^2$.

• Pairwise sums $a + b$, $b + c$, $c + a$: add all three to get $2(a + b + c)$, then subtract each.
• "$b + c$ exceeds $a$ by $k$" gives $s - a = \dfrac{k}{2}$ directly, which is exactly what Heron needs.
• Altitudes in a ratio: since $a\,h_a = b\,h_b = c\,h_c = 2\times\text{Area}$, the sides are in the inverse ratio of the altitudes.
• Sides written in letters: look for a coordinate triangle, or substitutions that make $s - a$, $s - b$, $s - c$ simple.