Triangles: Area & Pythagorean

What you need to know

• Area. Area $= \frac12 bh$, with $h$ the height to base $b$. With two sides $a$, $b$ and the angle $C$ between them, area $= \frac12 ab\sin C$. An equilateral triangle with side $s$ has area $\frac{\sqrt3}{4}s^2$. Example: sides $6$ and $8$ with a $30^\circ$ angle between them give area $\frac12\cdot6\cdot8\cdot\frac12 = 12$.
• Pythagorean Theorem. In a right triangle with legs $a$, $b$ and hypotenuse $c$, $a^2+b^2=c^2$. The converse is true too: if $a^2+b^2=c^2$, the angle across from $c$ is a right angle. Know the triples $3$-$4$-$5$, $5$-$12$-$13$, $8$-$15$-$17$, $7$-$24$-$25$, and their multiples, such as $9$-$12$-$15$.
• Special right triangles. A $45$-$45$-$90$ triangle has sides in the ratio $1:1:\sqrt2$. A $30$-$60$-$90$ triangle has sides $1:\sqrt3:2$, with the short leg across from $30^\circ$. Example: a $30$-$60$-$90$ triangle with hypotenuse $10$ has legs $5$ and $5\sqrt3$.
• Triangle Inequality. Any two sides add up to more than the third: $a+b>c$, strictly. Example: $4$, $5$, $9$ is not a triangle, since $4+5 = 9$.
• Acute, right, or obtuse? Let $c$ be the longest side. If $a^2+b^2 > c^2$, the triangle is acute; if equal, right; if less, obtuse. Example: $5$, $6$, $8$ is obtuse, since $25+36 = 61 < 64$.
• Heron's Formula. Area $=\sqrt{s(s-a)(s-b)(s-c)}$, where $s$ is the semiperimeter (half the perimeter). Example: sides $5$, $5$, $6$ give $s = 8$ and area $\sqrt{8\cdot3\cdot3\cdot2} = 12$.

How AMC 10 tests it

• Problems 3–10: a figure of squares and right triangles; find a length or area by dropping one perpendicular.
• "How many integers $x$ make a triangle with sides $a$, $b$, $x$?" or "...an obtuse triangle?"
• Mid-paper: two sides and an altitude, or a $60^\circ$ or $120^\circ$ angle. The altitude splits the triangle into two right triangles that share a leg.
• Shaded regions from overlapping triangles, or a triangle cut from a rectangle, found by subtraction.

Standard approaches

1. Drop an altitude and write a Pythagorean equation on each side of it; subtracting them removes the altitude.
2. Look for special right triangles or Pythagorean triples before working with square roots.
3. Find the area in two ways (two base–height pairs) to get an unknown height. Example: the $3$-$4$-$5$ triangle has area $6$, so the height $h$ to the hypotenuse has $\frac12\cdot5\cdot h = 6$, and $h = \frac{12}{5}$.
4. For shaded regions, subtract simple pieces from a rectangle or triangle around the region.

Worked example

In triangle $ABC$, $AB = 13$, $AC = 15$, and the altitude from $A$ to line $BC$ has length $12$. What is the sum of all possible values of $BC$?

(A) $14$ (B) $16$ (C) $18$ (D) $20$ (E) $22$

Solution. Let $D$ be the foot of the altitude, so $AD = 12$.

Step 1. Right triangle $ABD$ has hypotenuse $13$ and leg $12$, so it is a $5$-$12$-$13$ triangle and $BD = 5$.

Step 2. Right triangle $ACD$ has hypotenuse $15$ and leg $12$, so it is a $9$-$12$-$15$ triangle and $CD = 9$.

Step 3. If $D$ lies between $B$ and $C$, then $BC = 5 + 9 = 14$. If $D$ lies outside segment $BC$, beyond $B$ (so angle $B$ is obtuse), then $BC = 9 - 5 = 4$. Both triangles exist, so both values count.

Step 4. The sum is $14 + 4 = 18$.

The answer is $\boxed{\textbf{(C)}\ 18}$.

Pitfalls

• Forgetting the obtuse picture, where the foot of an altitude falls outside the triangle. In the example, that picture gives $BC = 4$.
• Mixing up the legs of a $30$-$60$-$90$ triangle. The longer leg is across from $60^\circ$ and is $\sqrt3$ times the shorter leg.
• Allowing $a+b=c$ (a flat triangle) when counting integer side lengths. Example: with sides $3$ and $7$, the third side can be $5$, $6$, $7$, $8$, or $9$, but not $4$ or $10$.

Traps that recur

• Counting 50 or 51 values per coordinate (including boundary points, which are not interior), or forgetting to subtract the points that are also 60-ray partitional. (2011 AMC 10A #25)
• Reading 'the altitude to side AB' as a segment drawn from B, which would put P at B; the altitude meant is the one from the opposite vertex C, and its foot H is what fixes BP. (2025 AMC 10A #23)
• Misreading the kite: its angles are 60°, 90°, 120°, 90° (the two √3 sides meet at 60°, the two unit sides at 120°), and its long diagonal is 2. (2024 AMC 10A #22)
• Guessing 1:1 because AC = CD and DE = EB look symmetric, or setting AD:DB = AC:DE = 4:3 as if it were an angle bisector. (2019 AMC 10B #16)