AMC 10 Step by Step
Printable problems

Triangles: Area & Pythagorean

Triangle area, Pythagorean theorem, special right triangles, triangle inequality

What you need to know

  • Area. Area =12bh= \frac12 bh, with hh the height to base bb. With two sides aa, bb and the angle CC between them, area =12absin⁡C= \frac12 ab\sin C. An equilateral triangle with side ss has area 34s2\frac{\sqrt3}{4}s^2. Example: sides 66 and 88 with a 30∘30^\circ angle between them give area 12⋅6⋅8⋅12=12\frac12\cdot6\cdot8\cdot\frac12 = 12.
  • Pythagorean Theorem. In a right triangle with legs aa, bb and hypotenuse cc, a2+b2=c2a^2+b^2=c^2. The converse is true too: if a2+b2=c2a^2+b^2=c^2, the angle across from cc is a right angle. Know the triples 33-44-55, 55-1212-1313, 88-1515-1717, 77-2424-2525, and their multiples, such as 99-1212-1515.
  • Special right triangles. A 4545-4545-9090 triangle has sides in the ratio 1:1:21:1:\sqrt2. A 3030-6060-9090 triangle has sides 1:3:21:\sqrt3:2, with the short leg across from 30∘30^\circ. Example: a 3030-6060-9090 triangle with hypotenuse 1010 has legs 55 and 535\sqrt3.
  • Triangle Inequality. Any two sides add up to more than the third: a+b>ca+b>c, strictly. Example: 44, 55, 99 is not a triangle, since 4+5=94+5 = 9.
  • Acute, right, or obtuse? Let cc be the longest side. If a2+b2>c2a^2+b^2 > c^2, the triangle is acute; if equal, right; if less, obtuse. Example: 55, 66, 88 is obtuse, since 25+36=61<6425+36 = 61 < 64.
  • Heron's Formula. Area =s(s−a)(s−b)(s−c)=\sqrt{s(s-a)(s-b)(s-c)}, where ss is the semiperimeter (half the perimeter). Example: sides 55, 55, 66 give s=8s = 8 and area 8⋅3⋅3⋅2=12\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 xx make a triangle with sides aa, bb, xx?" or "...an obtuse triangle?"
  • Mid-paper: two sides and an altitude, or a 60∘60^\circ or 120∘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 33-44-55 triangle has area 66, so the height hh to the hypotenuse has 12⋅5⋅h=6\frac12\cdot5\cdot h = 6, and h=125h = \frac{12}{5}.
  4. For shaded regions, subtract simple pieces from a rectangle or triangle around the region.

Worked example

In triangle ABCABC, AB=13AB = 13, AC=15AC = 15, and the altitude from AA to line BCBC has length 1212. What is the sum of all possible values of BCBC?

(A) 1414 (B) 1616 (C) 1818 (D) 2020 (E) 2222

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

Step 1. Right triangle ABDABD has hypotenuse 1313 and leg 1212, so it is a 55-1212-1313 triangle and BD=5BD = 5.

Step 2. Right triangle ACDACD has hypotenuse 1515 and leg 1212, so it is a 99-1212-1515 triangle and CD=9CD = 9.

Step 3. If DD lies between BB and CC, then BC=5+9=14BC = 5 + 9 = 14. If DD lies outside segment BCBC, beyond BB (so angle BB is obtuse), then BC=9−5=4BC = 9 - 5 = 4. Both triangles exist, so both values count.

Step 4. The sum is 14+4=1814 + 4 = 18.

The answer is (C) 18\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=4BC = 4.
  • Mixing up the legs of a 3030-6060-9090 triangle. The longer leg is across from 60∘60^\circ and is 3\sqrt3 times the shorter leg.
  • Allowing a+b=ca+b=c (a flat triangle) when counting integer side lengths. Example: with sides 33 and 77, the third side can be 55, 66, 77, 88, or 99, but not 44 or 1010.

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)

Problems © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. The topic note, tags and key insights are original to this site.