Triangles & Pythagorean Theorem | Math Learning Guide

📋 Prerequisites

Triangles are the building blocks of all geometry. Know their properties, the Pythagorean theorem, and the special right triangles, and you can solve half the geometry problems on any competition. This is the foundation everything else builds on.

📚 Key Concepts

Triangle Classification

By sides:

Scalene: all sides different lengths
Isosceles: at least two sides equal (the base angles are also equal)
Equilateral: all three sides equal (all angles = 60°)

By angles:

Acute: all angles < 90°
Right: one angle = 90° (the side opposite is the hypotenuse)
Obtuse: one angle > 90°

Angle Properties

Sum of angles: The three interior angles of any triangle add to 180°
Exterior angle: equals the sum of the two remote interior angles
Isosceles triangle: equal sides opposite equal angles (base angles theorem)
Equilateral triangle: all angles = 60°

Triangle Inequality

The sum of any two sides must be greater than the third side:

$ a + b > c $, $ a + c > b $, $ b + c > a $

Equivalently: the largest side must be less than the sum of the other two.

Given two sides, the third side is between $|a-b|$ and $a+b$ (strictly).

Area of a Triangle

Standard formula:

\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]

Height must be perpendicular to the base (altitude).

Equilateral triangle area: $ \frac{\sqrt{3}}{4} s^2 $ where $s$ = side length

Heron's formula: $ \sqrt{s(s-a)(s-b)(s-c)} $ where $s = \frac{a+b+c}{2}$ (semiperimeter)

Pythagorean Theorem

For a right triangle with legs $a, b$ and hypotenuse $c$:

\[ a^2 + b^2 = c^2 \]

Converse: If $a^2 + b^2 = c^2$, then the triangle is right-angled.

Pythagorean triples: Integer solutions like (3, 4, 5), (5, 12, 13), (7, 24, 25), (8, 15, 17)

Multiples work too: (6, 8, 10), (9, 12, 15), etc.

Special Right Triangles

45-45-90 (Isosceles right triangle):

Legs are equal: $ x, x $
Hypotenuse: $ x\sqrt{2} $
Ratio: $ 1 : 1 : \sqrt{2} $

30-60-90 triangle:

Short leg (opposite 30°): $ x $
Long leg (opposite 60°): $ x\sqrt{3} $
Hypotenuse: $ 2x $
Ratio: $ 1 : \sqrt{3} : 2 $

These come up constantly in competition problems — memorize the ratios!

Altitude-on-Hypotenuse Theorems

In a right triangle, drop an altitude from the right angle to the hypotenuse:

It creates two smaller right triangles, both similar to the original and to each other
Geometric mean theorem (leg rule): Each leg is the geometric mean of the hypotenuse and the adjacent segment
Altitude rule: The altitude is the geometric mean of the two hypotenuse segments

⚠️ Common Mistake: The Pythagorean theorem only works on right triangles! Don't use $a^2 + b^2 = c^2$ on acute or obtuse triangles. Also: the hypotenuse is always the longest side and always opposite the right angle — make sure you're squaring the right side.

💡 Key Insight: When you see a problem with side lengths that look like Pythagorean triples (or multiples), there's almost certainly a right triangle somewhere. Drop an altitude, draw a diagonal, or look for hidden right angles. The 3-4-5 and 5-12-13 families show up everywhere.

✏️ Example Problems

📝 Example 1 (Pythagorean theorem)

A right triangle has legs of length 9 and 12. What is the length of the hypotenuse?

Step 1: Apply the Pythagorean theorem: $ a^2 + b^2 = c^2 $

Legs: $ a = 9 $, $ b = 12 $. Find hypotenuse $c$.

\[ c^2 = 9^2 + 12^2 = 81 + 144 = 225 \]

Step 2: Take the square root.

$ c = \sqrt{225} = 15 $

Quick recognition: 9-12-15 is just the 3-4-5 triple multiplied by 3. ✓

Answer: 15

📝 Example 2 (Special right triangle 30-60-90)

In a 30-60-90 triangle, the side opposite the 60° angle has length $ 5\sqrt{3} $.
What is the length of the hypotenuse?

Recall 30-60-90 ratios: short leg : long leg : hypotenuse = $ 1 : \sqrt{3} : 2 $

Step 1: The side opposite 60° is the long leg = $ x\sqrt{3} $

Given: long leg = $ 5\sqrt{3} $

So $ x\sqrt{3} = 5\sqrt{3} $, which means $ x = 5 $.

Step 2: The short leg = $ x = 5 $.

Step 3: The hypotenuse = $ 2x = 2(5) = 10 $.

Check: sides are 5 (short), $5\sqrt{3}$ (long), 10 (hypotenuse). Ratio $1:\sqrt{3}:2$. ✓

Answer: 10

📝 Example 3 (Triangle inequality + perimeter)

A triangle has two sides of length 7 and 12. If the third side has integer length,
what is the maximum possible perimeter of the triangle?

Step 1: Apply triangle inequality.

The third side $x$ must satisfy:

$ 12 - 7 < x < 12 + 7 $

$ 5 < x < 19 $

Step 2: Find the maximum integer value of $x$.

$x$ must be strictly less than 19, so the largest integer is 18.

Step 3: Calculate the perimeter.

Perimeter = $ 7 + 12 + 18 = 37 $

Check: 7 + 12 > 18? 19 > 18 ✓. Triangle inequality satisfied.

Answer: 37