Pythagorean Theorem Basics

The Pythagorean Theorem is a fundamental principle in geometry, attributed to the ancient Greek mathematician Pythagoras (circa 570–495 BCE), though evidence suggests it was known to earlier civilizations like the Babylonians and Egyptians. It applies to right triangles—triangles with one 90-degree angle—and describes the relationship between the lengths of the sides. This theorem is not only a cornerstone of Euclidean geometry but also extends to three-dimensional space and coordinate systems, making it essential for fields like architecture, physics, and navigation.

The Formula

For a right triangle with legs \( a \) and \( b \) (the two sides forming the right angle) and hypotenuse \( c \) (the side opposite the right angle):

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

Where:

\( a \) and \( b \): Lengths of the legs
\( c \): Length of the hypotenuse
The theorem holds only for right triangles; for non-right triangles, other formulas like the Law of Cosines apply.

This equation can be rearranged to solve for any side: \( c = \sqrt{a^2 + b^2} \), \( a = \sqrt{c^2 - b^2} \), or \( b = \sqrt{c^2 - a^2} \), provided the expressions under the square root are non-negative. The theorem also underpins the concept of Pythagorean triples—sets of three positive integers that satisfy the equation (e.g., 3-4-5, 5-12-13).

Examples: Various Triangles, 3D Distance, Word Problems, Pythagorean Triples, Coordinate Geometry

Let’s explore the Pythagorean Theorem with diverse scenarios and detailed calculations.

Example 1: 3-4-5 Triangle

Given legs \( a = 3 \), \( b = 4 \), find the hypotenuse \( c \):

\[ a^2 + b^2 = c^2 \] \[ 3^2 + 4^2 = 9 + 16 \] \[ = 25 \] \[ c^2 = 25 \] \[ c = \sqrt{25} \] \[ c = 5 \]

Verification: A triangle with sides 3, 4, and 5 is a right triangle.

Example 2: Decimal Sides

Given legs \( a = 2.5 \), \( b = 6.2 \), find the hypotenuse \( c \):

\[ a^2 + b^2 = c^2 \] \[ (2.5)^2 + (6.2)^2 = 6.25 + 38.44 \] \[ = 44.69 \] \[ c^2 = 44.69 \] \[ c = \sqrt{44.69} \] \[ c \approx 6.69 \]

Hypotenuse is approximately 6.69 units.

Example 3: 3D Distance

Find the distance between points \( (0, 0, 0) \) and \( (3, 4, 5) \) in 3D space using the 3D Pythagorean Theorem (distance formula):

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \] \[ d = \sqrt{(3 - 0)^2 + (4 - 0)^2 + (5 - 0)^2} \] \[ = \sqrt{3^2 + 4^2 + 5^2} \] \[ = \sqrt{9 + 16 + 25} \] \[ = \sqrt{50} \] \[ \approx 7.07 \]

Distance is approximately 7.07 units.

Example 4: Word Problem

A ladder 10 m long leans against a wall. The base is 6 m from the wall. How high up the wall does the ladder reach?

\[ a = 6 \text{ (base)}, c = 10 \text{ (ladder)} \] \[ a^2 + b^2 = c^2 \] \[ 6^2 + b^2 = 10^2 \] \[ 36 + b^2 = 100 \] \[ b^2 = 100 - 36 \] \[ b^2 = 64 \] \[ b = \sqrt{64} \] \[ b = 8 \]

The ladder reaches 8 m up the wall.

Example 5: Verifying Non-Right Triangle

Check if sides 5, 6, 7 form a right triangle:

\[ 5^2 + 6^2 = 25 + 36 = 61 \] \[ 7^2 = 49 \] \[ 61 \neq 49 \]

Since \( 5^2 + 6^2 \neq 7^2 \), it is not a right triangle.

Example 6: Pythagorean Triples

Generate a triple: Start with 5-12-13 and scale by 2:

\[ a = 5 \cdot 2 = 10 \] \[ b = 12 \cdot 2 = 24 \] \[ c = 13 \cdot 2 = 26 \] \[ 10^2 + 24^2 = 100 + 576 = 676 \] \[ 26^2 = 676 \]

Triple 10-24-26 satisfies \( a^2 + b^2 = c^2 \).

Example 7: Coordinate Geometry

Find the distance between points \( (1, 2) \) and \( (4, 6) \):

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] \[ = \sqrt{(4 - 1)^2 + (6 - 2)^2} \] \[ = \sqrt{3^2 + 4^2} \] \[ = \sqrt{9 + 16} \] \[ = \sqrt{25} \] \[ = 5 \]

Distance is 5 units.

Proof by Area and Similarity

Two methods demonstrate the theorem’s validity.

Proof by Area

Construct a square with side \( a + b \). Place four right triangles (legs \( a \) and \( b \)) inside, with a smaller square of side \( c \) in the center. Total area:

\[ (a + b)^2 = a^2 + 2ab + b^2 \]

Area of four triangles plus inner square:

\[ 4 \cdot \frac{1}{2}ab + c^2 = 2ab + c^2 \]

Equate and simplify:

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

Proof by Similarity

Draw an altitude from the right angle to the hypotenuse, dividing the triangle into two smaller right triangles similar to the original. Let the hypotenuse be \( c \), legs \( a \) and \( b \), and altitude \( h \). By similarity:

\[ \frac{a}{c} = \frac{h}{b} \Rightarrow a \cdot b = c \cdot h \] \[ \frac{b}{c} = \frac{h}{a} \Rightarrow b \cdot a = c \cdot h \]

Area of original triangle \( = \frac{1}{2}ab \), also \( \frac{1}{2}ch \):

\[ \frac{1}{2}ab = \frac{1}{2}ch \] \[ ab = ch \]

Since \( h^2 = a^2 - (c - b)^2 \) (or \( b^2 - (c - a)^2 \)), but using areas:

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

Applications

The Pythagorean Theorem has wide-ranging uses. Here are detailed examples with calculations:

Distance in a Room: Find the diagonal of a room 5 m long, 4 m wide.
\[ a = 5, b = 4 \] \[ a^2 + b^2 = c^2 \] \[ 5^2 + 4^2 = 25 + 16 \] \[ = 41 \] \[ c = \sqrt{41} \] \[ \approx 6.40 \]

Diagonal is approximately 6.40 m.
Navigation: A ship travels 8 km east, then 6 km north. What’s the straight-line distance?
\[ a = 8, b = 6 \] \[ a^2 + b^2 = c^2 \] \[ 8^2 + 6^2 = 64 + 36 \] \[ = 100 \] \[ c = \sqrt{100} \] \[ = 10 \]

Distance is 10 km.
Construction - Right Angle Check: Verify a 3-4-5 triangle for a right angle with sides 6, 8, 10 (scaled by 2).
\[ 6^2 + 8^2 = 36 + 64 = 100 \] \[ 10^2 = 100 \]

Confirms a right angle.
3D Space - Box Diagonal: Find the space diagonal of a box 3 m long, 4 m wide, 5 m high.
\[ \text{Base diagonal: } d = \sqrt{3^2 + 4^2} \] \[ = \sqrt{9 + 16} \] \[ = \sqrt{25} \] \[ = 5 \] \[ \text{Space diagonal: } D = \sqrt{5^2 + 5^2} \] \[ = \sqrt{25 + 25} \] \[ = \sqrt{50} \] \[ \approx 7.07 \]

Space diagonal is approximately 7.07 m.
Surveying - Land Distance: A surveyor measures 15 m east and 20 m north. What’s the straight distance?
\[ a = 15, b = 20 \] \[ a^2 + b^2 = c^2 \] \[ 15^2 + 20^2 = 225 + 400 \] \[ = 625 \] \[ c = \sqrt{625} \] \[ = 25 \]

Distance is 25 m.
Coordinate Geometry - Line Segment: Distance between \( (-2, 3) \) and \( (4, -1) \).
\[ d = \sqrt{(4 - (-2))^2 + (-1 - 3)^2} \] \[ = \sqrt{(6)^2 + (-4)^2} \] \[ = \sqrt{36 + 16} \] \[ = \sqrt{52} \] \[ \approx 7.21 \]

Distance is approximately 7.21 units.