5.5 Pythagorean Theorem

Types: N/A
Examples: N/A
Constructions: 5.6 Law of Cosines
Generalizations: N/A

Properties: N/A
Sufficiencies: N/A
Questions: N/A

Pythagorean Theorem

Let $a$ , $b$ , and $c$ be real numbers representing the lengths of the base, height, and hypotenuse of a triangle respectively. Then,

a^{2} + b^{2} = c^{2}

\begin{proof}
Let $a, b, c$ be appropriate side lengths. Let $θ$ and $ϕ$ represent the two acute angles of our triangle. Then, consider the following shape:

We know the sum of all three interior angles of our triangle sum up to $180 ° = 90 ° + θ + ϕ$ . Using four of these triangles, let's construct a special quadrilateral:

Recall that a square's inner angles sum up to $360 °$ and all of its side lengths are equal. Because we know that $θ + ϕ = 90 °$ , we can immediately conclude that the inner quadrilateral created by these four triangles forms a square with area $c^{2}$ . We can also conclude the area of the large square is $(a + b) \cdot (a + b) = a^{2} + 2 a b + b^{2}$ .

In order to get the area of the small square, we can subtract the area of the large square by the area of the 4 triangles.

\begin{aligned} ⟹ c^{2} & = a^{2} + 2 a b + b^{2} - 4 (\frac{1}{2} a b) \\ = a^{2} + 2 a b + b^{2} - 2 a b \\ = a^{2} + b^{2} \end{aligned}

We get our Pythagorean Theorem.
\end{proof}