# Can someone explain pythagoras better than my teacher?

## all she said was a2 + b2=c2

Apr 21, 2018

Pythagoras' Theorem helps to find missing sides in any right angle triangle so long as you know the other two sides.

#### Explanation:

Pythagoras' Theorem helps to find missing sides in any right angle triangle so long as you know the other two sides. For example; In this triangle we can see that we are missing the side 'b'.

In the equation: ${a}^{2} + {b}^{2} = {c}^{2}$, the ${c}^{2}$ correlates to the hypotenuse - the longest side of the triangle, opposite the right angle. So in this example, the hypotenuse is the side 17cm.

We need to subsitute the numbers from the triangle into the equation so we work out:
${15}^{2} + {b}^{2} = {17}^{2}$.
We need to rearrange this equation to figure out, side b.
${b}^{2} = {17}^{2} - {15}^{2}$
We then solve this equation to find $b$
${b}^{2} = 289 - 225$
${b}^{2} = 64$
$b = \sqrt{64}$
$b = 8$

Apr 21, 2018 color(white)(wwwwwwwwwwwwwwwwwwwwwwwwwww[Source: Wikipedia]

I'll use the similar triangle method to prove here, there are a lot more :)

Consider $\triangle A B C$ and $\triangle B C H$,

$\frac{B C}{A B} = \frac{B H}{B C}$

=>color(red)( BC^2= AB xx BH

Consider $\triangle A B C$ and $\triangle A C H$,

$\frac{A C}{A B} = \frac{A H}{A C}$

=> color(magenta)(AC^2 =AB xx AH

$\implies \textcolor{red}{B {C}^{2}} + \textcolor{m a \ge n t a}{A {C}^{2}} = \left[\textcolor{red}{A B \times B H}\right] + \left[\textcolor{m a \ge n t a}{A B \times A H}\right]$

$\implies A B \left[B H + A H\right]$ color(white)(wwwwwww $\left[\text{from the fig. } B H + A H = A B\right]$

$\implies B {C}^{2} + A {C}^{2} = A {B}^{2}$

Note that $A B$ is the hypotenuse (side of the triangle opposite to the right angle).

Therefore, the Pythagoras statement, can be written as,

${\left[\text{side"_ 1 ]^2 + ["side"_2 ]^2 =["hypotenuse}\right]}^{2}$

Apr 21, 2018

#### Explanation:

We have a right triangle here: We also can draw the following: That looks like a square... but let's prove it.

First, we can see that the side lengths of the green quadrilateral are all equal.

Now, we label like the following: Using the fact that the angles of a triangle add up to 180 degrees, we see that the missing angle of the triangle, (the one opposite to side $b$) is:

$180 - 90 - x$

$90 - x$

We now have: Using the fact that a straight line has an angle of 180 degrees, the missing angle of the green quadrilateral is:

$180 - \left(90 - x\right) - x$

$\implies 180 - 90 + x - x$

$\implies 90$

You can use the same technique for the other three angles and see that the green quadrilateral is indeed a square.

Therefore, the area of the green square is:

$A = {c}^{2}$

Let's draw another sketch: You can use the same technique that I used for the previous square to prove that the green quadrilaterals there are squares.

We see that the area of the green squares is:

$A = {a}^{2}$ and $A = {b}^{2}$

The large square that contains both the squares and triangles for both cases have the same side length ($a + b$) and therefore the same area.

Since we have four congruent triangles for each sketch, the leftover areas (the green squares) must be equal to each other.

Therefore:

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

Which tells that the length of the longest side of a right triangle squared is equal to the sum of the other sides squared.

An example would be $3 - 4 - 5$ right triangle.

${3}^{2} + {4}^{2} = {5}^{2}$

Note that there are hundreds of ways to prove the theorem.