proof pythagoras theorem geometrically
Answers
Step-by-step explanation:
The Algebraic and Geometric Proofs of Pythagorean Theorem. The Pythagorean Theorem states that if a right triangle has side lengths and , where is the hypotenuse, then the sum of the squares of the two shorter lengths is equal to the square of the length of the hypotenuse. ... The converse of the theorem is also true.
First, we draw a triangle with side a b c lengths and as shown in Figure 1. Next, we create 4 triangles identical to it and using the triangles form a square with side lengths a+b as shown in Figure 4-A. Notice that the area of the white square in Figure 4-A is c^2.
Figure 4 – The Geometric proof of the Pythagorean theorem.
Rearranging the triangles, we can also form another square with the same side length as shown in Figure 4-B.This means that the area of the white square in the Figure 4-A is equal to the sum of the areas of the white squares in Figure 4-B (Why?). That is, c^2=a^2+b^2 which is exactly what we want to show. *And since we can always form a (big) square using four right triangles with any dimension (in higher mathematics, we say that we can choose arbitrary a and b as side lengths of a right triangle), this implies that the equation a square +b square=c^2 stated above is always true regardless of the size of the triangle.
