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sforiginals

Cantore Arithmetic is able to state word lettered ice. SFO Arithmetic:
a, words In Nature
b. words In Manufacturing
c. words Ware are the Wings
Words Wings is important as in the word Unwritten Word Book at word Genesis in word gentiles or word Gentile. Words Pay Gun for word Polytheism is as old or older than whom set word Lines to word Names word Man Pythagoras, and, words The Pythagoras theorem which is also referred to as the Pythagorean theorem explains the relationship between the three sides of a right-angled triangle. According to the Pythagorean theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides of a triangle. Let us learn more about the Pythagoras theorem, the Pythagoras theorem formula, and the proof of Pythagoras theorem along with examples.

Pythagoras Theorem Equation

In a right-angled triangle, the Pythagoras Theorem Formula is expressed as:
Proof of Pythagorean Theorem Formula using the Algebraic Method

- Step 1: This method is also known as the 'proof by rearrangement'. Take 4 congruent right-angled triangles, with side lengths 'a' and 'b', and hypotenuse length 'c'. Arrange them in such a way that the hypotenuses of all the triangles form a tilted square. It can be seen that in the square PQRS, the length of the sides is 'a + b'. The four right triangles have 'b' as the base, 'a' as the height and, 'c' as the hypotenuse.
- Step 2: The 4 triangles form the inner square WXYZ as shown, with 'c' as the four sides.
- Step 3: The area of the square WXYZ by arranging the four triangles is c2.
- Step 4: The area of the square PQRS with side (a + b) = Area of 4 triangles + Area of the square WXYZ with side 'c'. This means (a + b)2= [4 × 1/2 × (a × b)] + c2.This leads to a2 + b2 + 2ab = 2ab + c2. Therefore, a2 + b2 = c2. Hence, the Pythagoras theorem formula is proved.
Pythagorean Theorem Formula Proof using Similar Triangles
Derivation of Pythagorean Theorem Formula
Consider a right-angled triangle ABC, right-angled at B. Draw a perpendicular BD meeting AC at D.

Thus, △ABD ∼ △ACB (by AA similarity criterion)
Similarly, we can prove △BCD ∼ △ACB.
Thus △ABD ∼ △ACB, Therefore, AD/AB = AB/AC. We can say that AD × AC = AB2.
Similarly, △BCD ∼ △ACB. Therefore, CD/BC = BC/AC. We can also say that CD × AC = BC2.
Adding these 2 equations, we get AB2 + BC2 = (AD × AC) + (CD × AC)
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