# Pythagoras

Pythagoras was an ancient Greek philosopher. He was born in 570 BC, in Samos (Greece), and died in 495 BC, in Metapontum (Italy). His full name was Pythagoras of Samos.

He was credited with many discoveries in the field of science, mathematics, music, astronomy, and medicine. The discoveries done by him are, Pythagorean or Pythagoras theorem, Pythagorean tuning, Theory of proportions, Sphericity of the earth, the identity of planet Venus, and five regular solids. He also divided the globe into five climatic zones. He had given the main credit for the discovery and proof of Pythagoras theorem.

### Pythagoras or Pythagorean Theorem

Pythagoras theorem is based on the right-angled triangle or right triangle only. The theorem states that in a right triangle, the sum of the square of base and perpendicular is equal to the square of the hypotenuse.

In other words ,in a right triangle, the square of the hypotenuse is equal to the sum of the square of the two legs. The legs (base and perpendicular) are the sides of a triangle that forms the right-angle.

### Components of Right Triangle

The following figure represents a right triangle ∆ABC. • Base: It is a side of the right triangle that is adjacent to the perpendicular and hypotenuse. In ∆ABC, AB is base.
• Perpendicular: It is a side of the right triangle that is adjacent to the base and hypotenuse. It is called the height of the triangle. In ∆ABC, AC is perpendicular.
• Hypotenuse: The side opposite to the right angle is called the hypotenuse. In other words, the longest side of the right triangle is called the hypotenuse. In ∆ABC, BC is the hypotenuse.
• Right-angle: In geometry, the right-angle is an angle that makes an angle of 90°. In ∆ABC, ∠A is a right-angle .

### Pythagoras Triples

Pythagoras or Pythagorean triples is a set of three positive integers that satisfies the Pythagoras theorem. The least Pythagorean triple is (3, 4, 5).

In ∆ABC, (a, b, c) is the Pythagoras triples that represent positive integer value and satisfy the theorem.

a2+b2=c2 The following table enlists some Pythagorean triples.

 (3, 4, 5) (5, 12, 13) (7, 24, 25) (8, 15, 17) (9, 40, 41) (11, 60, 61) (12, 35, 37) (13, 84, 85) (16, 63, 65) (20, 21, 29) (28, 45, 53) (33, 56, 65) (36, 77, 85) (39, 80, 89) (48, 55, 73) (65, 72, 97)

• It always has one even number in a triple.
• The value of c will always be odd.
• It may have two prime numbers.

### Pythagoras Theorem Formula

Consider the following figure. In ∆ABC, AC is perpendicular or height, AB is base, and BC is the hypotenuse. The length of perpendicular, base, and hypotenuse is a, b, and c, respectively. According to the Pythagoras theorem, the Pythagoras theorem formula can be written as:

Perpendicular2 + Base2 = Hypotenuse2

Or

AC2+ AB2 = BC2

Or

a2+ b2 = c2

### Pythagoras Theorem Proof

To Prove: AC2 = AB2+ BC2

Given: A right triangle ∆ABC.

### Proof 1:

In the following figure, we have drawn a perpendicular (BD) from point B that meets at point D on the hypotenuse. The perpendicular divides the triangle into two triangles, i.e., ∆ADB and ∆BDC.

Remember: If we draw a perpendicular from the vertex of the right-angle, the triangles on both sides are equals to each other and also equal to the whole triangle.

According to the above statement, ∆ABC=∆ADB Adding the equations (i) and (ii), we get:

AB2+BC2=AC×AC

AB2+BC2=AC2

Hence, the Pythagoras theorem is proved.

Let's see the second way to prove the theorem.

### Proof 2:

In the following figure, we have drawn a square ABCD. Inside the square ABCD, we have drawn another square EFGH that forms four triangles ∆AEF, ∆FDG, ∆GCH and ∆HBE. Now, we will find the area of both squares and triangles, separately.

We know that, the area of square = a2 (where a is the side of the square)

Area of square ABCD=(a+b)2

We know that, area of triangle = bh

Area of a triangle= ab

There are a total of four triangles, so the area of four triangles will be:

Area of four triangles= 4× ab=2ab

Area of square EFGH=c2 (where c is the side of the square EFGH)

The total area of the square ABCD will be:

Area of ABCD = Area of Square EFGH + Area of four triangles

Putting the values, we get:

(a+b)2=c2+2ab
(a+b)(a+b)=c2+2ab
a2+2ab+b2=c2+2ab

Cancel out the 2ab on both sides, we get:

a2+b2=c2

Hence, the theorem is proved.

### Pythagoras Theorem Problems

Example 1: The three sides of a triangle are 5, 12, and 13 cm. Use the Pythagoras theorem and check that the triangle is a right triangle or not.

Solution:

Given, AB = 12 cm, BC = 5 cm, AC = 13 cm According to the Pythagoras theorem, AC2=BC2+AB2

132=52+122
169=25+144
169=169

Hence, the triangle is a right triangle.

Example 2: Find the value of AC if the length of the base is 3 cm, and the height of the triangle is 4 cm. Solution:

In the ∆ABC, given that BC = 3 cm and AB = 4 cm.

According to the Pythagoras theorem, BC2+AB2=AC2

Putting the values of AB and BC in the above formula, we get:

32+42=AC2
9+16=AC2
25=AC2
AC=√25
AC=5

Hence, the length of the hypotenuse is 5 cm.

Example 3: Find the value of the base. If the length of the hypotenuse 10 cm and height of the triangle is 8 cm. Solution:

In the ∆ABC, given that AC = 10 cm and BC = 8 cm.

According to the Pythagoras theorem, BC2+AB2=AC2

Putting the values of AC and BC in the above formula, we get:

102=82+AB2
100=64+AB2
100-64=AB2
36=AB2
AB=√36
AB=6

Hence, the length of the base is 6 cm.

Example 4: The length of the base and hypotenuse is 30 and 50 cm, respectively. Find the height of the triangle. Solution:

In the ∆ABC, given that AC = 50 m and AB = 30 m.

According to the Pythagoras theorem, AC2=BC2+AB2

Putting the values of AC and AB in the above formula, we get:

502=BC2+302
2500=BC2+900
2500-900=BC2
1600=BC2
BC=√1600
BC=40

Hence, the height of the triangle is 40 m.

Example 5: If a side of the squared stone is 9 m. Find the length of the diagonal.

Solution: In the above figure, we see that there are two triangles ∆ABC and ∆ADC. Let's take the triangle ∆ABC and find the diagonal.

According to the Pythagoras theorem, AC2=BC2+AB2

AC2=92+92
AC2=81+81
AC2=162
AC=√162
AC=9√2

Hence, the length of the diagonal is 9√2 m.

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