The “Pythagorean theorem,” was inspired by the ancient Greek Mathematician Pythagoras who invented the theorem during 500 B.C. It has been argued that the “Ancient Babylonians” already understood the theorem long before the invention by Pythagoras. They knew the relationship between the sides of the triangle and while solving for the hypotenuse of an isosceles triangle, they came up with the approximate value of up to 5 decimal places.

## What is Pythagoras theorem?

In a right-angled triangle, “The sum of squares of the lengths of the two sides is equal to the square of the length of the hypotenuse (or the longest side).”

In the given triangle, side “a” is called as “Perpendicular”, side “b” is called as “Base” and side “c” is called as “Hypotenuse.”

The side opposite to the right angle (\(90^{\circ}\)) is the longest side known as Hypotenuse, as we know that the side opposite to the greatest angle is longest.

**Note-**Pythagoras theorem is only applicable to a Right-Angled triangle.

### Another way of representation of the Pythagoras Theorem-

Another way of Pythagoras Theorem proof is described here. Consider a right-angled triangle having perpendicular as a, base as b, and hypotenuse as c. Consider three squares of sides a,b,c mounted on the three sides of a triangle having the same sides as shown.

Area of square A + Area of square B = Area of square C

**Example- Prove the Pythagoras Theorem for a right angle triangle having sides to be 3cm, 4cm and 5 cm.**Solution –

From Pythagoras Theorem we have,

\((Perpendicular)^{2} + (base)^{2} = (Hypotenuse)^{2}\)

Perpendicular = 3 cm

Base = 4 cm

Hypotenuse = 5 cm

\((3)^{2} + (4)^{2} = (5)^{2}\)

\(\Rightarrow 9 + 16 = 25\)

\(\Rightarrow 25 = 25\)

L.H.S. = R.H.S.

Therefore Pythagoras theorem is proved.

### Proof of Pythagoras Theorem-

To Prove- \(AC^{2} = AB^{2} + BC^{2}\)</p >

For this we drop a perpendicular BD onto the side AC

We know, \(\bigtriangleup ADB \sim \bigtriangleup ABC\)

Therefore, \(\frac{AD}{AB}=\frac{AB}{AC}\) (Condition for similarity)

Or, \(AB^{2}= AD \times AC\)……..(1)

Also, \(\bigtriangleup BDC \sim \bigtriangleup ABC\)

Therefore, \(\frac{CD}{BC}=\frac{BC}{AC}\) (Condition for similarity)

Or, \(BC^{2}= CD \times AC\)……..(2)

Adding the equations (1) and (2) we get,

\(AB^{2}+ BC^{2}= AD \times AC + CD \times AC\)

\(AB^{2}+ BC^{2}= AC (AD + CD)\)

Since, AD + CD = AC

Therefore, \(AC^{2} = AB^{2}+ BC^{2}\)

###
**Application-**

**(1st):** **To know the triangle is right-angled or not. It is to be noted that if the Pythagoras Theorem is proved then it must be a **right-angled triangle.

**Example- The sides of a triangle are 5,12 & 13 units. Check if it has a right angle or not.**

**Solution-**

To prove- Pythagoras theorem in order to find whether it has a right angle or not

From Pythagoras Theorem, we have-

\((Perpendicular)^{2} + (base)^{2} = (Hypotenuse)^{2}\)

Perpendicular = 12 units

Base = 5 units

Hypotenuse = 13 units

\((12)^{2} + (5)^{2} = (13)^{2}\)

\(\Rightarrow 144 + 25 = 169 \)

\(\Rightarrow 169 = 169 \)

L.H.S. = R.H.S.

Therefore the angles opposite to the 13 unit side will be at a right angle.

**(2nd):** **In a right-angled triangle, we can calculate the length of any side if other two sides are given.**

**Example- The two sides of a right angled as shown in the figure. Find the third side.**

**Solution-**

Given-

Perpendicular = 15cm

Base = b cm

Hypotenuse = 17 cm

From Pythagoras Theorem, we have

**\((Perpendicular)^{2} + (base)^{2} = (Hypotenuse)^{2}\)**

\( 15^{2} + b^{2} = 17^{2}\)

\(\Rightarrow 225 + b^{2} = 289 \)

\(\Rightarrow b^{2} = 289 – 225 \)

\(\Rightarrow b^{2} = 64 \)

\(\Rightarrow b = \sqrt{64}\)

Therefore \( b = 8 \)

**(3rd): To find the diagonal of a square**

**Example- Given the side of a square to be 4 cm. Find the length of the diagonal.**

**Solution-**

Given –

Sides of a square = 4 cm

To Find- The length of diagonal ac.

Consider triangle abc (or can also be acd)

\((ab)^{2} +(bc)^{2}= (ac)^{2}\)

\(\Rightarrow (4)^{2} +(4)^{2}= (ac)^{2}\)

\(\Rightarrow 16 + 16 = (ac)^{2} \)

\(\Rightarrow 32 = (ac)^{2}\)

\(\Rightarrow (ac)^{2} = 32 \)

or \( acl = 4 \sqrt{2} \)

## List of Pythagorean Triples:

**Here is the list of the most commonly used Pythagorean triples:**