## History

Pythagoras theorem is also known by another name as Pythagorean Theorem is the famous theorem in geometry. This theorem is based on three sides of a right angled triangle. The name of theorem was given after the name of the Greek philosopher Pythagoras. The brief history of the Pythagoras can be read from Pythagoras on Wikipedia.

There are many proofs the mathematicians have worked on till date. Some of the proofs of Pythagoras theorem are given at Pythagoras’s Theorem at Proof Wiki.

## Theorem Statement

**In a right angled triangle, the square of the hypotenuse is equal to the sum of the square of the other two
sides of the triangle.**

Let’s understand it with a diagram of the right angled triangle ABC, right angled at point B.

In the above right angle ΔABC, p is perpendicular, b is base and h is hypotenuse.

h^{2} = b^{2} + p^{2}

or
$h=\sqrt{{b}^{2}+{p}^{2}}$

Let’s solve the problem of finding the hypotenuse of a right angled triangle using Pythagoras theorem.

**
Find the length of hypotenuse of right angle ΔABC, where length of base =3cm and length of
perpendicular = 9cm.
**
**Solution:**

In ΔABC

b=3cm

p=4cm

∵, ΔABC is a right angle triangle (given)

∴ by applying Pythagoras theorem

$h=\sqrt{{b}^{2}+{p}^{2}}$

$h=\sqrt{{3}^{2}+{4}^{2}}$

$h=\sqrt{9+16}$

$h=\sqrt{25}$

∴ h = 5cm

## Applications of Pythagoras theorem in real life

As we have learnt above the Pythagoras theorem can calculate length of one side of a right angled triangle, if the length of the other two sides are known. The ability of theorem to find such lengths can be applied in many real life scenarios in distance calculations, where the three distances forms the shape of a right angled triangle.

Such applications of Pythagoras theorem in real life can be seen in while calculating the length of ladder needed to lie on the wall. Another example is to calculate the shortest distance or how far two cars are after some hours of travel in directions perpendicular to each other.

## 1. Ladder length problem

### Problem statement

**A ladder is placed against a wall such that its foot os at a distance of 6m from the wall and its top reaches
a window which is 8m above the ground. Find the length of the ladder needed to reach the window.
**

#### Solution

In the figure, window is placed at A, AC be the length of ladder and ladder is placed against the wall AB and C is the point where the foot of the ladder is lying.

The distance of window above the ground AB = 8m

The distance of the foot of the ladder from the wall BC = 8m

Length of the ladder AC is to be calculated.

ΔABC is a right angled triangle with right angle formed at point B.

Length of ladder AC is the hypotenuse of ΔABC.

AB the height of the window above the ground is the height of the triangle ΔABC.

BC the distance of foot of ladder from the wall is the base of the triangle ΔABC.

**By using Pythagoras theorem**

(AC)^{2} = (AB)^{2} + (BC)^{2}

(AC)^{2} = (8)^{2} + (6)^{2}

(AC)^{2} = 64 + 36

(AC)^{2} = 100

(AC)^{2} = 10^{2}

∴ AC = 10 m

∴ AC, which is the length of the ladder is 10 m.

## 2. The shortest distance between two cars calculations

### Problem statement

**Two cars P and Q are travelling in directions perpendicular to each other. Car P is moving towards north and car Q is moving towards the east direction. After one hour car P travels a distance of 50 km and car Q travels a distance of 100 km. Find how far cars P and Q will be after one hour of travel from
each other.
**

#### Solution

In the figure, the east direction is taken towards the right and north direction towards up.

Both cars P and Q start moving from the same point A. Car P travels in the north direction. Car Q travels in direction.

After one hour of travel car P reaches point C in north direction and travels a distance of AC which is
80 km.

After one hour of travel car Q reaches point B in east direction and travels a distance of AB which is 60
km.

As per problem statement, BC is the distance that needs to be calculated, which is the distance between cars. BC will also be the shortest distance between cars P and Q after one hour of travel.

ΔBAC forms a right angled triangle, right angled at A.
**By using Pythagoras theorem**

(BC)^{2} = (AB)^{2} + (AC)^{2}

(BC)^{2} = (60)^{2} + (80)^{2}

(BC)^{2} = 3600 + 6400

(BC)^{2} = 10000

(BC)^{2} = 100^{2}

∴ BC = 100 km

∴ BC, which is the shortest distance between two cars P and Q after 1 hour is 100 km.