# Pythagoras Theorem

Pythagoras theorem is applicable in case of right-angled triangle. It says that, the square of the hypotenuse is equal to the sum of the squares of the other two sides.^{2}= (Base)

^{2}+ (Perpendicular)

^{2}

*a*

^{2}+

*b*

^{2}=

*c*

^{2}

*a*= 3,

*b*= 4 and

*c*= 5. You can check that

3

^{2}+ 4

^{2}= 9 + 16 = 25 = 5

^{2}.

Sometimes we use the notation (

*a*,

*b*,

*c*) to denote such a triple.

Notice that the greatest common divisor of the three numbers 3, 4 and 5 is 1. Pythagorean triples with this property are called

*primitive*.# Proofs of Pythagoras Theorem

**Proof 1**

Now we start with four copies of the same triangle. Three of these have been rotated at 90Â°, 180Â°, and 270Â°, respectively. Each has the area ab/2. Letâ€™s put them together without additional rotations so that they form a square with side c.

*a*âˆ’

*b*). By summing up its area (a âˆ’ b)

^{2}and 2ab, the area of the four triangles (4Â·ab/2), we get

C

^{2}= (

*a*âˆ’

*b*)

^{2 }+ 2

*ab*=

*a*

^{2 }+

*b*

^{2}. QED

**Proof 2**

ABC is a right-angled triangle at B

To Prove: AC

^{2}= AB

^{2}+ BC

^{2}

Construction: Draw BD âŠ¥ AC

âˆ´ (Sides are proportional)

Or AB

^{2}= AD Ã— AC (1)

Also, âˆ†CDB âˆ¼âˆ†CBA

âˆ´

âˆ´ (Sides are proportional)

or BC

^{2}= CD Ã— CA

Adding (1) and (2)

AB

^{2}+ BC

^{2 }= AD Ã— AC + CD Ã— AC

= AC [AD + CD]

= AC Ã— AC = AC

^{2 }QED.

# Pythagorean Triplets

A*Pythagorean triplet*is a set of three positive whole numbers

*a*,

*b*and

*c*that are the lengths of the sides of a right triangle.

*a*

^{2}+

*b*

^{2}=

*c*

^{2}

It is noteworthy to see here that all of

*a*,

*b*and

*c*cannot be odd simultaneously. Either of

*a*or

*b*has to be even and

*c*can be odd or even.

The various possibilities for

*a*,

*b*and

*c*are tabled below.

a |
b |
c |

Odd | Odd | Even |

Even | Odd | Odd |

Odd | Even | Odd |

Even | Even | Even |

Some Pythagoras triplets are:

3 = 4 = 5 = (3

^{2}+ 4

^{2}= 5

^{2})

^{2}+ 12

^{2}= 13

^{2})

^{2}+ 24

^{2}= 25

^{2})

^{2}+ 15

^{2}= 17

^{2})

^{2}+ 40

^{2}= 41

^{2})

^{2}+ 60

^{2}= 61

^{2})

^{2}+ 21

^{2}= 29

^{2})

**Note :**If each term of any pythagorean triplet is multiplied or divided by a constant (say P, P > 0) then the triplet so obtained will also be a pythagorean triplet. This is because if a2 + b2 = c2, then (Pa)2 + (Pb)2 = (Pc)2, where P > 0.

**For example,**

3 Ã— 2 4 Ã— 2 5 Ã— 2 gives

6 8 10 (6

^{2}+ 8^{2}= 102)Using Pythagoras theorem to determine the nature of triangle

*c*

^{2}=

*a*

^{2}+

*b*

^{2}, then the triangle is right-angled triangle.

*c*

^{2}>

*a*

^{2}+

*b*

^{2}, then the triangle is an obtuse-angled triangle.

*c*

^{2}<

*a*

^{2}+

*b*

^{2}, then the triangle is an acute-angled triangle.

Mechanism to derive a Pythagorean triplet If the length of the smallest side is odd, assume the length of the smallest side = 5

*Step 1*Take the square of 5 (length of the smallest side) = 25

*Step 2*Break 25 into two parts P and Q, where P â€“ Q = 1. In this case, P = 13 and Q = 12. Now these two parts P and Q along with the smallest side constitute pythagorean triplet.

However, there is another general formula for finding out all the primitive pythagorean triplets:

*a*=

*r*

^{2 }âˆ’

*s*,

^{2}*b*= 2

*rs*,

*c*=

*r*

^{2}+

*s*

^{2},

*r*>

*s*> 0 are whole numbers,

*r*âˆ’

*s*is odd, and

The greatest common divisor of

*r*and

*s*is 1.

Table of small primitive Pythagorean triplets Here is a table of the first few primitive Pythagorean triplets:

r |
s |
a |
b |
c |

2 | 1 | 3 | 4 | 5 |

3 | 2 | 5 | 12 | 13 |

4 | 1 | 15 | 8 | 17 |

4 | 3 | 7 | 24 | 25 |

5 | 2 | 21 | 20 | 29 |

5 | 4 | 9 | 40 | 41 |

6 | 1 | 35 | 12 | 37 |

6 | 5 | 11 | 60 | 61 |

7 | 2 | 45 | 28 | 53 |

# Perimeter, area, inradius and shortest side

The perimeter P and area K of a Pythagorean triple triangle are given by

P =

*a*+*b*+*c*= 2*r*(*r*+*s*)*d*.K =

*ab*/2 =*rs*(*r*^{2}â€“*s*^{2})*d*^{2}.Example-1

Two sides of a plot measure 32 m and 24 m and the angle between them is a right angle. The other two sides measure 25 m each and the other three angles are not right angles.

What is the area of the plot (in m

^{2})?- 768
- 534
- 696
- 684

Solution

The figure given above can be seen as

Since ABD is a right-angled triangle, so it will satisfy the Pythagoras theorem. And the triplet used here is â€“ 3(Ã—8), 4(Ã—8) and 5(Ã—8). Similarly the other part of the figure can also be bifurcated by drawing a perpendicular from C on BD.

So, the area of the plot is:

Area (âˆ†ABD) + Area (âˆ†CBD) Ã— 24 Ã— 32 + 2 Ã— (1/2 Ã— 20 Ã— 15) = 684 m

^{2}Example-2

A ladder of length 65 m is resting against a wall. If it slips 8 m down the wall, then its bottom will move away from the wall by N m. If it was initially 25 m away from it, what is the value of

*x*?- 60 m
- 39 m
- 14 m
- 52 m

Solution

Using Pythagorean triplets, (5, 12, 13),

â‡’ h = 60

After it has slipped by 8 m, the new height = 52 m, and the length of the ladder = 65 m.

So 25 +

*x*= 39 (3, 4, 5 triplet)â‡’

*x*= 14 m