Pythagorean theorem

Pythagorean theorem. The sum of two squares whose sides are the two legs (blue and red) is equal to the area of the square whose side is the hypotenuse (purple).

The Pythagorean Theorem is an important mathematical theorem that explains the final side of a right-angled triangle when two sides are known. In any right triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle), so we can take out the length of any third side of a right angled triangle, if the other two sides are given.

Another corollary of the theorem is that in any right triangle, the hypotenuse is greater than any one of the legs, but less than the sum of them. Some of the most common right triangles with integer sides are the 3-4-5, 5-12-13, 7-24-25, and the 8-15-17 triangles. The Pythagorean theorem is named after the Greek mathematician Pythagoras, who by tradition is credited with its discovery and proof. The theorem is:

a 2 + b 2 = c 2 {\displaystyle a^{2}+b^{2}=c^{2}\\!\,}

where c represents the length of the hypotenuse, and a and b represent the lengths of the other two sides. For example, we consider a triangle whose two sides, except the hypotenuse, are of lengths 12 and 5 cm. We can find the hypotenuse using the Pythagorean Theorem.

a 2 + b 2 = c 2 {\displaystyle a^{2}+b^{2}=c^{2}\\!\,}

= 12 2 + 5 2 = x 2 {\displaystyle =12^{2}+5^{2}=x^{2}\\!\,}

= 144 + 25 = x 2 {\displaystyle =144+25=x^{2}\\!\,}

= 169 = x 2 {\displaystyle =169=x^{2}\\!\,}

= 169 = x , {\displaystyle ={\sqrt {169}}=x,}

∴ x = 13 {\displaystyle \therefore x=13} c m {\displaystyle cm}

## Proofs

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Proof by rearrangement of four identical right triangles | Animation showing another proof by rearrangement | Proof using an elaborate rearrangement | Diagram of Garfield's proof

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## Examples

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1: Find the hypotenuse of a right angle triangle whose base is 3 meters and perpendicular is 4 meters in length.
Solution: Let the hypotenuse be x meters.

x 2 = 4 2 + 3 2 {\displaystyle x^{2}=4^{2}+3^{2}}

x = 4 2 + 3 2 {\displaystyle x={\sqrt {4^{2}+3^{2}}}}

∴ x = 5 {\displaystyle \therefore x=5}

So, the hypotenuse will be 5 meters.

2: Find the length of a leg of a right angled triangle whose hypotenuse is 17 cm and the other leg is 15 cm.

###### Solution: Method 1:

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Let the length of the unknown leg = x centimetres

x 2 + 15 2 = 17 2 {\textstyle x^{2}+15^{2}=17^{2}}

By subtracting 152 from both sides, we get:

x 2 = 17 2 − 15 2 {\textstyle x^{2}=17^{2}-15^{2}}

x 2 = ( 17 − 15 ) ( 17 + 15 ) ∵ A 2 − B 2 = ( A − B ) ( A + B ) {\textstyle x^{2}=(17-15)(17+15)\because A^{2}-B^{2}=(A-B)(A+B)}

x 2 = 2 × 32 = 64 {\textstyle x^{2}=2\times 32=64}

x = 64 {\textstyle x={\sqrt {64}}}

∴ x = 8 {\textstyle \therefore x=8}

So, the length of the unknown leg is 8 centimetres.

Method 2:

Let the length of the unknown leg = x centimetres

x 2 + 15 2 = 17 2 {\textstyle x^{2}+15^{2}=17^{2}}

By subtracting 152 from both sides, we get:

x 2 = 17 2 − 15 2 {\textstyle x^{2}=17^{2}-15^{2}}

x 2 = 289 − 225 = 64 {\displaystyle x^{2}=289-225=64}

x = 64 {\textstyle x={\sqrt {64}}}

∴ x = 8 {\textstyle \therefore x=8}

So, the length of the unknown leg is 8 centimeters.

## Composition of each value a, b, c

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a 2 + b 2 = c 2 . . . ( 1 ) {\displaystyle a^{2}+b^{2}=c^{2}...(1)}

a 2 = c 2 − b 2 {\displaystyle a^{2}=c^{2}-b^{2}}

a 2 = ( c + b ) ( c − b ) {\displaystyle a^{2}=(c+b)(c-b)}

The left hand side is a perfect square. Therefore, the right hand side must be a perfect square.

Let

c + b = K m 2 . . . ( 2 ) {\displaystyle c+b=Km^{2}...(2)}

and let

c − b = K n 2 . . . ( 3 ) {\displaystyle c-b=Kn^{2}...(3)}

The right hand side  = ( c + b ) ( c − b ) = ( K m 2 ) ( K n 2 ) = K 2 m 2 n 2 {\displaystyle =(c+b)(c-b)=(Km^{2})(Kn^{2})=K^{2}m^{2}n^{2}} a perfect square.

a 2 = K 2 m 2 n 2 . . . ( 4 ) {\displaystyle a^{2}=K^{2}m^{2}n^{2}...(4)}

( 2 ) + ( 3 ) , 2 c = K m 2 + K n 2 {\displaystyle (2)+(3),2c=Km^{2}+Kn^{2}}

c = K ( m 2 + n 2 ) 2 {\displaystyle c={\frac {K(m^{2}+n^{2})}{2}}}

c 2 = K 2 ( m 2 + n 2 ) 2 4 . . . ( 5 ) {\displaystyle c^{2}={\frac {K^{2}(m^{2}+n^{2})^{2}}{4}}...(5)}

( 2 ) − ( 3 ) , 2 b = K m 2 − K n 2 {\displaystyle (2)-(3),2b=Km^{2}-Kn^{2}}

b = K ( m 2 − n 2 ) 2 {\displaystyle b={\frac {K(m^{2}-n^{2})}{2}}}

b 2 = K 2 ( m 2 − n 2 ) 2 4 . . . ( 6 ) {\displaystyle b^{2}={\frac {K^{2}(m^{2}-n^{2})^{2}}{4}}...(6)}

Substitute (4), (5) and (6) in (1)

K 2 m 2 n 2 + K 2 ( m 2 − n 2 ) 2 4 = K 2 ( m 2 + n 2 ) 2 4 {\displaystyle K^{2}m^{2}n^{2}+{\frac {K^{2}(m^{2}-n^{2})^{2}}{4}}={\frac {K^{2}(m^{2}+n^{2})^{2}}{4}}}

Simplify

4 m 2 n 2 + ( m 2 − n 2 ) 2 = ( m 2 + n 2 ) 2 {\displaystyle 4m^{2}n^{2}+(m^{2}-n^{2})^{2}=(m^{2}+n^{2})^{2}}

Therefore

a = 2 m n , b = m 2 − n 2 , c = m 2 + n 2 {\displaystyle a=2mn,b=m^{2}-n^{2},c=m^{2}+n^{2}}

Check:

a 2 = 4 m 2 n 2 {\displaystyle a^{2}=4m^{2}n^{2}}

b 2 = m 4 − 2 m 2 n 2 + n 4 {\displaystyle b^{2}=m^{4}-2m^{2}n^{2}+n^{4}}

a 2 + b 2 = m 4 + 2 m 2 n 2 + n 4 {\displaystyle a^{2}+b^{2}=m^{4}+2m^{2}n^{2}+n^{4}}

c 2 = m 4 + 2 m 2 n 2 + n 4 = a 2 + b 2 {\displaystyle c^{2}=m^{4}+2m^{2}n^{2}+n^{4}=a^{2}+b^{2}}

For example, consider the triple  5 , 12 , 13 {\displaystyle 5,12,13} .

m = 3 , n = 2 {\displaystyle m=3,n=2}

a = 2 m n = 2 ( 3 ) ( 2 ) = 12 {\displaystyle a=2mn=2(3)(2)=12}

b = m 2 − n 2 = 3 2 − 2 2 = 9 − 4 = 5 {\displaystyle b=m^{2}-n^{2}=3^{2}-2^{2}=9-4=5}

c = m 2 + n 2 = 3 2 + 2 2 = 9 + 4 = 13 {\displaystyle c=m^{2}+n^{2}=3^{2}+2^{2}=9+4=13}

When we have a Pythagorean triple a,b,c:

a = 2 m n , b = m 2 − n 2 , c = m 2 + n 2 {\displaystyle a=2mn,b=m^{2}-n^{2},c=m^{2}+n^{2}} .

## Topics related to Pythagoras' Theorem

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Any integer may be expressed as the sum of two squares.

( 3 7 5 ) 2 + ( 4 7 5 ) 2 = ( 7 ) 2 ( ( 3 5 ) 2 + ( 4 5 ) 2 ) = 7 3 2 + 4 2 5 2 = 7 5 2 5 2 = 7 {\displaystyle ({\frac {3{\sqrt {7}}}{5}})^{2}+({\frac {4{\sqrt {7}}}{5}})^{2}=({\sqrt {7}})^{2}(({\frac {3}{5}})^{2}+({\frac {4}{5}})^{2})=7{\frac {3^{2}+4^{2}}{5^{2}}}=7{\frac {5^{2}}{5^{2}}}=7}

Angle in a semi-circle is a right angle.

Definition of a circle in the cartesian system depends on Pythagoras' Theorem.

r = ( x − p ) 2 + ( y − q ) 2 {\displaystyle r={\sqrt {(x-p)^{2}+(y-q)^{2}}}}

In electronics, in a series circuit containing resistance and capacitance and supplied with a sine-wave voltage (E),
voltage across resistance (Er) is in phase with current, voltage across capacitance (Ec) lags current by 90°.

E = ( E r ) 2 + ( E c ) 2 {\displaystyle E={\sqrt {(Er)^{2}+(Ec)^{2}}}}

A complex number may be expressed in "Eulerian" format.

3 + 3 i = 2 3 ( 1 2 + 3 2 i ) = 2 3 ( cos ⁡ ( 60 ) + sin ⁡ ( 60 ) i ) {\displaystyle {\sqrt {3}}+3i=2{\sqrt {3}}({\frac {1}{2}}+{\frac {\sqrt {3}}{2}}i)=2{\sqrt {3}}(\cos(60)+\sin(60)i)}

The familiar Theorem of Pythagoras is a special case of the general identity applied to any triangle.

c 2 = a 2 + b 2 − 2 ( a b ) cos ⁡ ( C ) {\displaystyle c^{2}=a^{2}+b^{2}-2(ab)\cos(C)}

In aeronautical navigation, if an aircraft is heading North at 100 knots and the wind at altitude is blowing from the East
at 10 knots, the groundspeed of the aircraft is

100 2 + 10 2 {\displaystyle {\sqrt {100^{2}+10^{2}}}} knots

Some perfect squares can be expressed as the sum of two squares in more ways than one:

65 2 = 39 2 + 52 2 = 25 2 + 60 2 {\displaystyle 65^{2}=39^{2}+52^{2}=25^{2}+60^{2}}

The expression  m 2 + 1 {\displaystyle m^{2}+1} is a perfect square if  m {\displaystyle m} has a value like  3 4 {\displaystyle {\frac {3}{4}}} or  12 5 {\displaystyle {\frac {12}{5}}} .

The expression  m 2 − 1 {\displaystyle m^{2}-1} is a perfect square if  m {\displaystyle m} has a value like  5 3 {\displaystyle {\frac {5}{3}}} or  17 8 {\displaystyle {\frac {17}{8}}} .

The expression  1 − m 2 {\displaystyle 1-m^{2}} is a perfect square if  m {\displaystyle m} has a value like  3 5 {\displaystyle {\frac {3}{5}}} or  24 25 {\displaystyle {\frac {24}{25}}} .

The equation of the line  a x + b y + c = 0 {\displaystyle ax+by+c=0} can be put into normal form:  a x + b y + c a 2 + b 2 = 0 {\displaystyle {\frac {ax+by+c}{\sqrt {a^{2}+b^{2}}}}=0} .
The normal form of the line  4 x + 3 y + 10 = 0 {\displaystyle 4x+3y+10=0} is  4 5 x + 3 5 y + 2 = 0 {\displaystyle {\frac {4}{5}}x+{\frac {3}{5}}y+2=0} where  a 2 + b 2 = 1 {\displaystyle a^{2}+b^{2}=1} .

In the ellipse, the major axis (length  2 a {\displaystyle 2a} ), the minor axis (length  2 b {\displaystyle 2b} ) and the distance between foci (length  2 c {\displaystyle 2c} ) have the relationship  a 2 = b 2 + c 2 {\displaystyle a^{2}=b^{2}+c^{2}} .

## See also

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* School:Mathematics