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Algebraic Proof of Heron's Formula

Heron's formula provides an algebraic proof for calculating the area of a triangle using only the lengths of its three sides. It expresses the height of the triangle in terms of the sides and semi-perimeter, then uses the formula for area of a triangle (1/2 x base x height) to derive Heron's formula: the area equals the square root of the semi-perimeter multiplied by the semi-perimeter minus each side, in turn.

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Chun Hao Theo
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242 views2 pages

Algebraic Proof of Heron's Formula

Heron's formula provides an algebraic proof for calculating the area of a triangle using only the lengths of its three sides. It expresses the height of the triangle in terms of the sides and semi-perimeter, then uses the formula for area of a triangle (1/2 x base x height) to derive Heron's formula: the area equals the square root of the semi-perimeter multiplied by the semi-perimeter minus each side, in turn.

Uploaded by

Chun Hao Theo
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Heron’s Formula – An algebraic proof

Procedure: (1) Express the height h of triangle ABC in terms of the sides a, b, c and
semi-perimeter s.
1
(2) Use × height × base to obtain the area of the triangle.
2

By Pythagoras’ Theorem,
b 2 =p 2 + h 2 ⇒ h 2 =b2 − p 2
a 2 =(c − p ) 2 + h 2
= c 2 − 2cp + p 2 + h 2
=c 2 − 2cp + b 2
c2 + b2 − a 2
p=
2c
h= b − p 2
2 2

=(b + p )(b − p )
 c2 + b2 − a 2   c2 + b2 − a 2 
=
 b +  b − 
 2c  2c 

=
( 2bc + c 2
+ b 2 − a 2 )( 2bc − c 2 − b 2 + a 2 )
4c 2
(b + c) 2 − a 2   a 2 − (b − c) 2 
=
4c 2
( b + c + a )( b + c − a )  ( a − b + c ) ( a + b − c ) 
=
4c 2
( a + b + c )( −a + b + c )  ( a − b + c )( a + b − c ) 
=
4c 2
a+b+c
Semi-perimeter, s = ⇒ 2s = a + b + c
2
2( s − a ) =−a + b + c
2( s − b) = a − b + c
2( s − c) = a + b − c
Hence,
2 s ⋅ 2( s − a ) ⋅ 2( s − b) ⋅ 2( s − c) 4 s ( s − a )( s − b)( s − c)
h2 =
4c 2 c2
2 s ( s − a )( s − b)( s − c)
h=
c
1
Area of ∆ABC = ch
2
1  2 s ( s − a )( s − b)( s − c) 
= c 
2  c 
= s ( s − a )( s − b)( s − c)

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