Harder applications of Heron's formula
Yue Kwok Choy
Question 1
You are given a rope of length s , find the largest area of the triangle formed by this
rope. What is its area ?
Solution
We are going to prove that the equilateral triangle has the maximum area for any fixed
perimeter s .
Let a , b , c be the sides of the triangle .
The perimeter is fixed and a + b + c = 2s , so 2s is a constant .
By Heron's formula :
Now,
  ss  a s  b s  c 
s
,
abc
2
….
(1)
….
(2)
by Arithmetic mean – Geometric mean inequality (for three variables) ,
3
3
3
s  a s  bs  c   s  a   s  b  s  c   3s  a  b  c    3s  2s  


3

3


3

s3
27
and the equality occurs when s – a = s – b = s – c (the variables are all equal) , that is, a = b = c .
The triangle is therefore equilateral .
By (2), since the product is always less than a constant
and by Heron's formula,
s3
, this constant is the maximum for the product
27
the maximum area of a triangle with fixed perimeter s is
 max  ss  a s  b s  c 
 
s s3
s2


27
3 3
3s 2
9
Question 2
You are given the medians of the triangle ma , mb
and mc . Find the area of ABC .
Solution
As in the diagram, D, E, F are mid-points of BC, CA and AB.
G
is the centroid .
Now,

AG 
AGF 
ma = AD , mb = BE and mc = CF .
2
AD
3
1
1
ABG  ABC
2
6
Let P be the mid-point of AG , we have
FPG 
PG 
1
ma ,
3
FP 
1
mb ,
3
FG 
1
mc
3
1
1
AFG 
ABC
2
12
By Heron's formula,
1

ABC  12FPG  12 

1
mm  m a m  m b m  m c 
9
where
m
1
m a  m b  m c 
2
4
mm  m a m  m b m  m c 
3
Question 3
You are given the altitudes of the triangle ha , hb and hc . Find the area of ABC .
Solution
As in the diagram,
G
AD BC, BE CA, CF AB.
is the orthocentre ha = AD , hb = BE and hc = CF .
Since

1
1
1
h a a  h b b  h c c , we have :
2
2
2
a
2
2
2
,b 
,c 
ha
hb
hc
….
(3)
s
 1
abc
1
1
  
 
2
 ha hb hc 
….
(4)
Put (3), (4) in Heron's formula :
  ss  a s  b s  c 

1
1 
1
1 
1
1 
1
1 
  1

  1

  1

  1

2  

  

 

 

 

  h a h b h c 

  h a h b h c 

  h a h b h c 

  h a h b h c 

1
 1
1
1  1
1
1  1
1
1  1
1
1 

 











 h a h b h c  h a h b h c  h a h b h c  h a h b h c 


Question 4
You are given the altitudes of the triangle hb ,
hc
and side a . Show that the area of ABC ,  ,
  1

 1
1 
1 
satisfies the equation : 16 2  2  4  8a 2  2  2   2 2  a 4  0 .
hc 
hc 
  h b
 hb

2
If
h b  4, h c 
7 2
and
2
a = 5 , find  .
Solution
As in the diagram, AD BC, BE CA, CF AB.
G
is the orthocentre ha = AD , hb = BE and hc = CF .
We have
b
2
2
,c 
hb
hc
Put (5) in Heron's formula :
….
(5)
  ss  a s  b s  c 
2
1  1
1  
  a  
 2
4   h b h c  
 1
1
16  

 h b h c
2
1/ 2

 1
1  
 2
 a  
 h b h c  

2


 42  a 2 


1/ 2

 1
1
a 2  


 hb hc
1/ 2
  1
1  
 2
a  
  h b h c  
2


 42 


1/ 2
  1
1  
 2
a  
  h b h c  
1/ 2
1/ 2
  1

 1
1 
1 
16 2  2  4  8a 2  2  2   2 2  a 4  0
hc 
hc 
  h b
 hb

2
After expansion, we can get an equation :
This is a bi-quadratic equation in
 , we can solve for unique positive real root .
3