Question

derivation of heron`s formula

Answer 1

From triangle ADB
x2+h2=c2
x2=c2–h2
x=
√
c2−h2



From triangle CDB
(b–x)2+h2=a2
(b–x)2=a2–h2
b2–2bx+x2=a2–h2


Substitute the values of x and x2
b2–2b
√
c2−h2
+(c2–h2)=a2–h2
b2+c2–a2=2b
√
c2−h2



Square both sides
(b2+c2–a2)2=4b2(c2−h2)
(b2+c2–a2)2
4b2

=c2−h2
h2=c2−
(b2+c2–a2)2
4b2

h2=
4b2c2−(b2+c2–a2)2
4b2

h2=
(2bc)2−(b2+c2–a2)2
4b2

h2=
[2bc+(b2+c2–a2)][2bc−(b2+c2–a2)]
4b2

h2=
[2bc+b2+c2–a2][2bc−b2−c2+a2]
4b2

h2=
[(b2+2bc+c2)–a2][a2−(b2−2bc+c2)]
4b2

h2=
[(b+c)2–a2]⋅[a2−(b−c)2]
4b2

h2=
[(b+c)+a][(b+c)−a]⋅[a+(b−c)][a−(b−c)]
4b2

h2=
(b+c+a)(b+c−a)(a+b−c)(a−b+c)
4b2

h2=
(a+b+c)(b+c−a)(a+c−b)(a+b−c)
4b2

h2=
(a+b+c)(a+b+c−2a)(a+b+c−2b)(a+b+c−2c)
4b2

h2=
P(P−2a)(P−2b)(P−2c)
4b2

note: P = perimeter

h=
√
P(P−2a)(P−2b)(P−2c)
2b




Substitute h to equation (1)
A=
1
2

b
√
P(P−2a)(P−2b)(P−2c)
2b

A=
1
4

√
P(P−2a)(P−2b)(P−2c)
A=
√
1
16

P(P−2a)(P−2b)(P−2c)
A=
√
P
2

(
P−2a
2

)(
P−2b
2

)(
P−2c
2

)
A=
√
P
2

(
P
2

−a)(
P
2

−b)(
P
2

−c)



Recall that P/2 = s. Thus,

A=
√
s(s−a)(s−b)(s−c)