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Derivation of Heron's Formula

Area of triangle ABC

A=1 /2bh → equation (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

b2–2bc2−h2−−−−−−√+(c2–h2)=a2–h2

b2+c2–a2=2bc2−h2−−−−−−√

Square both sides

(b2+c2–a2)2=4B2(c2−h2)

(b2+c2–a2)24b2=c2−h2

h2=c2−(b2+c2–a2)24b2

h2=4b2c2−(b2+c2–a2)24b2

h2=(2BC)2−(b2+c2–a2)24b2

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=12bP(P−2a)(P−2b)(P−2c)−−−−−−−−−−−−−−−−−−−−−−√2b

A=14P(P−2a)(P−2b)(P−2c)−−−−−−−−−−−−−−−−−−−−−−√

A=116P(P−2a)(P−2b)(P−2c)−−−−−−−−−−−−−−−−−−−−−−−−√

A=P2(P−2a2)(P−2b2)(P−2c2)−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√

A=P2(P2−a)(P2−b)(P2−c)−−−−−−−−−−−−−−−−−−−−−−−−−−√

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=12bP(P−2a)(P−2b)(P−2c)−−−−−−−−−−−−−−−−−−−−−−√2b

A=14P(P−2a)(P−2b)(P−2c)−−−−−−−−−−−−−−−−−−−−−−√

A=116P(P−2a)(P−2b)(P−2c)−−−−−−−−−−−−−−−−−−−−−−−−√

A=P2(P−2a2)(P−2b2)(P−2c2)−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√

A=P2(P2−a)(P2−b)(P2−c)−−−−−−−−−−−−−−−−−−−−−−−−−−√

Hence proved

Recall that P/2 = s. Thus,

A = \sqrt{s(s - a)(s - b)(s - c)}

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