How does heron derive the formula of area of triangle?
Answers
For a triangle of given three sides, say a, b, and c, the formula for the area is given by
where s is the semi perimeter equal to P/2 = (a + b + c)/2.
Derivation of Heron's Formula
Area of triangle ABC
A=12bhA=12bh → equation (1)
From triangle ADB
x2+h2=c2x2+h2=c2
x2=c2–h2x2=c2–h2
x=c2−h2−−−−−−√x=c2−h2
From triangle CDB
(b–x)2+h2=a2(b–x)2+h2=a2
(b–x)2=a2–h2(b–x)2=a2–h2
b2–2bx+x2=a2–h2b2–2bx+x2=a2–h2
Substitute the values of x and x2
b2–2bc2−h2−−−−−−√+(c2–h2)=a2–h2b2–2bc2−h2+(c2–h2)=a2–h2
b2+c2–a2=2bc2−h2−−−−−−√b2+c2–a2=2bc2−h2
Square both sides
(b2+c2–a2)2=4b2(c2−h2)(b2+c2–a2)2=4b2(c2−h2)
(b2+c2–a2)24b2=c2−h2(b2+c2–a2)24b2=c2−h2
h2=c2−(b2+c2–a2)24b2h2=c2−(b2+c2–a2)24b2
h2=4b2c2−(b2+c2–a2)24b2h2=4b2c2−(b2+c2–a2)24b2
h2=(2bc)2−(b2+c2–a2)24b2h2=(2bc)2−(b2+c2–a2)24b2
h2=[2bc+(b2+c2–a2)][2bc−(b2+c2–a2)]4b2h2=[2bc+(b2+c2–a2)][2bc−(b2+c2–a2)]4b2
h2=[2bc+b2+c2–a2][2bc−b2−c2+a2]4b2h2=[2bc+b2+c2–a2][2bc−b2−c2+a2]4b2
h2=[(b2+2bc+c2)–a2][a2−(b2−2bc+c2)]4b2h2=[(b2+2bc+c2)–a2][a2−(b2−2bc+c2)]4b2
h2=[(b+c)2–a2]⋅[a2−(b−c)2]4b2h2=[(b+c)2–a2]⋅[a2−(b−c)2]4b2
h2=[(b+c)+a][(b+c)−a]⋅[a+(b−c)][a−(b−c)]4b2h2=[(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)4b2h2=(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)4b2h2=(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)4b2h2=(a+b+c)(a+b+c−2a)(a+b+c−2b)(a+b+c−2c)4b2
h2=P(P−2a)(P−2b)(P−2c)4b2h2=P(P−2a)(P−2b)(P−2c)4b2 note: P = perimeter
h=P(P−2a)(P−2b)(P−2c)−−−−−−−−−−−−−−−−−−−−−−√2bh=P(P−2a)(P−2b)(P−2c)2b
Substitute h to equation (1)
A=12bP(P−2a)(P−2b)(P−2c)−−−−−−−−−−−−−−−−−−−−−−√2bA=12bP(P−2a)(P−2b)(P−2c)2b
A=14P(P−2a)(P−2b)(P−2c)−−−−−−−−−−−−−−−−−−−−−−√A=14P(P−2a)(P−2b)(P−2c)
A=116P(P−2a)(P−2b)(P−2c)−−−−−−−−−−−−−−−−−−−−−−−−√A=116P(P−2a)(P−2b)(P−2c)
A=P2(P−2a2)(P−2b2)(P−2c2)−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√A=P2(P−2a2)(P−2b2)(P−2c2)
A=P2(P2−a)(P2−b)(P2−c)−−−−−−−−−−−−−−−−−−−−−−−−−−√A=P2(P2−a)(P2−b)(P2−c)
Recall that P/2 = s. Thus,
A=s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−