Question

Can anyone prove Heron's Formula ( I will mark him as BRAINLIEST and FOLLOW HIM ☺☺☺)​

Answer 1

Heron's Formula Derivation

A = 1/2 bh _ _ _ _ (i)

Draw a perpendicular BD on AC

Consider a ∆ADB

x2 + h2 = c2

x2 = c2 − h2—(ii)

⇒x = √(c2−h2)−−−−−−—(iii)

Consider a ∆CDB,

(b−x)2 + h2 = a2

(b−x)2 = a2 − h2

b2 − 2bx + x2 = a2–h2

Substituting the value of x and x2 from equation (ii) and (iii), we get

b2 – 2b√(c2−h2)+ c2−h2 = a2 − h2

b2 + c2 − a2 = 2b√(c2 − h2)

Squaring on both sides, we get;

(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 = [(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 = (a+b+c)(b+c−a)(a+c−b)(a+b−c)4b2

The perimeter of a ∆ABC is

P= a+b+c

⇒ h2 = P(P–2a)(P–2b)(P−2c)4b2

⇒ h =P(P–2a)(P–2b)(P−2c)−−−−−−−−−−−−−−−−−−−−√2b

Substituting the value of h in equation (i), we get;

A = 12bP(P–2a)(P–2b)(P−2c)√2b

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

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

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

Semi perimeter(s) = perimeter2 = P2

⇒ A = s(s–a)(s–b)(s–c)−−−−−−−−−−−−−−√