Heron's Formula

Hi, can you help me prove Heron's Formula? Just put your proof down below this comment.

#Algebra #Heron'sFormula

Note by Jonathan Hsu
5 years, 11 months ago

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Comments

The perimeter of the\large{\text{The perimeter of the}} ΔABC\Delta ABC is given by P=a+b+c\large{\text{is given by P=a+b+c}}

Area of\large{\text{Area of}} ΔABC=12bh.........................1\large{\Delta ABC = \dfrac{1}{2}bh}.........................\boxed{1}

From\large{\text{From}} ΔADB, \large{\Delta ADB} ,

x2+h2=c2\large{x^2 + h^2 = c^2} [Using Pythagoras theorem]

x2=c2−h2\large{x^2 = c^2 - h^2}

x=c2−h2\large{x = \sqrt{c^2 - h^2}}.

From\large{\text{From}} ΔCDB \large{\Delta CDB}

(b−x)2+h2=a2\large{(b - x)^2 + h^2 = a^2} [Using Pythagoras theorem]

(b−x)2=a2−h2\large{(b - x)^2 = a^2 - h^2}

b2−2bx+x2=a2−h2\large{b^2 - 2bx + x^2 = a^2 - h^2 }

Substitute the values of \large{\text{Substitute the values of }} x\large{x} and\large{\text{and}} x2\large{x^2}

b2−2bc2−h2+(c2−h2)=a2−h2\large{b^2 - 2b\sqrt{c^2 - h^2} + (c^2 - h^2) = a^2 - h^2}

b2+c2−a2=2bc2−h2\large{b^2 + c^2 - a^2 = 2b\sqrt{c^2 - h^2}}

Squaring on both sides,\large{\text{Squaring on both sides,}}

(b2+c2−a2)2=4b2(c2−h2)\large{(b^2 + c^2 - a^2)^2 = 4b^2(c^2-h^2)}

(b2+c2−a2)24b2=(c2−h2)\large{\dfrac{(b^2+c^2-a^2)^2}{4b^2} = (c^2-h^2)}

h2=c2−(b2+c2−a2)24b2\large{h^2=c^2 -\dfrac{(b^2+c^2-a^2)^2}{4b^2}}

h2=4b2c2−(b2+c2−a2)22b2\large{h^2 = \dfrac{4b^2c^2 -(b^2+c^2-a^2)^2}{2b^2} }

h2=(2bc)2−(b2+c2−a2)4b2\large{h^2 = \dfrac{(2bc)^2 - (b^2+c^2-a^2)}{4b^2}}

h2=[2bc+(b2+c2−a2)][2bc−(b2+c2−a2)]4b2\large{h^2 = \dfrac{[2bc +(b^2+c^2-a^2)][2bc -(b^2+c^2-a^2)]}{4b^2}}

h2=[2bc+b2+c2−a2][2bc−b2−c2+a2]4b2\large{h^2=\dfrac{[2bc+b^2+c^2-a^2][2bc -b^2-c^2+a^2]}{4b^2}}

h2=[(b2+c2+2bc)−a2][2bc−(b2+c2−2bc)]4b2\large{h^2=\dfrac{[(b^2+c^2+2bc)-a^2][2bc-(b^2+c^2-2bc)]}{4b^2}}

h2=[ (b+c)2−a2 ][ a2−(b−c)2 ]4b2\large{h^2 = \dfrac{[ \, (b + c)^2 - a^2 \, ] [ \, a^2 - (b - c)^2 \, ]}{4b^2}}

h2=[ (b+c)+a ][ (b+c)−a ][ a+(b−c) ][ a−(b−c) ]4b2\large{h^2 = \dfrac{[ \, (b + c) + a \, ][ \, (b + c) - a \, ] [ \, a + (b - c) \, ][ \, a - (b - c) \, ]}{4b^2}}

h2=(b+c+a)(b+c−a)(a+b−c)(a−b+c)4b2\large{h^2 = \dfrac{(b + c + a)(b + c - a)(a + b - c)(a - b + c)}{4b^2}}

h2=(a+b+c)(b+c−a)(a+c−b)(a+b−c)4b2\large{h^2 = \dfrac{(a + b + c)(b + c - a)(a + c - b)(a + b - c)}{4b^2}}

h2=(a+b+c)(a+b+c−2a)(a+b+c−2b)(a+b+c−2c)4b2\large{h^2 = \dfrac{(a + b + c)(a + b + c - 2a)(a + b + c - 2b)(a + b + c - 2c)}{4b^2}}

h2=P(P−2a)(P−2b)(P−2c)4b2\large{h^2 = \dfrac{P(P - 2a)(P - 2b)(P - 2c)}{4b^2}}

h=P(P−2a)(P−2b)(P−2c)2b\large{h = \dfrac{\sqrt{P(P - 2a)(P - 2b)(P - 2c)}}{2b}}

Substitute h to\large{\text{Substitute h to}} 1\boxed{1}

A=12bP(P−2a)(P−2b)(P−2c)2b\large{A = \dfrac{1}{2}b\dfrac{\sqrt{P(P - 2a)(P - 2b)(P - 2c)}}{2b}}

A=14P(P−2a)(P−2b)(P−2c)\large{A = \dfrac{1}{4}\sqrt{P(P - 2a)(P - 2b)(P - 2c)}}

A=116P(P−2a)(P−2b)(P−2c)\large{A = \sqrt{\dfrac{1}{16}P(P - 2a)(P - 2b)(P - 2c)}}

A=P2(P−2a2)(P−2b2)(P−2c2)\large{A = \sqrt{\dfrac{P}{2} \left( \dfrac{P - 2a}{2} \right)\left( \dfrac{P - 2b}{2} \right)\left( \dfrac{P - 2c}{2} \right)}}

A=P2(P2−a)(P2−b)(P2−c)\large{A = \sqrt{\dfrac{P}{2} \left( \dfrac{P}{2} - a \right)\left( \dfrac{P}{2} - b \right)\left( \dfrac{P}{2} - c \right)}}

We know that s =\large{\text{We know that s =}}P2\large{\dfrac{P}{2}}

∴A=s(s−a)(s−b)(s−c)\Large{\boxed{\therefore A=\sqrt{s(s-a)(s-b)(s-c)}}}

Sai Ram - 5 years, 5 months ago

This is the diagram :

file:///C:/Users/Abhi/Desktop/Capture.JPG

Sai Ram - 5 years, 5 months ago
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