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Heron's Formula

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

Note by Jonathan Hsu
1 year, 6 months ago

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This is the diagram :

file:///C:/Users/Abhi/Desktop/Capture.JPG Sai Ram · 1 year ago

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

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

\(\large{\text{From}}\) \( \large{\Delta ADB} ,\)

\(\large{x^2 + h^2 = c^2}\) [Using Pythagoras theorem]

\(\large{x^2 = c^2 - h^2}\)

\(\large{x = \sqrt{c^2 - h^2}}\).

\(\large{\text{From}}\) \( \large{\Delta CDB}\)

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

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

\(\large{b^2 - 2bx + x^2 = a^2 - h^2 }\)

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

\(\large{b^2 - 2b\sqrt{c^2 - h^2} + (c^2 - h^2) = a^2 - h^2}\)

\(\large{b^2 + c^2 - a^2 = 2b\sqrt{c^2 - h^2}}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(\Large{\boxed{\therefore A=\sqrt{s(s-a)(s-b)(s-c)}}}\) Sai Ram · 1 year ago

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