Defeating Heron

Let a ∆ABC

Now area of the triangle =\(\frac{\sqrt{a^{2} × b^{2} - q^{2}}}{2}\)

Where q = c2a2b22\frac{c^{2} - a^{2} - b^{2}}{2}

Proof:

x + y = c

h = AD

x = b2h2\sqrt{b^{2} - h^{2}}

y = a2h2\sqrt{a^{2} - h^{2}}

c = b2h2\sqrt{b^{2} - h^{2}} + a2h2\sqrt{a^{2} - h^{2}}

c2c^{2} = (b2h2+a2h2)2 (\sqrt{b^{2} - h^{2}} + \sqrt{a^{2} - h^{2}})^{2}

c2c^{2} = a2+b22h2+2(a2h2)(b2h2)a^{2} + b^{2} - 2h^{2} + 2\sqrt{(a^{2} - h^{2})(b^{2} - h^{2})}

2(a2h2)(b2h2)2\sqrt{(a^{2} - h^{2})(b^{2} - h^{2})} = c2a2b2+2h2c^{2} - a^{2} - b^{2} +2h^{2}

(a2h2)(b2h2)(a^{2} - h^{2})(b^{2} - h^{2}) = (c2a2b22+h2)2(\frac{c^{2} - a^{2} - b^{2} }{2} + h^{2})^{2}

Let q = c2a2b22\frac{c^{2} - a^{2} -b^{2}}{2} and solve left side further and cancelling h4h^{4} from both sides we get

a2×b2(a2+b2)h2a^{2} × b^{2} - (a^{2} + b^{2})h^{2} = q2+2qh2q^{2} +2qh^{2}

a2×b2q2a^{2} × b^{2} - q^{2} = h2(2q+a2+b2)h^{2}(2q + a^{2} + b^{2}) ...(1)

Now as 2q2q = c2a2b2c^{2} - a^{2} - b^{2}

Therefore 2q+a2+b22q + a^{2} + b^{2} = c2c^{2}. Putting this into (1)

h2×c2h^{2} × c^{2} = a2×b2q2a^{2} × b^{2} - q^{2}

hh = a2×b2q2c\frac{\sqrt{a^{2} × b^{2} - q^{2}}}{c}

Now as area = 12ch\frac{1}{2}ch

Area = a2×b2q22\frac{\sqrt{a^{2} × b^{2} - q^{2}}}{2}

Hence proved

Try this new formula and check if you find it useful

Note by Sahar Bano
6 months, 2 weeks ago

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1 vote

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Where did you find this, @Sahar Bano? This is really good!

A Former Brilliant Member - 5 months, 3 weeks ago

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Just figured it out myself

Sahar Bano - 5 months, 3 weeks ago

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Nice!, @Sahar Bano

A Former Brilliant Member - 5 months, 3 weeks ago

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