# Defeating Heron

Let a ∆ABC

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

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

Proof:

x + y = c

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

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

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

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

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

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

$(a^{2} - h^{2})(b^{2} - h^{2})$ = $(\frac{c^{2} - a^{2} - b^{2} }{2} + h^{2})^{2}$

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

$a^{2} × b^{2} - (a^{2} + b^{2})h^{2}$ = $q^{2} +2qh^{2}$

$a^{2} × b^{2} - q^{2}$ = $h^{2}(2q + a^{2} + b^{2})$ ...(1)

Now as $2q$ = $c^{2} - a^{2} - b^{2}$

Therefore $2q + a^{2} + b^{2}$ = $c^{2}$. Putting this into (1)

$h^{2} × c^{2}$ = $a^{2} × b^{2} - q^{2}$

$h$ = $\frac{\sqrt{a^{2} × b^{2} - q^{2}}}{c}$

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

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

Hence proved

Try this new formula and check if you find it useful

Note by Sahar Bano
1 month, 1 week ago

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

- 2 weeks ago

Just figured it out myself

- 2 weeks ago

Nice!, @Sahar Bano

- 1 week, 6 days ago

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