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Question Number 49280 by mr W last updated on 05/Dec/18

Commented by mr W last updated on 05/Dec/18

$T h e d i s t a n c e s f r o m i t s c e n t r o i d t o$ $i t s v e r t i c e s a r e g i v e n . F i n d t h e a r e a$ $o f t h e t r i a n g l e .$
Commented by behi83417@gmail.com last updated on 05/Dec/18

$g_{a} = \frac{2}{3} m_{a} = \frac{1}{3} \sqrt{2 (b^{2} + c^{2}) - a^{2}} =$ $= \frac{1}{3} \sqrt{2 (a^{2} + b^{2} + c^{2}) - 3 a^{2}} =$ $= \frac{1}{3} \sqrt{2 \times \frac{4}{3} Σ m_{a}^{2} - 3 a^{2}} =$ $= \frac{1}{3} \sqrt{\frac{8}{3} . \frac{9}{4} Σ g_{a}^{2} - 3 a^{2}} = \frac{1}{3} \sqrt{6 Σ g_{a}^{2} - 3 a^{2}}$ $\Rightarrow 3 g_{a}^{2} = 2 (g_{a}^{2} + g_{b}^{2} + g_{c}^{2}) - a^{2}$ $\Rightarrow a^{2} = 2 (g_{b}^{2} + g_{c}^{2}) - g_{a}^{2} \Rightarrow$ $a = \sqrt{2 (g_{b}^{2} + g_{c}^{2}) - g_{a}^{2}}$ $b = \sqrt{2 (g_{a}^{2} + g_{c}^{2}) - g_{b}^{2}}$ $c = \sqrt{2 (g_{a}^{2} + g_{b}^{2}) - g_{c}^{2}}$ $now we can find area from : {H e r o n}^{'} s formula .$ $note : Σ m_{a}^{2} = \frac{3}{4} Σ a^{2} .$
Commented by mr W last updated on 05/Dec/18

$t h a n k s s i r!$
	Answered by MJS last updated on 05/Dec/18

	$A = (\begin{matrix} 0 \\ 0 \end{matrix}) B = (\begin{matrix} c \\ 0 \end{matrix}) C = (\begin{matrix} \frac{- a^{2} + b^{2} + c^{2}}{2 c} \\ \frac{δ (a, b, c)}{2 c} \end{matrix})$ $δ (a, b, c) = \sqrt{(a + b + c) (a + b - c) (a - b + c) (- a + b + c)}$ $G = (\begin{matrix} \frac{- a^{2} + b^{2} + 3 c^{2}}{6 c} \\ \frac{δ (a, b, c)}{6 c} \end{matrix})$ $g_{a}^{2} = \frac{- a^{2} + 2 b^{2} + 2 c^{2}}{9}$ $g_{b}^{2} = \frac{2 a^{2} - b^{2} + 2 c^{2}}{9}$ $g_{c}^{2} = \frac{2 a^{2} + 2 b^{2} - c^{2}}{9}$ $\Rightarrow$ $a^{2} = - g_{a}^{2} + 2 g_{b}^{2} + 2 g_{c}^{2}$ $b^{2} = 2 g_{a}^{2} - g_{b}^{2} + 2 g_{c}^{2}$ $c^{2} = 2 g_{a}^{2} + 2 g_{b}^{2} - g_{c}^{2}$ $\Rightarrow$ $area (a, b, c) = \frac{δ (a, b, c)}{4} = \frac{3 δ (g_{a}, g_{b}, g_{c})}{4}$
	Commented by mr W last updated on 05/Dec/18

	$t h a n k y o u s i r!$ $t h e l a s t l i n e i s n e w t o m e, v e r y n i c e!$
	Commented by MJS last updated on 05/Dec/18

	${it}^{'} s just {Heron}^{'} s sentence$
	Commented by mr W last updated on 05/Dec/18

	Commented by mr W last updated on 05/Dec/18

	$A r e a o f h e x a g o n = 2 \times A r e a o f Δ A B C$ $A r e a o f h e x a g o n = 6 \times A r e a o f t r i a n g l e w i t h s i d e s g_{a}, g_{b}, g_{c}$ $\Rightarrow A r e a o f Δ A B C = 3 \times A r e a o f t r i a n g l e w i t h s i d e s g_{a}, g_{b}, g_{c}$
	Commented by mr W last updated on 05/Dec/18

	$N o w I u n d e r s t a n d i t g e o m e t r i c a l l y .$
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