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Question Number 15828 by Tinkutara last updated on 14/Jun/17

In a Δ A B C, let M_{a}, M_{b} and M_{c} denote

the length of medians,

s = \frac{M_{a} + M_{b} + M_{c}}{2} and Δ = ar (Δ A B C) .

Prove that

Δ = \frac{4}{3} \sqrt{s (s - M_{a}) (s - M_{b}) (s - M_{c})}

Answered by ajfour last updated on 14/Jun/17

Commented by ajfour last updated on 14/Jun/17

M_{a} = A P M_{b} = B Q M_{c} = C R

A t r i a n g l e w i t h s i d e s e q u a l t o

\frac{2}{3} M_{a}, \frac{2}{3} M_{b}, \frac{2}{3} M_{c} i s f o r

e x a m p l e Δ G B L . (A l l s i x t r i a n g l e s

Δ A G N, Δ B G N, Δ B G L, Δ C G L,

Δ C G M, a n d Δ A G M o f

h e x a g o n A N B L C M h a v e t h e s e

l e n g t h s a s t h e i r s i d e s .

s = \frac{M_{a} + M_{b} + M_{c}}{2}

l e t l = \frac{1}{2} (\frac{2}{3} M_{a} + \frac{2}{3} M_{b} + \frac{2}{3} M_{c})

t h e n l = \frac{2}{3} s

A r e a o f a n y s u c h Δ o f t h e h e x a g o n

= \sqrt{l (l - \frac{2}{3} M_{a}) (l - \frac{2}{3} M_{b}) (l - \frac{2}{3} M_{c})}

a s l = \frac{2}{3} s w e h a v e

\frac{1}{6} (h e x a g o n a r e a)

= {(\frac{2}{3})}^{2} \sqrt{s (s - M_{a}) (s - M_{b}) (s - M_{c})}

h e x a g o n a r e a = 2 (a r e a o f Δ A B C)

= 2 Δ, t h e r e f o r e

\frac{1}{6} (2 Δ) = \frac{4}{9} \sqrt{s (s - M_{a}) (s - M_{b}) (s - M_{c})}

Δ = \frac{4}{3} \sqrt{s (s - M_{a}) (s - M_{b} (s - M_{c})} .

Commented by Tinkutara last updated on 14/Jun/17

How are the sides of Δ G B L two - third

of the medians of Δ A B C ?

Commented by ajfour last updated on 14/Jun/17

l i n e s a r e c o n s t r u c t e d p a r a l l e l t o

t h e m e d i a n s . .

G B = \frac{2}{3} B Q = \frac{2}{3} M_{b}

B L = \frac{2}{3} C G = \frac{2}{3} M_{c}

G L = B N = A G = \frac{2}{3} A P = \frac{2}{3} M_{a} .

Commented by Tinkutara last updated on 14/Jun/17

Thanks Sir!

Commented by RasheedSoomro last updated on 14/Jun/17

V N i C E S ir!