two-medians-of-a-triange-are-3-and-4-cm-respectively-find-the-maximum-area-of-the-triangle-

Question Number 181485 by mr W last updated on 25/Nov/22

t w o m e d i a n s o f a t r i a n g e a r e 3 a n d

4 c m r e s p e c t i v e l y . f i n d t h e m a x i m u m

a r e a o f t h e t r i a n g l e .

Answered by Acem last updated on 26/Nov/22

Commented by mr W last updated on 26/Nov/22

g r e a t!

Commented by Acem last updated on 26/Nov/22

Commented by Acem last updated on 26/Nov/22

T h a n k y o u S i r!

B e c a u s e m y E n g l i s h, p l e a s e w r i t e a n e x p l a n a t i o n

o f t h e i d e a o f t h a t r a t i o \frac{4}{3}, w i t h m y t h a n k s

Commented by mr W last updated on 26/Nov/22

Answered by mr W last updated on 26/Nov/22

Commented by mr W last updated on 26/Nov/22

s a y t h e r e d m e d i a n s a r e g i v e n .

w e k n o w [Δ A F E] = \frac{[Δ A B C]}{4}

\Rightarrow [E F B C] = \frac{3 [Δ A B C]}{4}

\Rightarrow [Δ A B C] = \frac{4}{3} [E F B C]

= \frac{4}{3} \times \frac{m_{b} m_{c} \sin θ}{2} = \frac{2 m_{b} m_{c} \sin θ}{3}

w e s e e t h e a r e a o f t r i a n g l e A B C

i s m a x i m u m w h e n θ = 90 ° .

{[Δ A B C]}_{m a x} = \frac{2 m_{b} m_{c}}{3} = \frac{2 \times 3 \times 4}{3} = 8 {c m}^{2}

i n t h i s c a s e Δ A B C i s r i g h t a n g l e d .

Commented by SEKRET last updated on 26/Nov/22

\boldsymbol beatiful \boldsymbol answer

Commented by Acem last updated on 26/Nov/22

I l i k e d y o u r m e t h o d m o r e t h a n m i n e, t h a n k y o u

Commented by Acem last updated on 26/Nov/22

m i n e g i v e s t h e v a l u e o f t h e 3 r d m i d i a n a n d i t

s h o w h o w t o c o n v e r t t h e m i d i a n s i n t o s i d e s o f

t h e n e w t r i a n g l e “ r i g h t o n e = {m a x}^{″}

{t h a t}^{'} s g o o d, b u t i f e l t t h a t {i t}^{'} s a l i t t l e u n c o n v i n c i n g .

Commented by mr W last updated on 26/Nov/22

Commented by mr W last updated on 26/Nov/22

m a n y r o a d s l e a d t o R o m e!

m a n y r o a d s l e a d t o R o m e!

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