The-medians-of-a-triangle-are-m-1-m-2-m-3-Find-the-length-of-each-sides-the-triangle-

Question Number 2458 by alib last updated on 20/Nov/15

T h e m e d i a n s o f a t r i a n g l e

a r e m_{1}, m_{2}, m_{3} .

F i n d t h e l e n g t h o f e a c h s i d e s

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

Answered by prakash jain last updated on 20/Nov/15

m_{1} = \frac{1}{2} \sqrt{2 c^{2} + 2 b^{2} - a^{2}} \Rightarrow a^{2} = - 4 m_{1}^{2} + 2 c^{2} + 2 b^{2} (1)

m_{2} = \frac{1}{2} \sqrt{2 a^{2} + 2 c^{2} - b^{2}} \Rightarrow b^{2} = - 4 m_{2}^{2} + 2 a^{2} + 2 c^{2} (2)

m_{3} = \frac{1}{2} \sqrt{2 a^{2} + 2 b^{2} - c^{2}} \Rightarrow c^{2} = - 4 m_{3}^{2} + 2 a^{2} + 2 b^{2} (3)

(2) + (3)

b^{2} + c^{2} = - 4 (m_{2}^{2} + m_{3}^{2}) + 2 (b^{2} + c^{2}) + 4 a^{2}

(b^{2} + c^{2}) = 4 [(m_{3}^{2} + m_{2}^{2}) - a^{2}]

s u b t i t u t e i n (1)

a^{2} = - 4 m_{1}^{2} + 8 (m_{3}^{2} + m_{2}^{2}) - 8 a^{2}

9 a^{2} = 8 m_{3}^{2} + 8 m_{2}^{2} - 4 m_{1}^{2}

a = \frac{2}{3} \sqrt{2 m_{3}^{2} + 2 m_{2}^{2} - m_{1}^{2}}

s i m i l a r l y

b = \frac{2}{3} \sqrt{2 m_{3}^{2} + 2 m_{1}^{2} - m_{2}^{2}}

c = \frac{2}{3} \sqrt{2 m_{1}^{2} + 2 m_{2}^{2} - m_{3}^{2}}