The medians of a triangle are m_1 , m_2 , m_3 . Find the length of each sides the triangle.


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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

	$m_{1} = \frac{1}{2} \sqrt{2 c^{2} + 2 b^{2} - a^{2}} \Rightarrow a^{2} = - 4 m_{1}^{2} + 2 c^{2} + 2 b^{2} (1)$ $m_{2} = \frac{1}{2} \sqrt{2 a^{2} + 2 c^{2} - b^{2}} \Rightarrow b^{2} = - 4 m_{2}^{2} + 2 a^{2} + 2 c^{2} (2)$ $m_{3} = \frac{1}{2} \sqrt{2 a^{2} + 2 b^{2} - c^{2}} \Rightarrow c^{2} = - 4 m_{3}^{2} + 2 a^{2} + 2 b^{2} (3)$ $(2) + (3)$ $b^{2} + c^{2} = - 4 (m_{2}^{2} + m_{3}^{2}) + 2 (b^{2} + c^{2}) + 4 a^{2}$ $(b^{2} + c^{2}) = 4 [(m_{3}^{2} + m_{2}^{2}) - a^{2}]$ $s u b t i t u t e i n (1)$ $a^{2} = - 4 m_{1}^{2} + 8 (m_{3}^{2} + m_{2}^{2}) - 8 a^{2}$ $9 a^{2} = 8 m_{3}^{2} + 8 m_{2}^{2} - 4 m_{1}^{2}$ $a = \frac{2}{3} \sqrt{2 m_{3}^{2} + 2 m_{2}^{2} - m_{1}^{2}}$ $s i m i l a r l y$ $b = \frac{2}{3} \sqrt{2 m_{3}^{2} + 2 m_{1}^{2} - m_{2}^{2}}$ $c = \frac{2}{3} \sqrt{2 m_{1}^{2} + 2 m_{2}^{2} - m_{3}^{2}}$
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