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The lengths of the sides of a traingle are c=5, b=6 and a=10 cm respectively. Find the square of the length of the median of the greatest side.

Hello Guest!

$m_a^2=\frac{1}{4}\cdot (2(b^2+c^2)-a^2)\\ m_a^2=\frac{1}{4}\cdot (2(36+25)-100)cm^2$

$m_a^2=5.5cm^2$

  !

The lengths of the sides of a triangle are 5, 6 and 10 cm respectively.

Find the square of the length of the median of the greatest side.

cos-rule:

$\begin{array}{|lrcll|} \hline (1) & 6^2 &=& 5^2+10^2-2*5*10*\cos(A) \\ & 36 &=& 125-100\cos(A) \\ & 100\cos(A) &=& 125-36 \\ & 100\cos(A) &=& 89 \\ &\mathbf{ \cos(A) } &=& \mathbf{\dfrac{89}{100}} \\ \hline \end{array}$

$\begin{array}{|lrcll|} \hline (2) & x^2 &=& 5^2+5^2-2*5*5*\cos(A) \\ & x^2 &=& 50-50\cos(A) \\ & x^2 &=& 50-50*\dfrac{89}{100} \\ & x^2 &=& 50-\dfrac{89}{2} \\ & x^2 &=& 50-44.5 \\ &\mathbf{ x^2 } &=& \mathbf{5.5} \\ \hline \end{array}$

The square of the length of the median of the greatest side is $\mathbf{5.5}\ \text{cm}$

The lengths of the sides of a triangle are 5, 6, and 10 cm respectively. Find the square of the length of the median of the greatest side.

a = 5           b = 6        c = 10

(m_c)² = {[ sqrt( 2a² + 2b² - c² )] / 2}²  = 5.5 cm²